2025-12-30 07:00:00
Throughout, let $P(x,y)=x+\frac{1}{x}+y+\frac{1}{y}+1$. This Laurent polynomial is one of the Boyd-Deninger polynomials, which have been extensively studied for their connections to Mahler measures and special values of $L$-functions.
The polynomial $P(x,y)$ is special for several reasons. First of all, the projective closure of $P(x,y)=0$ is an elliptic curve $E$ of conductor $15$ and the curve does not admit complex multiplication (we will show that in this post). Second, this polynomial evoke Beilinson’s conjecture (we will justify in another post). As one can prove, there is a satisfyingly simple relation between the Mahler measure of $P$ and the $L$-function of the curve $E$:
In this post, we will do the preparation: the geometry of $P$ and $E$, and the determination of the Deninger path of $P$, an important subset of $E$, which will be useful for the future study.
It is expected that the reader know the basic arithmetic of elliptic curves and the basics of projective curves.
Let $Z_P$ be the zero locus of $P(x,y)=x+\frac{1}{x}+y+\frac{1}{y}+1$ in $(\mathbb{C}^\times)^2=\operatorname{Spec}(\mathbb{C}[X_1,X_1^{-1},X_2,X_2^{-1}])$. We would like to know its properties in terms of a curve, before moving on to the study of the projective closure of $Z_P$ in $\mathbb{P}^2(\mathbb{C})$.
Proposition 1.1 The algebraic set $Z_P$ is a smooth algebraic curve.
Proof. We notice that $\frac{\partial P}{\partial x}(x,y)=1-\frac{1}{x^2}$, $\frac{\partial P}{\partial y}(x,y)=1-\frac{1}{y^2}$. A singular point of $P$ must satisfy $\frac{\partial P}{\partial x}(x,y)=\frac{\partial P}{\partial y}(x,y)=P(x,y)=0$, and the only possible candidates of $(x,y) \in (\mathbb{C}^\times)^2$ are $(1,1)$, $(1,-1)$, $(-1,-1)$ and $(-1,1)$, where the values of $P$ are never $0$, therefore $P$ does not admit singular points. $\square$
Let $E=\overline{Z_P}$ be the projective closure of $Z_P$ in $\mathbb{P}^2(\mathbb{C})$. We will show that $E$ is an elliptic curve of conductor $15$.
Proposition 1.2. The following statements are true:
- $E$ is a non-singular cubic defined over $\mathbb{Q}$.
- $E$ has a rational point, hence is an elliptic curve defined over $\mathbb{Q}$.
- $E \setminus Z_P$ is a subgroup of $E(\mathbb{Q})$ isomorphic to $\mathbb{Z}/4\mathbb{Z}$.
- $E$ has conductor $15$.
- $E$ does not admit complex multiplication.
A few comments are in order before the proof. As a matter of fact we are looking at the compactification of $Z_P$, which is, according to the third point, $Z_P$ with $4$ points added. Besides, we are allowed to use the modularity theorem to construct a map $X_1(15) \to E$ to enable us to study the polynomial in the domain of modular forms. The modularity theorem is nevertheless an overkill; we can thereafter give a more explicit construction.
Proof. We notice that the homogenization of $P(x,y)$ is
In other words, $E$ is the zero locus of $\widetilde{P}(X,Y,Z)$ in $\mathbb{P}^2(\mathbb{C})$. We can see indeed that $E$ is a cubic curve defined over $\mathbb{Q}$. We need to see if $E$ is smooth. To do this, we search for singular points.
We have
We can verify the first statement on each affine cover $U_X=\{X \ne 0\}$, $U_Y=\{Y \ne 0\}$ and $U_Z=\{Z\ne 0\}$.
On $U_X$, a singular point $[1:Y:Z]$ on $E$ satisfies
We will show that the system of equation has no solution. By combining the first two equations, we get
If $Y=0$, then $2YZ+2Z+Y=0$ becomes $Z=0$, which is absurd because then $1+Z^2+2Y+Z=1=0$. Therefore we can only have $Z^2=1$, i.e. $Z=1$ or $Z=-1$. The case where $Z=1$ implies that $2Y+3=0$ and $3Y+2=0$, which is again a contradiction. Finally, the case where $Z=-1$ implies that $2Y+1=0$ and $-2-Y=0$, which is still not possible. Conclusion: the first two equations cannot hold at the same time, therefore on $U_X$ there is no singular point. The singularity on $U_Y$ and $U_Z$ is verified in the same manner so we omit the proof. To conclude, $E$ is a smooth curve, hence irreducible (see proposition 5.1 of this document).
To see that $E$ is an elliptic curve, we notice that $\widetilde{P}(1,0,0)=0$ so $E$ is a smooth cubic curve that admits a rational point $P_1=[1:0:0]$ thus an elliptic curve.
One finds also some other simple points that lies on $E$. That is, $P_2=[0:0:1]$, $P_3=[1:-1:0]$ and $P_4=[0:1:0]$. In fact, we have $E \setminus Z_P = \{P_1,P_2,P_3,P_4\}$. To see this, we notice that $Z_P \subset U_Z$ and for all points on $Z_P$, the coordinates of $X$ and $Y$ cannot be zero. Thus $P_2 \in U_Z$ is the only point of $E \cap U_Z$ that is not on $Z_P$. On $\mathbb{P}^2(\mathbb{C}) \setminus U_Z$, the points on $E$ satisfy $X^2Y+XY^2=XY(X+Y)=0$, the only solution of which is $P_1,P_3$ and $P_4$.
Moving on, we study the structure of $E \setminus Z_P$ as a group. Indeed, the unit element of $E(\mathbb{Q})$ can be set as $P_1=[1:0:0]$. To calculate the sum $P_2+P_2$, we first of all calculate the tangent line of $E$ at $P_2$. Indeed, the gradient of $\widetilde{P}$ at $P_2$ is $(1,1,0)$, therefore the tangent line is $L_2=\{X+Y=0\}$, and we have $E \cap L_2 = \{X^2Z=0\} \cap \{X+Y = 0\}=\{[1:-1:0],[0:0:1]\}$. Therefore the other point on $E$ passing through $L_2$ is $P_3=[1:-1:0]$. The point $2P_2$ is therefore the intersection of $E$ and the line containing $P_1$ and $P_3$. We notice that the line containing $P_1$ and $P_3$ is $L_{13}=\{Z=0\}$. We see that $L_{14} \cap E=\{X^2Y+XY^2=0\} \cap \{Z=0\}=\{P_1,P_3,P_4\}$, which implies that $2P_2=P_4$.
We can calculate $2P_4$ in the same manner. Indeed, the tangent line of $E$ at $P_4$ is $L_4=\{X=0\}$, and $L_4 \cap E = \{YZ^2=0\} \cap \{X=0\}=\{P_4,P_2\}$. Therefore $2P_4$ is the intersection of $E$ and the line passing through $P_2$ and $P_1$, which is $L_{12}=\{Y=0\}$. We see that $L_{12} \cap E = \{XZ^2=0\}\cap \{Y=0\}=\{P_1,P_2\}$. Notice that $L_{12}$ is tangent at $P_1$ therefore we have $2P_4 = P_1$.
The calculation of $P_3$ can be tricky so we circumvent it by considering $P_2+P_4$. Indeed, the line passing through $P_2$ and $P_4$ is $L_{24}=L_4=\{X=0\}$, which is tangent at $P_4$. Therefore the sum $P_2+P_4$ is collinear with $P_1$ and $P_4$, which is exactly $P_3$ since we have shown that $L_{14} \cap E = \{P_1,P_3,P_4\}$.
To conclude, with $P_1$ being chosen as the unit, we have $2P_2 = P_4$, $3P_2 = P_2+P_4 = P_3$ and $4P_2 = 2P_4 = P_1$. Therefore $E \setminus Z_P$ is a cyclic group of order $4$ generated by $P_2$ as expected.
Finally, we see why the conductor of $E$ is $15$. The Weierstrass form of $E$ is $y^2 + xy + y = x^3 + x^2$. This form can be found by a change of variable $(X,Y,Z) \mapsto (X+Y+Z,-Y,-X)$ as we have
From the Weierstrass form it is easy to see that the discriminant $\Delta = -15$ and therefore the conductor $N=15$. One can simply use the first two steps of Tate’s algorithm working on prime numbers $p=3$ and $5$. One can also simply use SageMath to carry out the calculation. The code can be found after the proof.
For the last statement, as one can compute (with the help of SageMath), that $j(E)=-\frac{1}{15}$ is not an algebraic integer, therefore $E$ does not admit complex multiplication. See Theorem II.6.1. of this book. $\square$
The following code block contains a minimal implementation of the elliptic curve $E$ in SageMath. If the reader does not have SageMath on their device, it is recommended to run the code on an online server like SageMathCell. It is also recommended to check the graph of $P(x,y)$ with the help of for instance Geogebra.
1 |
# Sage implementation on finding the Weierstrass form and the conductor |
The Deninger path of a polynomial $Q(x,y) \in \mathbb{C}[x^{\pm},y^{\pm}]$ is defined to be
The reason why we are interested in such a circle is that we want to learn the value of the Mahler measure of a polynomial, which is defined in the following manner for $Q$:
This integral is not easy to study. To make our lives easier, we try to integrate $\log|Q(x,y)|$ with one variable at a time. Let $Q^\ast(x)$ be the leading coefficient of $Q(x,y)$ viewed as a Laurent polynomial with coefficients in $\mathbb{C}[x^{\pm}]$. Then whenever $Q^\ast(x) \ne 0$, we have, by Jensen’s formula),
We then integrate the value above with respect to $x$, only to find
Therefore the integration path $\gamma(Q)$ appears naturally. In this part, we want to find the Deninger path of $P(x,y)=x+\frac{1}{x}+y+\frac{1}{y}+1$.
Proposition 2. The Deninger path $\gamma(P)$ is given by
Proof. The determination of the path is surprisingly elementary. It is routine to verify that such $(e^{i\theta},Y(\theta))$ lie on $\gamma(P)$ so it suffices to verify the inverse. Let $(x,y) \in \gamma(P)$. Since $|x|=1$, there exists $\theta \in \mathbb{R}/2\pi\mathbb{Z}$ such that $x=e^{i\theta}$. We can pick $\theta$ in such a manner that $-\pi \le \theta \le \pi$. As a result,
which implies that
and $y$ has to be one of the solutions. Before finding the solutions, we notice that in order that $(x,y) \in \gamma(P)$, we must have $\operatorname{Im}(y)=0$. If not, then $y \ne \overline{y}$ and both $y$ and $\overline{y}$ are roots of $y^2+(2\cos\theta+1)y+1$, which implies that $|y|^2=y\overline{y}=1$ according to Vieta’s formula.
It follows that $\Delta = (2\cos\theta+1)^2-4 \ge 0$, which implies that $-\frac{\pi}{3}\le \theta \le \frac{\pi}{3}$. The boundary cases have to be excluded because if $\theta=\pm \frac{\pi}{3}$ then $2\cos\theta+1=2$ and the double solution to $y^2+2y+1=0$ is $y=1$.
Therefore we must have $-\frac{\pi}{3}< \theta < \frac{\pi}{3}$ and solving $y^2+(2\cos\theta+1)y+1=0$ yields
and another root is omitted because its absolute value is smaller than $1$. To conclude, points on $\gamma(P)$ are exactly of the form $(e^{i\theta},Y(\theta))$. $\square$
We terminate the post before it becomes way too long. In a future post, we will explain how to compute $m(P)$ using the method of modular forms.
Other than the linked documents in text, this post is based on Many Variations of Mahler Measures, A Lasting Symphony by François Brunault and Wadim Zudilin.
2025-10-13 22:06:01
Let $B$ be a commutative ring with unity. We say that $B$ is a Boolean ring if $x^2=x$ for all $x \in B$. The name “Boolean” certainly rings a bell of the idea of bool values in programming, or in general, the Boolean algebra that is frequently used in logic, digital electronics and computer science.
In this post, we will examine Boolean rings on a level of commutative algebra, followed by an explicit example in algebraic number theory.
Throughout, let $B$ be a Boolean ring.
Proposition 1. In the Boolean ring $B$, we have
- $2x=0$ for all $x \in B$.
- Every prime ideal $\mathfrak{p} \subset B$ is maximal, and $B/\mathfrak{p}$ is a field with two elements.
- Every finitely generated ideal in $A$ is principal.
Proof. For 1, notice that
For 2, it suffices to show that for every prime ideal $\mathfrak{p} \subset B$, we have $B/\mathfrak{p} \cong \mathbb{Z}/2\mathbb{Z}$.
Pick $x \in B \setminus \mathfrak{p}$. Then in $B/\mathfrak{p}$ we have $\overline{x}^2=\overline{x}$, where $\overline{x}=x+\mathfrak{p} \in B/\mathfrak{p}$. Therefore $\overline{x}(\overline{x}-\overline{1})=0$. However, since $B/\mathfrak{p}$ is entire, we see that we must have $\overline{x}=\overline{1}$ since $x \not\in \mathfrak{p}$. Therefore there are exactly two elements in $B/\mathfrak{p}$, namely $\overline{0}$ and $\overline{1}$.
For 3, we use the induction. If $\mathfrak{a}$ is generated by one element, there is nothing to prove. If $\mathfrak{a}=(x,y)$, then we set
This element is interesting because
Therefore for all elements $a=rx+sy$, we have
Therefore we have $\mathfrak{a}=(u)=(x+y+xy)$.
Suppose now we have proved that all ideals generated by $n$ elements are principal. Then for an ideal generated by $n+1$ elements, let’s say $\mathfrak{a}=(x_1,\dots,x_n,x_{n+1})$, for an element
there is an element $y_{n+1} \in \mathfrak{a}$ such that $a_1x_1+\dots+a_nx_n=b_{n+1}y_{n+1}$, and if we set $u_{n+1}=x_{n+1}+y_{n+1}+x_{n+1}y_{n+1}$, then
and therefore $\mathfrak{a}=(u_{n+1})$ as expected. $\square$
Indeed, if $B$ is noetherian, then we see immediately that $\dim B = 0$, where $\dim$ denotes the Krull dimension. Besides, in this case, $B$ is automatically a PID. We should notice however $B$ is not necessarily noetherian. For example
is Boolean but not noetherian because we can consider the chain of ideals
and $I_1 \subset I_2 \subset \cdots$ is not a stationary chain.
Next we see the topology of $\operatorname{Spec}B$. It is required to have the basic knowledge of the Zariski topology.
Proposition 2. Let $X=\operatorname{Spec}B$ and $X_f=X \setminus V(f)$, where $V(f)=\{\mathfrak{p} \in X:f \in \mathfrak{p}\}$ be the basic open sets of $X$ [recall that an open subset of $X$ is quasi-compact if and only if it is a finite union of sets $X_f$]. Then
- For each $f \in B$, the set $X_f$ is both open and closed in $X$.
- Let $f_1,\dots,f_n \in B$, then $X_{f_1} \cup \cdots \cup X_{f_n}=X_f$ for some $f \in B$.
- The sets $X_f$ are the only subsets of $X$ which are both open and closed.
- $X$ is a compact Hausdorff space.
Proof. By definition $X_f$ is indeed open. To show that $X_f$ is closed, it suffices to show that $V(f)$ is always open. To do this, we use the fact that $B$ is Boolean, i.e. $f^2=f$ for all $f \in B$. We see immediately that
and on the other hand,
This is to say we have $X_f = X \setminus V(f) = V(1-f)$ to be closed all the time.
For 2, we can simply use the identity $X_f=V(1-f) proved above. Indeed,
where $F(f_1,\dots,f_n) \in B$ is a finite sum and product of $f_1,\dots,f_n$ and is the element $f$ that we were looking for.
For 3, we pick a open and closed set $Y \subset X$. Since $Y$ is open, we can write $Y = \bigcup_{i \in I}X_{f_i}$ for some index set $I$. Since $Y$ is closed in $X$, we see that $Y$ is quasi-compact, and therefore the index set $I$ can be chosen to be finite. By 2, there is therefore a $f \in B$ such that $Y=X_f$.
Finally, we show that $X$ is Hausdorff. Indeed, if $\mathfrak{p},\mathfrak{q} \in X$ with $\mathfrak{p} \ne \mathfrak{q}$, then without loss of generality we can assume that there exists $x \in \mathfrak{p}$ such that $x \not\in \mathfrak{q}$. We see then $\mathfrak{p} \in V(x)$ and $\mathfrak{q} \in V(1-x)$, and both $V(x)$ and $V(1-x)$ are open, while $V(x) \cap V(1-x)=\varnothing$. $\square$
Let $K$ be a number field and let $\mathcal{O}_K$ be the ring of integers of $K$. We would expect that $\mathcal{O}_K=\mathbb{Z}[x]$ for some $x \in \mathcal{O}_K$. For example, if $d \in \mathbb{Z}\setminus\{0,1\}$ is a care-free integer, and if we set $K=\mathbb{Q}(\sqrt{d})$, then
So now we pose a question : if we consider $K=\mathbb{Q}(\sqrt{-7},\sqrt{17})$, then does there exist $x \in \mathcal{O}_K$ such that $\mathcal{O}_K = \mathbb{Z}[x]$?
Instead of trying to find such a $x$ manually, we will solve this question with a general setting.
Proposition 3. Let $m,n \in \mathbb{Z}\setminus\{0,1\}$ be distinct integers, square-free, such that $m \equiv n \equiv 1 \pmod{4}$. If we put $K = \mathbb{Q}(\sqrt{m},\sqrt{n})$, then $\mathcal{O}_K = \mathbb{Z} \oplus \mathbb{Z}\alpha \oplus \mathbb{Z}\beta\oplus\mathbb{Z}\alpha\beta$, where $\alpha=\frac{1+\sqrt{n}}{2}$ and $\beta = \frac{1+\sqrt{m}}{2}$.
Proof. First of all we notice that $[K:\mathbb{Q}]=4$ and that $\{1,\alpha,\beta,\alpha\beta\}$ as well as $\{1,\sqrt{m},\sqrt{n},\sqrt{mn}\}$ are two $\mathbb{Q}$-basis of $K$. The Galois group $G(K/\mathbb{Q})$ is isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, generated by $\sigma:\sqrt{n} \mapsto -\sqrt{n}$ and $\gamma:\sqrt{m} \mapsto -\sqrt{m}$. Therefore
We next want to show that
On one hand, $\alpha,\beta \in \mathcal{O}_K$ should be clear. Notice that
and
Therefore if we put $f(X)=X^2-X-\frac{n-1}{4}$, then $f(\alpha)=0$. Likewise, if we put $g(X)=X^2-X-\frac{m-1}{4}$, then $g(\beta)=0$. The first inclusion $\mathbb{Z} \oplus \mathbb{Z}\alpha \oplus \mathbb{Z}\beta\oplus\mathbb{Z}\alpha\beta \subset \mathcal{O}_K$ is then proved.
On the other hand, pick an arbitrary $x = a+b\sqrt{n}+c\sqrt{m}+d\sqrt{mn} \in \mathcal{O}_K \subset K$. We know on the first place that $a,b,c,d\in\mathbb{Q}$. However, we notice that
is an algebraic integer as it is the root of a monic polynomial $(X-2a)^2-4b^2n$. At the same time, we have $2a+2b\sqrt{n}$. Therefore $2a+2b\sqrt{n} \in \mathcal{O}_{\mathbb{Q}(\sqrt{n})}=\mathbb{Z}\left[\frac{1+\sqrt{n}}{2}\right]$ (since $n\equiv 1\pmod{4}$). Therefore there exists $a’,b’\in\mathbb{Z}$ such that
which implies that
Likewise, we see
and in the same way we can prove that $4c \in \mathbb{Z}$.
Finally,
from which it follows that $4d \in \mathbb{Z}$. We have therefore proved that
Finally, since
we have
If we consider the discriminant of $\mathcal{O}_K$, noted by $\Delta_K$, then
and at the same time,
However, since $m^2n^2$ is impair (as $m\equiv n \equiv 1 \pmod{4}$), we can only have
which forces $\{1,\alpha,\beta,\alpha\beta\}$ to be a $\mathbb{Z}$-basis of $\mathcal{O}_K$. $\square$
To answer our question, we restrict ourselves to the case $m \equiv n \equiv 1 \pmod{8}$. In this question we will see that the Boolean ring arises naturally.
Proposition 4. Let $m,n \in \mathbb{Z}\setminus\{0,1\}$ be distinct integers, square-free, such that $m \equiv n \equiv 1 \pmod{8}$. If we put $K = \mathbb{Q}(\sqrt{m},\sqrt{n})$, then there does not exist $t \in \mathcal{O}_K$ such that $\mathcal{O}_K = \mathbb{Z}[t]$.
Proof. The proposition invites us to try to write $\mathcal{O}_K$ as a polynomial ring over $\mathbb{Z}$. As one can see easily,
where $\alpha=\frac{1+\sqrt{n}}{2}$ and $\beta=\frac{1+\sqrt{m}}{2}$ as above with the isomorphism induced by the map
Since in our question, $m\equiv n \equiv 1 \pmod{8}$, we see that $\frac{1-n}{4},\, \frac{1-m}{4} \in 2\mathbb{Z}$. Therefore by a modulo of $2$, we obtain
where $\mathbf{F}_2$ is the finite field of $2$ elements. Here, the ring $B=\mathcal{O}_K/2\mathcal{O}_K$ is a Boolean ring. Indeed, we can now even explicitly write down $\mathcal{O}_K/2\mathcal{O}_K$ as $\mathbf{F}_2[x,y]$ with $x^2=x$ and $y^2=y$. All elements of $\mathcal{O}_K/2\mathcal{O}_K$ can be identified as $a+bx+cy+dxy$ with $a,b,c,d\in\mathbf{F}_2$. There are $2^4=16$ elements in total, and it can be easily seen that $(a+bx+cy+dxy)^2=a+bx+cy+dxy$.
Since $B\cong \mathbf{F}_2[x,y]$ is Boolean, all prime ideals are maximal. There are exactly $4$ maximal ideals:
For a homomorphism $\varphi:B \to \mathbf{F}_2$, we have $\varphi(0)=0$, $\varphi(1)=1$ so $\varphi$ is surjective. The kernel $\ker\varphi$ is therefore a maximal ideal. There are thus exactly $4$ homomorphisms $B \to \mathbf{F}_2$, which correspond to, sending $x$ to $0$ and $y$ to $1$, sending $x$ to $1$ and $y$ to $0$, sending $x$ and $y$ to $1$ and finally sending $x$ and $y$ to $0$, respectively.
Now we show that we cannot pick $t \in \mathcal{O}_K$ such that $\mathcal{O}_K=\mathbb{Z}[t]$. To reach a contradiction, we suppose that such a $t$ exist. It follows that
where $P$ is a polynomial of degree $4$. However this is absurd because for a homomorphism
we can only have two possibilities: $\psi(X)=1$ or $\psi(X)=0$. However we have shown that $\mathcal{O}_K/2\mathcal{O}_K$ can be mapped onto $\mathcal{F}_2$ in $4$ ways. A contradiction. $\square$
Therefore unfortunately, for the number field $K=\mathbb{Q}(\sqrt{-7},\sqrt{17})$, we cannot find $x \in \mathcal{O}_K$ such that $\mathcal{O}_K=\mathbb{Z}[x]$.
2025-05-16 22:55:17
Throughout, let $K$ be a field of characteristic $p\ne 0$ and $E/K$ a cyclic extension of order $p^{m-1}$ with $m >1$. The algebraic closure $\overline K^\mathrm{a}$, the separable algebraic closure $\overline K^{\mathrm{s}}$ are always fixed. We use $\mathbf{F}_p$ to denote the finite field of $p$ elements.
For proposition 2 in the post, let $G$ be the Galois group of the extension of $\overline K^\mathrm{s}/K$ (which is, the projective limit of $\mathrm{Gal}(K’/K))$, with $K’$ running over all finite and separable extension of $K$; see this post for the definition of projective limit). The reader is expected to know how to induce a long exact sequence from a short exact sequence, for example from this post.
In this post (the reader is urged to make sure that he or she has understood the concept of characters and more importantly Hilbert’s theorem 90), we have shown that if $[E:K]=p$, then $E=K(x)$ where $x$ is the zero of a polynomial of the form $X^p-X-\alpha$ where $\alpha \in K$. In this belated post, we want to show that, whenever it comes to an extension of order $p^{m-1}$, we are running into the a polynomial of the form $X^p-X-\alpha$. The theory behind is called Artin-Schreier theory, which has its own (highly non-trivial) nature.
Definition 1. An Artin-Schreier polynomial $A_\alpha(X) \in K[X]$ is of the form
An immediate property of Artin-Schreier polynomials that one should notice is the equation
To see this, one should notice that for $x,y \in K$ we have $(x+y)^p=x^p+y^p$.
With this equation we can easily show that
Proposition 1. If $A_\alpha(X)$ has a root in $K$, then all roots of $A_\alpha(X)$ is in $K$. Otherwise, $A_\alpha(X)$ is irreducible over $K$. In this case, let $x$ be a root of $A_\alpha(X)$, then $K(x)/K$ is a cyclic extension of degree $p$.
Proof. We suppose that $x \in K$ is a root of $A_\alpha(X)$. Then
Therefore, by induction, we see easily that $x, x+1, \cdots, x+p-1$ are roots of $A_\alpha(X)$, all of which are in $K$.
Now we suppose that $A_\alpha(X)$ has no root in $K$. Let $x \in \overline K$ be a root of $A_\alpha(X)$. Then in $\overline K[X]$, the polynomial will be written in the form
because, again due to the equation $A_\alpha(X+Y)=A_\alpha(X)+A_\alpha(Y)-A_\alpha(0)$, we can see that $x,x+1,\dots,x+p-1$ are roots of $A_\alpha$.
By contradiction we suppose that $A_\alpha$ is reducible, say $A_\alpha(X)=f(X)g(X)$ where $1 \le d=\deg f < p$ and $f,g \in K[X]$. It follows that
where $\{n_1,\dots,n_d\} \subset \{1,2,\cdots,p\}$. If we expand the polynomial above, we see
Therefore $\left(\sum_{j=1}^{d}n_j-dx\right) \in K$ which is absurd because we then have $x \in K$. Therefore we see that $A_\alpha$ is irreducible.
To see that $K(x)/K$ is Galois, we first notice that this extension is normal : $K(x)$ contains all roots of $A_\alpha(X)$. This extension is separable because all roots of $A_\alpha(X)$, namely $x,x+1,\dots,x+p-1$, are pairwise distinct, i.e. $A_\alpha(X)$ has no multiple roots.
Finally, to see why the Galois group of $K(x)/K$ is cyclic, we notice the action of the Galois group $G$ over the roots of $A_\alpha(X)$. Since $A_\alpha(X)$ is irreducible, there exists $\sigma \in G$ such that $\sigma(x)=x+1$. We see easily that $\sigma^j(x)=x+j$ so $\sigma$ generates $G$ which has period $p$. $\square$
The correspondence between extensions of degree $p$ and polynomials of the form $X^p-X-\alpha$ inspires us to consider them in a distinguished manner.
Definition 2. The field extension $E/K$ is called an Artin-Schreier extension if $E=K(x)$ for some $\alpha \in L \setminus K$ such that $x^p-x\in K$.
Consider the map $\wp:\overline K^\mathrm{s} \to \overline K^\mathrm{s}$ defined by $u \mapsto u^p-u$. We certainly want to find the deep relation between Artin-Schreier extensions of a given field $K$ and the map $\wp$. One of the key information can be found through the following correspondence.
Proposition 2. There is an isomorphism $\operatorname{Hom}(G,\mathbf{F}_p) \cong K/\wp(K)$.
Proof. We first notice that $\wp$ is a $G$-homomorphism, that is, it commutes with the action of $G$ on $\overline K^\mathrm{s}$. Indeed, for any $x \in \overline K^\mathrm{s}$ and $g \in G$, we have
On the other hand, $\wp$ is surjective. Indeed, for any $a \in \overline{K}^\mathrm{s}$, the equation $X^p-X=a$ always has a solution in $\overline K^\mathrm{s}$ because the polynomial $X^p-X-a$ is separable.
We can also see that the kernel of $\wp$ is $\mathbf{F}_p$. This is because the splitting field of $X^p-X$ is the field of $p^1$ elements, which has to be $\mathbf{F}_p$ itself. Therefore we have obtained a short exact sequence
where $\iota$ is the embedding. Taking the long exact sequence of cohomology, noticing that, by Hilbert’s Theorem 90, $H^1(G,\overline{K}^\mathrm{s})=0$, we have another exact sequence
where the first arrow is induced by $\wp$ and the second by $\iota$. Therefore we have $\operatorname{Hom}(G,\mathbf{F}_p) \cong K/\wp(K)$. One can explicitly show that there is a surjective map $K \to \operatorname{Hom}(G,\mathbf{F}_q)$ with kernel $\wp(K)$ that defines the isomorphism. For $c \in K$, one solves $x^p-x=c$, then $\varphi_c:g\mapsto g(x)-x$ is the desired map. The key ingredient of the verification involves the (infinite) Galois correspondence, but otherwise the verification is very tedious. We remark that for any $\varphi \in \operatorname{Hom}(G,\mathbf{F}_p)\setminus\{0\}$ and put $H=\ker\varphi$. Then $K^H/K$ is an Artin-Schreier extension with Galois group $G/H$ and on the other hand $H=\mathrm{Gal}(\overline K^\mathrm{s}/K^H)$. $\square$
We conclude this post by showing that, under a certain condition, one can find an Artin-Schreier extension $L/E$ such that $L/K$ is cyclic of order $p^m$.
Lemma 1. Let $\beta \in E$ be an element such that $\operatorname{Tr}_K^E(\beta)=1$, then there exists $\alpha \in K$ such that $\sigma(\alpha)-\alpha = \beta^p-\beta$, where $\sigma$ is the generator of $\operatorname{Gal}(E/K)$.
Proof. Notice that $\operatorname{Tr}_K^E(\beta^p)=\operatorname{Tr}_K^E(\beta)^p=1$, which implies that $\operatorname{Tr}_K^E(\beta^p-\beta)=0$. By Hilbert’s theorem 90, such $\alpha$ exists. $\square$
Lemma 2. The polynomial $f(X)=X^p-X-\alpha$ is irreducible over $E$; that is, let $\theta$ be a root of $f$, then $E(\theta)$ is an Artin-Schreier extension of $E$.
Proof. By contradiction, we suppose that $\theta \in E$. By Artin-Schreier, all roots of $f$ lie in $E$. In particular, $\sigma(\theta)$ is a root of $f$. Therefore
which implies that
It follows that $\sigma\theta-\theta-\beta$ is a root of $g(X)=X^p-X$. This implies that $\sigma\theta-\theta-\beta\in\mathbf{F}_p \subset K$ and therefore
However, by assumption and Artin-Schreier, $\sigma\theta-\theta \in \mathbf{F}_p \subset K$ we therefore have $\operatorname{Tr}_K^E(\sigma\theta-\theta)=0$ and finally
which is absurd. $\square$
Proposition 3. The field extension $K(\theta)/K$ is Galois, cyclic of degree $p^m$ of $f$, whose Galois group is generated by an extension $\sigma^\ast$ of $\sigma$ such that
Proof. First of all we show that $K(\theta)=E(\theta)$. Indeed, since $K \subset E$, we have $K(\theta) \subset E(\theta)$. However, since $\theta \not \in E$, we must have $K \subset E \subsetneq K(\theta)$. Therefore $p=[E(\theta):K(\theta)][K(\theta):E]$, which forces $E(\theta)$ to be exactly $K(\theta)$.
Let $h(X)$ be the minimal polynomial of $\theta$ over $K$ of degree $p^m$. Then we give an explicit expression of $h$. Notice that since $f(X)$ is the polynomial of $\theta$ over $E$ of degree $p$, we must have $f(X)|h(X)$. For any $k$, we see that $f^{\sigma^k}(X)|g^{\sigma^k}(X)$ too. However, since $\sigma$ fixes $K$, we must have $g^{\sigma^k}(X)=g(X)$, from which it follows that $f^{\sigma^k}(X)|g(X)$ for all $0 \le k \le p^{m-1}-1$. Since the degree of each $f^{\sigma^k}(X)$ is $p$, we obtain
Knowing that $\theta$ is a root of $g$, we see that $\theta+\beta$ is a root of $g(X)$ too because
and by induction we see that for $0 \le k \le p^{m-1}-1$, $f^{\sigma^k}(X)$ has a root in the form
By Artin-Schreier, all roots of $f^{\sigma^k}(X)$ lie in $E(\theta)$ and therefore $h(X)$ splits in $E(\theta)$. Since $E(\theta)/E$ is separable, $E/K$ is separable, we see also $E(\theta)/K$ is separable, which means that $E(\theta)=K(\theta)$ is Galois over $K$.
To see why $K(\theta)/K$ is cyclic, we consider an homomorphism $\sigma^\ast$ of $K(\theta)$ such that $\sigma^{\ast}|_E=\sigma$ and that $\sigma^\ast(\theta)=\theta+\beta$. It follows that $\sigma^\ast \in \operatorname{Gal}(K(\theta)/K)$ because its restriction on $K$, which is the restriction of $\sigma$ on $K$, is the identity. We see then for all $0 \le n \le p^{m}$, one has
In particular,
from which it follows that $(\sigma^\ast)^{p^{m-1}}$ has order $p$, which implies that $\sigma^\ast$ has order $p^m$, thus the Galois group is generated by $\sigma^\ast$. $\square$
2025-05-11 07:33:19
Regular local rings are important objects in modern algebra, number theory and algebraic geometry. Therefore it would be way too ambitious to try to briefly justify the motivation of studying regular local rings. In this post, we try to collect equivalent conditions of being a regular local ring of dimension $1$ and prove them. There are plenty of equivalent conditions and it is difficult to find a book that collects as many as them as possible, let alone giving a detailed proof. The reader is also encouraged to prove the conditions himself, after knowing that the most important tool in the proof is Nakayama’s lemma.
The reader may have come up with the definition of discrete valuation rings, without knowing the motivation. Indeed, one way to interpret discrete valuation rings is to see them as “Taylor expansions”. The analogy after the definition may explain why.
Definition 1. Let $F$ be a field. A surjective function $F:\mathbb{Z} \to \{\infty\}$ is called a discrete valuation if
- $v(\alpha)=\infty \iff \alpha = 0$;
- $v(\alpha\beta)=v(\alpha)+v(\beta)$;
- $v(\alpha+\beta)\ge\min(v(\alpha),v(\beta))$.
The ring $R_v=\{\alpha \in F:v(\alpha) \ge 0\}$ is called a discrete valuation ring. It is a local ring with maximal ideal $\mathfrak{m}_v=\{\alpha \in F:v(\alpha) > 0\}$.
We should not compare $R_v$ with a polynomial ring, as all polynomial rings are not local. Let $t \in \mathfrak{m}_v$ be an element such that $v(t)=1$. We will show that $\mathfrak{m}_v = (t)$. Indeed, for any $u \in \mathfrak{m}_v$, we see that
and as a result we can write $u=(ut^{-1})t$. If we look further, suppose that $v(u)=m$. Then $\alpha = ut^{-m} \in R_v$ is a unit and thus we have $u=\alpha t^m$. In other words, every element can be expressed as a monomial of $t$.
The analogy or even example to bring about here is the order of zero at origin of (rational) functions over $\mathbb{R}$. For a rational function $F(x)=f(x)/g(x)$, we see that if we define $v(F)=\deg{f}-\deg{g}$, then $\lim_{x\to 0}\frac{F(x)}{x^m}$ is non-zero and finite. The degree of zero polynomial depends on the context, and in our context we make it infinite as no matter how big $m$ is, we are never reaching a point that $\lim_{x \to 0}\frac{0}{x^m}$ is non-zero and finite. Therefore the discrete valuation ring in our story is the polynomials where the function is equivalent to a monomial of positive degree, and the generator of the maximal ideal is the “identity” map. In short, one way of imagining the discrete valuation ring is the space of “smooth” functions at a point that converge to $0$ with the evaluation being the degree of approximation.
For a ring $R$, we use $\dim(R)$ to denote the Krull dimension and for a vector space $V$ over a field $K$, $\dim_K(V)$ is used to denote the dimension of $V$ as a vector space over $K$.
Theorem 2. Let $R$ be a commutative noetherian local ring with unit and maximal ideal $\mathfrak{m}$ with the residue field $\kappa=R/\mathfrak{m}$. Then the following conditions are equivalent.
- $R$ is a discrete valuation ring in its field of fraction;
- $\dim_\kappa(\mathfrak{m}/\mathfrak{m}^2)=\dim(R)=1$, i.e., $R$ is a regular local ring of dimension $1$;
- $R$ is a unique factorization domain of Krull dimension $1$;
- $\mathfrak{m}$ is a principal ideal and $\dim(R)=1$.
- $R$ is a principal ideal domain which is not a field;
- $R$ is an integrally closed domain of Krull dimension $1$.
（N. B. - We have to assume the axiom of choice by all means, otherwise none of these makes sense. In fact, without assuming the axiom of Choice, it is unprovable that a principal ideal domain has a maximal ideal or the ring has a prime element when it is not a field. See this article for more details.)
Proof. Suppose first that $R$ is a discrete valuation ring with a discrete valuation $v$. Then $\mathfrak{m}=\{a\in R:v(a)>0\}$ is the maximal ideal of $R$ that can be generated by an element $t \in \mathfrak{m}$ such that $v(t)=1$. Let $\mathfrak{a}$ be another ideal of $R$ and let $k=\min v(\mathfrak{a})$. There is an element $x \in \mathfrak{a}$ such that $v(x)=k$ and we can write $x=ut^k$ where $u$ is a unit of $R$. For any other element $y \in \mathfrak{a}$, we have $\ell=v(y)\ge k$ and therefore $y=vt^{\ell}=vu^{-1}t^{\ell-k}ut^{k}=vu^{-1}t^{\ell-k}x$. In other words, we have $\mathfrak{a}=(x)=(t^k)$ for some $k \ge 1$. When $k>1$, the ideal $(t^k)$ is not prime let alone maximal, so we have shown that when $R$ is a discrete valuation ring, the maximal ideal $\mathfrak{m}$ is principal, the Krull dimension of $R$ is $1$ and $R$ is principal but not a field because the maximal ideal is not zero.
This is to say, we have $1 \implies 4,5$. Since a principal ideal domain is also a unique factorization domain, we also get $3$. Besides, we have shown that in all 6 scenarios, the ring $R$ is of Krull dimension $1$. Therefore from now on we assume that $R$ is a commutative noetherian local ring of Krull dimension $1$ a priori. This condition implies that the maximal ideal $\mathfrak{m}$ is not nilpotent because $\mathfrak{m}$ is nilpotent if and only if the dimension of $R$ would be $0$ (hint: Nakayama’s lemma; consider the possibility that $\mathfrak{m}^n=\mathfrak{m}^{n+1}$).
Now assume that $\mathfrak{m}$ is principal and we write $\mathfrak{m}=(t)$ for some $t\in\mathfrak{m}$. For any $a \in R \setminus \{0\}$, if $a$ is invertible, then we can write $a=at^{0}$. Otherwise we have $a\in\mathfrak{m}$ and therefore $a=a_1t$ for some $a_1 \in R\setminus\{0\}$. We show that there exists a unique $n \ge 0$ such that $a = a_n t^n$ where $a_n$ is a unit in $R$.
When $a$ is a unit, as shown above, there is nothing to prove. Therefore, to reach a contradiction, we suppose that such $n$ does not exist when $a$ is not a unit. Then by induction, for each $j>0$, there exists $a_j \in R\setminus\{0\}$ such that $a=a_jt^j$, which means that $a \in (t^j)=\mathfrak{m}^j$ for all $j$. By Krull’s intersection theorem, we have $\bigcap_{j=1}^{\infty}\mathfrak{m}^j=\{0\}$ (this is a consequence of Nakayama’s lemma and Artin-Rees lemma), and therefore $a=0$, which is absurd. Therefore the desired $n$ always exists.
Next we show that such $n$ is unique. Suppose that $a = a_m t^m=a_nt^n$ where $a_m,a_n \in R^\times$ and without loss of generality we assume that $m \ge n$. Then $a - a = (a_mt^{m-n}-a_n)t^n=0$. Since $t$ is not nilpotent, we must have $a_mt^{m-n}-a_n=0$. In this case we must have $m=n$ and $a_m=a_n$ because otherwise $a_mt^{m-n}$ would not be a unit in $R$.
Therefore for all $a\in R \setminus\{0\}$, we can always uniquely write $a = ut^{v(a)}$ where $v(a) \ge 0$ is an integer. Since $t$ is not nilpotent, we see that $R$ is an integral domain and it is a discrete valuation ring in its field of fraction. Besides, $R$ is a principal ideal domain because for any ideal $\mathfrak{a} \subset \mathfrak{m}$, the ideal is generated by the element $a=v^{-1}(\min v(\mathfrak{a}))$.
Next we study the dimension of $\mathfrak{m}/\mathfrak{m}^2$ over $\kappa$, where $\mathfrak{m}=(t)$. Notice that $\dim_\kappa \mathfrak{m}/\mathfrak{m}^2\ge 1$ because otherwise $t=1$ or $0$. We show that $\dim_\kappa\mathfrak{m}/\mathfrak{m}^2 <2$ under the assumption of 4. Let $u,v\in \mathfrak{m}/\mathfrak{m}^2$ be two distinct non-zero vectors. We show that there exists $\alpha \in \kappa$ such that $\alpha u = -v$. Suppose that $u = rt \pmod{\mathfrak{m}^2}$ and $v = st \pmod{\mathfrak{m}^2}$. Then $r,s \not\in \mathfrak{m}$ because otherwise $u=v=0$. If we choose $\alpha = -\frac{s}{r}\pmod{\mathfrak{m}}$, we see that $\alpha u = -st\pmod{\mathfrak{m}^2}=-v$ as desired.
To conclude, we have shown that $4 \implies 1,2,5$.
Moving on, we assume 5 and see what we can get. First of all every principal ideal domain is a unique factorisation ring so we get $3$ (axiom of choice is indispensable here). Besides since every ideal is principal then in particular the maximal ideal is principal so we get $4$. To conclude, we get $5 \implies 3,4$.
Finally we need to study the points 2,3 and 6. To begin with, we assume 3. Then by an elementary verification we see that $R$ is integrally closed (see ProofWiki). Next we show that $\mathfrak{m}$ is principal. Let $\mathscr{P}$ be the family of proper principal ideals of $R$ (they are contained in $\mathfrak{m}$ since $R$ is local). Then the set $\mathscr{P}$ is ordered by inclusion and every chain has a maximal element given by the union. By Zorn’s lemma, in $\mathscr{P}$ there is a maximal element $\mathfrak{M} \in\mathscr{P}$ that contains all proper principal ideals. Next we show that $\mathfrak{M}$ is maximal hence it is equal to $\mathfrak{m}$. To see this, assume that $a \in R \setminus \mathfrak{M}$. Then $(a)$ is not a proper principal ideal of $R$ because otherwise $(a) \subset \mathfrak{M} \implies a \in \mathfrak{M}$. Therefore $a$ is a unit and $\mathfrak{M}$ is the maximal ideal of the local ring $R$, which means $\mathfrak{M}=\mathfrak{m}$. This shows that $3 \implies 4,6$.
Next we assume 2. We use proposition 2 of this old post, only need to notice that the dimension of $\mathfrak{m}/\mathfrak{m}^2$ is exactly the number of generators of $\mathfrak{m}$. Therefore we obtain $2 \implies 4$.
For the last part we assume that $R$ is integrally closed. Choose an arbitrary non-unit $a \in R$. If $a=0$ then $a \in \mathfrak{m}$. Otherwise, consider the ring $\widetilde{R}=R_\mathfrak{m}/aR_\mathfrak{m}$ which is not a field. Then $\tilde{R}$ is of Krull dimension $0$ therefore the maximal ideal $\tilde{\mathfrak{m}}=\mathfrak{m}R_\mathfrak{m}/aR_\mathfrak{m}$, is nilpotent. There exists $n>0$ such that $\tilde{\mathfrak{m}}^n\ne 0$ but $\tilde{\mathfrak{m}}^{n+1}=0$, which implies that $\mathfrak{m}^n \not \subset (a)$ but $\mathfrak{m}^{n+1} \subset (a)$. Choose $b\in (a) \setminus \mathfrak{m}^n$. Then we claim that $\mathfrak{m}=(x)$ where $x=a/b \in K(R)$, the field of fraction of $R$. To see this, notice that $x^{-1}\mathfrak{m} \subset R$ because $b\mathfrak{m} \subset \mathfrak{m}^{n+1} \subset (a)$ so every element of $b\mathfrak{m}$ is of the form $ua$ where $u \in R$ and consequently every element of $\frac{b}{a}\mathfrak{m}$ is of the form $u$ where $u \in R$. Therefore $x^{-1}\mathfrak{m}$ can be considered as an ideal of $R$. However, we also have $x^{-1}\mathfrak{m} \not\subset \mathfrak{m}$ which is because, otherwise, $\mathfrak{m}$, as a finitely generated $R$-module, would be a faithful $R[x^{-1}]$-module, and therefore $x^{-1}$ is integral over $R$, thus lies in $R$. Hence we must have $x^{-1}\mathfrak{m}=R$, which implies that $\mathfrak{m}=(x)$. Therefore we obtain $6 \implies 4$.
We have established all necessary implications to obtain the equivalences. $\square$
2023-11-12 06:12:19
Let $\mathbb{F}_3$ be the field of three elements and $SL_2(\mathbb{F}_3)$ be the group of $2 \times 2$ matrices with determinant $1$. In this post we show that $SL_2(\mathbb{F}_3)$ is the semi-direct product of $H_8$ and $\mathbb{Z}/3\mathbb{Z}$.
First of all we determine the cardinality of $SL_2(\mathbb{F}_3)$. To do this, we consider $GL_2(\mathbb{F}_3)$ and notice that $SL_2(\mathbb{F}_3)$ is the kernel of $\det$ function.
To determine $GL_2(\mathbb{F}_3)$, fix a basis of $\mathbb{F}_3 \oplus \mathbb{F}_3$ and let $A$ be a matrix representation of an element in $GL_2(\mathbb{F}_3)$. The first column of $A$ has $3^2-1$ number of choices: we only exclude the case of $(0,0)^T$. The second column has $3^2-3$ choices. We exclude $3$ possibilities given by the scalar multiplication of the first column to prevent linear dependence. Therefore $|GL_2(\mathbb{F}_3)|=(3^2-1)(3^2-3)=48$. Next we consider the exact sequence
We get $|SL_2(\mathbb{F}_3)|=|GL_2(\mathbb{F}_3)|/(\mathbb{F}_3)^\ast|=48/2=24$.
We immediately think about the possibility that $SL_2(\mathbb{F}_3)\cong \mathfrak{S}_4$. Is that the case?
As a group of order 24, we immediately consider the elements of order $2$, $3$ and $4$ in order to know the structure of the group we are looking at.
There are ${4 \choose 2}/2!=3$ elements of order $2$ in $\mathfrak{A}_4$, i.e. those being products of two $2$-cycles. However, how many elements of order $2$ are there in $SL_2(\mathbb{F}_3)$? Let $A$ be such an element, then $A^2 = I$. Therefore all elements of order $2$ is nullified by the polynomial
If $A \in SL_2(\mathbb{F}_3)$ is of order $2$, then the minimal polynomial of $A$ divides $f(X)$, hence is either $X+1$ or $X^2-1$. The second case is impossible because then $f(X)$ will be the characteristic polynomial of $A$ and therefore $A$ has eigenvalue $1$ and $-1$ thus determinant $-1$. We get
Proposition 1. The element in $SL_2(\mathbb{F}_3)$ of order $2$ is only $A=-I$. In particular, $SL_2(\mathbb{F}_3)$ is not isomorphic to $\mathfrak{S}_4$.
Checking elements of order $2$ is not out of nowhere. Since $24=2^3 \cdot 3$, it makes sense to look at $2$-Sylow and $3$-Sylow subgroups of $SL_2(\mathbb{F}_3)$. Sylow’s theorem ensures that there is a subgroup of order $3$, which can only be $\mathbb{Z}/3\mathbb{Z}$. We have also determined that the subgroup of order $2$ is $\{-I,I\}$. Next we determine the group of order $8$.
To study elements of order $4$, we immediately consider the polynomial
Let $A \in SL_2(\mathbb{F}_3)$ be an element of order $4$. Then $g(A)=0$. But since $A+I \ne 0$ and $A-I \ne 0$, we will be considering $h(X)=X^2+1$ instead. Notice that $h(X)$ is irreducible in $\mathbb{F}_3[X]$ and therefore it is minimal polynomial of $A$. Since the degree of $h$ is $2$, we also see $h(X)$ is the characteristic polynomial of $A$.
From this polynomial we see that $\mathrm{tr}(A)=0$. Combining with the fact that $|A|=1$, we can easily deduce that elements of order $4$ consists of
We in particular have $i^3=i^{-1}=-i$, $j^3=j^{-1}=-j$ and $k^3=k^{-1}=-k$. Furthermore, $k=ij=-ji$. These identities rings a bell of quaternion number. We therefore have the quaternion group lying in $SL_2(\mathbb{F}_3)$ as a $2$-Sylow subgroup:
Is there any other $2$-Sylow subgroup? The answer is no. To see this, let $H’$ be another $2$-Sylow group. Then there exists some $g \in SL_2(\mathbb{F}_3)$ such that $H’=gH_8 g^{-1}$, which is equal to $H_8$ because all elements in $K$ will have order $4$.
Proposition 2. The quaternion group $H_8$ can be embedded into $SL_2(\mathbb{F}_3)$ as the unique $2$-Sylow group. In particular, $SL_2(\mathbb{F}_3)$ has no element of order $8$.
Let $A \in SL_2(\mathbb{F}_3)$ be an element of order $3$. Then its minimal polynomial $m(X)$ divides $X^3-1=(X-1)^3=(X-1)^2(X-1)$. Since $A-I \ne 0$, we must have $m(X)=(X-1)^2=X^2+X+1$. We can also see that the characteristic polynomial of $A$ is also $X^2+X+1$. In particular, we see the trace of $A$ is $-1=2$. We can then choose
Therefore $K=\{I,A,A^2\}$ is a $3$-Sylow subgroup of $SL_2(\mathbb{F}_3)$, which is not unique, because for example one can also consider the group generated by the transpose of $A$.
Notice that $H \cap K = \{1\}$ because $\gcd(3,4)=1$. Therefore the map $H \times K \to HK$ given by $(x,y) \mapsto xy$ is bijective. Since $H$ is also normal, we are safe to write $G=H\ltimes K$ because $|HK|=|H||K|=24=|G|$.
2023-10-21 16:39:01
Polynomial is of great interest in various fields, such as analysis, geometry and algebra. Given a polynomial, we try to extract as many information as possible. For example, given a polynomial, we certainly want to find its roots. However this is not very realistic. Abel-Ruffini theorem states that it is impossible to solve polynomials of degree $\ge 5$ in general. For example, one can always solve the polynomial $x^n-1=0$ for arbitrary $n$, but trying to solve $x^5-x-1=0$ over $\mathbb{Q}$ is not possible. Galois showed that the flux of solvability lies in the structure of the Galois group, depending on whether it is solvable group-theoretically.
In this post, we will explore the theory of solvability in the modern sense, considering extensions of arbitrary characteristic rather than solely number fields over $\mathbb{Q}$.
Definition 1. Let $E/k$ be a separable and finite field extension, and $K$ the smallest Galois extension of $k$ containing $E$. We say $E/k$ is solvable if $G(K/k)$ (the Galois group of $K$ over $k$) is solvable.
Throughout we will deal with separable extensions because without this assumption one will be dealing with normal extensions instead of Galois extensions. Although we will arrive at a similar result.
Proposition 1. Let $E/k$ be a separable extension. Then $E/k$ is solvable if and only if there exists a solvable Galois extension $L/k$ such that $k \subset E \subset L$.
Proof. If $E/k$ is solvable, it suffices to take $L$ to be the smallest Galois extension of $k$ containing $E$. Conversely, Suppose $L/k$ is a solvable and Galois such that $k \subset E \subset L$. Let $K$ be the smallest Galois extension of $k$ containing $E$, i.e. we have $k \subset E \subset K \subset L$. We see $G(K/k) \cong G(L/k)/G(L/K)$ is a homomorphism image of $G(L/k)$ and it has to be solvable. $\square$
Next we introduce an important concept concerning field extensions.
Definition 2. Let $\mathcal{C}$ be a certain class of extension fields $F \subset E$. We say that $\mathcal{C}$ is distinguished if it satisfies the following conditions:
- Let $k \subset F \subset E$ be a tower of fields. The extension $k \subset E$ is in $\mathcal{C}$ if and only if $k \subset F$ is in $\mathcal{C}$ and $F \subset E$ is in $\mathcal{C}$.
- If $k \subset E$ is in $\mathcal{C}$ and if $F$ is any given extension of $k$, and $E,F$ are both contained in some field, then $F \subset EF$ is in $\mathcal{C}$ too. Here $EF$ is the compositum of $E$ and $F$, i.e. the smallest field that contains both $E$ and $F$.
- If $k \subset F$ and $k \subset E$ are in $\mathcal{C}$ and $F,E$ are subfields of a common field, then $k \subset FE$ is in $\mathcal{C}$.
When dealing with several extensions at the same time, it can be a great idea to consider the class of extensions they are in. For example, Galois extension is not distinguished because normal extension does not satisfy 1. That’s why we need to have the fundamental theorem of Galois theory, a.k.a. Galois correspondence, because not all intermediate subfields are Galois. Separable extension is distinguished however. We introduce this concept because:
Proposition 2. Solvable extensions form a distinguished class of extensions. (N.B. these extensions are finite and separable by default.)
Proof. We verify all three conditions mentioned in definition 2. To make our proof easier however, we first verify 2.
Step 1. Let $E/k$ be solvable. Let $F$ be a field containing $k$ and assume $E, F$ are subfields of some algebraically closed field. We need to show that $EF/F$ is solvable. By proposition 1, there is a Galois solvable extension $K/k$ such that $K \supset E \supset k$. Then $KF$ is Galois over $F$ and $G(KF/F)$ is a subgroup of $G(K/k)$. Therefore $KF/F$ is a Galois solvable extension and we have $KF \supset EF \supset F$, which implies that $EF/F$ is solvable.
Step 2. Consider a tower of extensions $E \supset F \supset k$. Assume now $E/k$ is solvable. Then there exists a Galois solvable extension $K$ containing $E$, which implies that $F/k$ is solvable because $K \supset F$. We see $E/F$ is also solvable because $EF=E$ and we are back to step 1.
Conversely, assume that $E/F$ is solvable and $F/k$ is solvable. We will find a solvable extension $M/k$ containing $E$. Let $K/k$ be a Galois solvable extension such that $K \supset F$, then $EK/K$ is solvable by step 1. Let $L$ be a Galois solvable extension of $K$ containing $EK$. If $\sigma$ is any embedding of $L$ over $k$ in a given algebraic closure, then $\sigma K = K$ and hence $\sigma L$ is a solvable extension of $K$. [This sentence deserves some explanation. Notice that $L/k$ is not necessarily Galois, therefore $\sigma$ is not necessarily an automorphism of $L$ and $\sigma L \ne L$ in general . However, since $K/k$ is Galois, the restriction of $\sigma$ on $K$ is an automorphism so therefore $\sigma K = K$. The extension $\sigma L / \sigma K$ is solvable because $\sigma L$ is isomorphic to $L$ and $\sigma K = K$.]
We let $M$ be the compositum of all extensions $\sigma L$ for all embeddings $\sigma$ of $L$ over $k$. Then $M/k$ is Galois and so is $M/K$ [note: this is the property of normal extension; besides, $M/k$ is finite]. We have $G(M/K) \subset \prod_{\sigma}G(\sigma L/K)$ which is a product of solvable groups. Therefore $G(M/K)$ is solvable, meaning $M/K$ is a solvable extension. We have a surjective homomorphism $G(M/k) \to G(K/k)$ (given by $\sigma \mapsto \sigma|_K$) and therefore $G(M/k)$ has a normal subgroup whose factor group is solvable, meaning $G(M/k)$ is solvable. Since $E \subset M$, we are done.
Step 3. If $F/k$ and $E/k$ are solvable and $E,F$ are subfields of a common field, we need to show that $EF$ is solvable over $k$. By step 1, $EF/F$ is solvable. By step 2, $EF/k$ is solvable. $\square$
Definition 2. Let $F/k$ be a finite and separable extension. We say $F/k$ is solvable by radicals if there exists a finite extension $E$ of $k$ containing $F$, and admitting a tower decomposition
such that each step $E_{i+1}/E_i$ is one of the following types:
- It is obtained by adjoining a root of unity.
- It is obtained by adjoining a root of a polynomial $X^n-a$ with $a_i \in E_i$ and $n$ prime to the characteristic.
- It is obtained by adjoining a root of an equation $X^p-X-a$ with $a \in E_i$ if $p$ is the characteristic $>0$.
For example, $\mathbb{Q}(\sqrt{-2})/\mathbb{Q}$ is solvable by radicals. We consider the polynomial $f(x)=x^2-2x+3$. We know its roots are $x_1=-1-\sqrt{-2}$ and $x_2=-1+\sqrt{-2}$. However let’s see the question in the sense of field theory. Notice that
Therefore $f(x)=0$ is equivalent to $(x-1)^2=-2$. Then $x-1=\sqrt{-2}$ and $x-1=-\sqrt{-2}$ in $\mathbb{Q}(\sqrt{-2})$ are two equations that make perfect sense. Thus we obtain our desired roots. The field gives us the liberty of basic arithmetic, and the radical extension gives us the method to look for a radical root.
It is immediate that the class of extensions solvable by radicals is a distinguished class.
In general, we are adding “$n$-th root of something”. However, when the characteristic of the field is not zero, there are some complications. For example, talking about the $p$-th root of an element in a field of characteristic $p>0$ will not work. Therefore we need to take good care of that. The second and third types are nods to Kummer theory and Artin-Schreier theory respectively, which are deduced from Hilbert’s theorem 90’s additive and multiplicative form. We interrupt the post by introducing the respective theorems.
Let $K/k$ be a cyclic extension of degree $n$, that is, $K/k$ is Galois and $G(K/k)$ is cyclic. Suppose $G(K/k)$ is generated by $\sigma$. Then we have the celebrated “Theorem 90”:
Theorem 1 (Hilbert’s theorem 90, multiplicative form). Notation being above, let $\beta \in K$. The norm $N_{k}^{K}(\beta)=1$ if and only if there exists an element $\alpha \ne 0$ in $K$ such that $\beta = \alpha/\sigma\alpha$.
To prove this, we need Artin’s theorem of independent characters. With this, we see the second type of extension in definition 2 is cyclic.
Theorem 2. Let $k$ be a field, $n$ an integer $>0$ prime to the characteristic of $k$, and assume that there is a primitive $n$-th root of unity in $k$.
- Let $K$ be a cyclic extension of degree $n$. Then there exists $\alpha \in K$ such that $K = k(\alpha)$ and $\alpha$ satisfies an equation $X^n-a=0$ for some $a \in k$.
- Conversely, let $a \in k$. Let $\alpha$ be a root of $X^n-a$. Then $k(\alpha)$ is cyclic over $k$ of degree $d|n$, and $\alpha^d$ is an element of $k$.
All in all, theorem 2 states that a $n$-th root of $a$ yields a cyclic extension. However we don’t drop the assumption that $n$ is prime to the characteristic of $k$. When this is not the case, we will use Artin-Schreier theorem.
Theorem 3 (Hilbert’s theorem 90, additive form). Let $K/k$ be a cyclic extension of degree $n$. Let $\sigma$ be the generator of $G(K/k)$. Let $\beta \in K$. The trace $\mathrm{Tr}_k^K(\beta)=0$ if and only if there exists an element $\alpha \in K$ such that $\beta = \alpha-\sigma\alpha$.
This theorem requires another application of the independence of characters.
Theorem 4 (Artin-Schreier). Let $k$ be a field of characteristic $p$.
- Let $K$ be a cyclic extension of $k$ of degree $p$. Then there exists $\alpha \in K$ such that $K=k(\alpha)$ and $\alpha$ satisfies an equation $X^p-X-a=0$ with some $a \in k$.
- Conversely, given $a \in k$, the polynomial $f(X)=X^p-X-a$ either has one root in $k$, in which case all its roots are in $k$, or it is irreducible. In the latter case, if $\alpha$ is a root then $k(\alpha)$ is cyclic of degree $p$ over $k$.
In other words, instead of looking at the $p$-th root of unity in a field of characteristic $p$, we look at the root of $X^p-X-a$, which still yields a cyclic extension.
Now we are ready for the core theorem of this post.
Theorem 5. Let $E$ be a finite separable extension of $k$. Then $E$ is solvable by radicals if and only if $E/k$ is solvable.
Proof. First of all we assume that $E/k$ is solvable. Then there exists a finite Galois solvable extension of $k$ containing $E$ and we call it $K$. Let $m$ be the product of all primes $l$ such that $l \ne \operatorname{char}k$ but $l|[K:k]$. Let $F=k(\zeta)$ where $\zeta$ is a primitive $m$-th root of unity. Then $F/k$ is abelian and is solvable by radical by definition.
Since solvable extensions form a distinguished class, we see $KF/F$ is solvable. There is a tower of subfields between $F$ and $KF$ such that each step is cyclic of prime order, because every solvable group admits a tower of cyclic groups, and we can use Galois correspondence. By theorem 2 and 4, we see $KF/F$ is solvable by radical because extensions of prime order have been determined by these two theorems. It follows that $E/k$ is solvable by radicals: $KF/F$ is solvable by radicals, $F/k$ is solvable by radicals $\implies$ $KF/k$ is solvable by radicals $\implies$ $E/k$ is solvable by radicals because $KF \supset E \supset k$.
The elaboration of the “if” part is as follows. In order to prove $E/k$ is solvable by radicals, we show that there is a much bigger field $KF$ containing $E$ such that $KF/k$ is solvable by radical. First of all there exists a finite Galois solvable extension $K/k$ containing $E$. Next we define a cyclotomic extension $F/k$ with the following intentions
To reach these two goals, we decide to put $F=k(\zeta)$ where $\zeta$ is a $m$-th root of unity and $m$ is the radical of $[K:k]$ divided by the characteristic of $k$ when necessary. This field $F$ certainly ensures that $F/k$ is solvable by radical. For the second goal, we need to take a look of the subfield between $F$ and $KF$. Let $k = K_0 \subset K_1 \subset \dots \subset K_n = K$ be a tower of field extensions such that every step $K_{i+1}/K_i$ is of prime degree [this is possible due to the solvability of $K/k$]. These prime numbers can only be factors of $[K:k]$ Then in the lifted field extension $F=K_0F \subset K_1F \subset \dots \subset K_nF=KF$ we do not introduce new prime numbers. Why do we consider prime factors of $[K:k]$? Let’s say $[K_{i+1}F:K_iF] = \ell$ is a prime number. If $\ell=\operatorname{char}k$ then we can use theorem 4. Otherwise we still have $\ell|[K:k]$ so we use theorem 2. However this theorem requires a primitive $\ell$-th root to be in $K_{i}F$. Our choice of $m$ and $\zeta$ guaranteed this to happen because $\ell|m$ and therefore a primitive $\ell$-th root of unity exists in $F$. We can make $m$ bigger but there is no necessity. The “only if” part does nearly the same thing, with an alternation of logic chain.
Conversely, assume that $E/k$ is solvable by radicals. For any embedding $\sigma$ of $E$ in $E^{\mathrm{a}}$ over $k$, the extension $\sigma E/k$ is also solvable by radicals. Hence the smallest Galois extension $K$ of $E$ containing $k$, which is a composite of $E$ and its conjugates is solvable by radicals. Let $m$ be the product of all primes unequal to the characteristic dividing the degree $[K:k]$ and again let $F=k(\zeta)$ where $\zeta$ is a primitive $m$-th root of unity. It will suffice to prove that $KF$ is solvable over $F$, because it follows that $KF$ is solvable by $k$ and hence $G(K/k)$ is solvable because it is a homomorphic image of $G(KF/k)$. But $KF/F$ can be decomposed into a tower of extensions such that each step is prime degree and of the type described in theorem 2 and theorem 4. The corresponding root of unity is in the field $F$. Hence $KF/F$ is solvable, proving the theorem. $\square$