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Let $a,b,c,d \in \mathbb{Z}$ such that $n = \left| ad - bc \right| \neq 0$. Prove that $\mathbb{C}(x,y)/\mathbb{C}(x^a y^b, x^c y^d)$ is an extension of degree $n$ and describe the Galois group of this extension.

I literally have no ideas, all the things I know is that this is a finite extension. For example, $x$ is algebraic over $L = \mathbb{C}(x^a y^b, x^c y^d)$ because it is root of the following polynomial $$f(t) = t^{n/l} - \frac{(x^a y^b)^{v}}{(x^c y^d)^u}$$ in which $l = \mathrm{gcd}(b,d), lv = d, lu = b$. Of course I know if I keep trying then I could show that the polynomial above is irreducible but that is not enough to determine the Galois group. I wonder if there exists certain systematic theory behind this problem or at least some famous theorem I do not know? (I do not major in number theory)

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Let $K = \mathbb C (x^ay^b,x^cy^d)$ and $L = K(x,y)$.

Observe that by closure, $K$ contains any element of the form $$(x^ay^b)^{a'}(x^cy^d)^{b'} = x^{a'a+b'c} y^{a'b + b'd}$$ and hence any pair of elements $$x^{a'a+b'c} y^{a'b + b'd}, x^{c'a+d'c} y^{c'b + d'd}$$

if you identify a pair $(x^ay^b,x^cy^d)$ with the matrix $\begin{bmatrix} a&b\\c&d\end{bmatrix}$ we see that this new pair corresponds to a product $$\begin{bmatrix} a'&b'\\c'&d'\end{bmatrix}\begin{bmatrix} a&b\\c&d\end{bmatrix} = \begin{bmatrix} a'a + b'c & a'b + b'd\\ c'a + d'c& c'b + d'd\end{bmatrix}$$

This means if we modify our generators of $K$ by "multiplying by the left by invertible (over $\mathbb Z$) matrices", we still get generators of $K$ (under the matrix interpretation above). Also recognize that $n$ is the (absolute value of the) determinant of the matrix corresponding to the generators and so these invertible matrices also preserve that quantity.

In particular, we can exchange our original matrix for one which is triangular by taking of
$$c' = c/\gcd(a,c)$$ $$d' = a/\gcd(a,c)$$ which are coprime because of $p$ divides both quotients then it should have appeared in the $\gcd$! Since they are coprime there exist $a',b'$ such that $a'd' - b'c' = 1$ by the euclidean algorithm, and hence the matrix with those entries has determinant 1.

Now we are very nearly finished. Replace and rename the generators so that we have $K = \mathbb C(x^ay^b,y^d)$ and $ad = n$. It is now evident that $y$ has degree at most $d$ over $K$ and that $x$ has degree at most $a$ over $K(y)$, so multiplication in towers says that $L=K(x,y)$ has degree at most $ad$ over $K$.

That the degrees are exact is tricker. I believe there is an exercise in Lang which does the job or something close(#20 in Chapter V I think, or somewhere around there). But let's try ourselves.

Let's look at the first one: $y$ over $\mathbb C(x^ay^b, y^d)$, $a,d$ nonzero. A polynomial in $K[t]$ with $y$ as a root is $$f(t) = t^d - y^d$$ By Gauss's lemma, it suffices to prove irreducibility in $\mathbb C[x^ay^b,y^d][t]$. This will be the case by Eisenstein's criterion if the ideal $(y^d)$ is prime in $\mathbb C[x^ay^b,y^d]$, which is obvious by looking at the $x$-degree; if $f(x^ay^b,y^d)g(x^ay^b,y^d)$ is in $(y^d)$ then both $f$ and $g$ are actually degree $0$ in the first coordinate so really we just have $f(y^d)g(y^d)$ is in $(y^d)$ and we want to argue that at least one of the two constant terms is zero, since then that element will be in $(y^d)$. That is, of course, obvious, since if both constant terms were nonzero, so too would be the constant term of the product but then it couldn't be in $(y^d)$.

Now the last step. We want the degree of $x$ over $K(y) = \mathbb C(x^ay^b,y^d,y) = \mathbb C(x^a,y)$ - evidently we can just exchange the roles of $x$ and $y$ in the last proof to verify that the degree of $x$ over this field is $a$.

So it remains to compute the Galois group. Using the matrix interpretation, we can kill our matrix by its adjoint to get $x^n$ and $y^n$ in the ground field. This tells us that the only possible conjugates of $x$ and $y$ are of the form $\zeta x$ and $\zeta y$ where $\zeta$ is an $n$th root of unity. In particular we get all of these after adjoining just $x$ and $y$, so our extension really is normal.

Now, we've already replaced our generators by $x^a y^b, y^d$. Since the Galois group has exactly $n$ elements we will find them by simply counting all the admissible possibilities. In particular, every element corresponds to a pair $(i,j) \in \mathbb Z/n\mathbb Z \times \mathbb Z/n\mathbb Z$ by $$\sigma_{(i,j)} (x) = \zeta^i x$$ $$\sigma_{(i,j)} (y) = \zeta^j y$$ since they are determined by their action on $x$ and $y$.

So the Galois group can be identified with a subgroup of order $n$ in $(\mathbb Z/n\mathbb Z)^2$. From the degree $a$ extension at the top you can find an explicit collection of elements in the Galois group, namely $(i*n/a,0)$ since that Galois group fixes $y$ (hence the second component is 0) and sends $x$ to $\zeta^i x$ for all $0\leq i \leq a-1$ from the general theory of extensions of the form $k(x)/k(x^a)$ (Lang is a good reference).

On the other hand, the Galois group fixes $y^d$ and so the second coordinate of every element under this identification must be of the form $jn/d$.

Now count the number of elements of the form $(in/a, jn/d)$: there are exactly $a*d$ of these, and any member of the Galois group must be one of these admissible elements. At the same time, the Galois group has $n$ elements, and so all of these must be realized. Thus the Galois group is isomorphic to the subgroup of those elements in $\mathbb Z/n\mathbb Z \times \mathbb Z/n\mathbb Z$, which one can see is generated by $(n/a, 0)$ and $(0, jn/d)$.

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  • $\begingroup$ Sorry, I left out computing the Galois group! I think it's probably straightforward so I'll edit it in... $\endgroup$
    – user208649
    Apr 19 '20 at 0:44
  • $\begingroup$ Now the answer is complete, let me know if any part is unclear. $\endgroup$
    – user208649
    Apr 19 '20 at 1:29
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    $\begingroup$ Dear @TokenToucan, in the paragraph where you use Eisenstein's Criterion, there is a claim: "if $f(x^a y^b,y^d)g(x^a y^b,y^d)$ is in $(y^d)$ then both $f$ and $g$ are actually degree $0$ in the first coordinate". What if $f=g=y^d x^ay^b$ for example? I know my example doesn't contradict later conclusion anyway, but things get complicated if you can't reduce to degree $0$. $\endgroup$ Apr 19 '20 at 3:56
  • $\begingroup$ Oh you’re completely right, I was thinking the ideal had only polynomials in $y$. It is certainly irreducible (at least if any bit of math is sane) but needs a better argument. I’ll go over it later and think about things. It follows for instance if the two generators are algebraically independent, and for that a degree argument actually works (I think). $\endgroup$
    – user208649
    Apr 19 '20 at 4:43
  • $\begingroup$ One more comment that might be more significant to the proof. The Galois group you have computed is $\mathbb{Z}/a\mathbb{Z}\times \mathbb{Z}/d\mathbb{Z}$, where $a,d$ are from the triangularized matrix. From your proof, it is evident that $a$ is $\gcd(a,c)$ (the former $a$). But if that is the case, then the roles of $a,c$ and $b,d$ is not symmetric in the final result. (Also, you could check that you must change the sign of $c'$). $\endgroup$ Apr 19 '20 at 16:16
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Through trial and error, I found out that any attempts to attack this problem directly lead to this system of linear equation in $\mathbb{Z}/n\mathbb{Z}$: $$\begin{aligned}ai+bj &\equiv 0 \pmod{n}\\ci+dj &\equiv 0\pmod{n}\end{aligned} $$ I couldn't find out any way to solve this linear system; even the number of solutions is uncomputable. TokenToucan's answer done this (up to a change of coordinate), but I'm not very sure about some of his argument. Later I realize that the "right" approach is to look at its Pontryagin duality.

For brevity, let $L=\mathbb{C}(x,y)$ and $K=\mathbb{C}(x^ay^b,x^cy^d)$. It's easy to see that $x^n,y^n\in K$. Therefore $L$ is the splitting field of $(X^n-x^n)(X^n-y^n)$ over $K$, so $L/K$ is Galois. The hard part is to compute the Galois group.

Let $w=e^{2\pi i/n}$ and $G=\mathrm{Gal}(L/K)$. For $\sigma\in G$, there exists $i,j$ such that $\sigma(x)=xw^i,\sigma(y)=yw^j$. One can deduce that $G$ is abelian and $\sigma^n=1$, so the exponent of $G$ divides $n$. Since $K$ contains every $n$-root of unity, $L/K$ is an $n$-Kummer extension. Consider the set $M(L)=\{a\in L^{\times}:a^n\in K^{\times}\}$, then $M(L)$ is a subgroup of $L^{\times}$ that contains $K^{\times}$. Denote $M(L)/K^{\times}$ by $\mathrm{kum}(L/K)$ and the group of $n$-root of unity by $\mu_n$, we can define the Kummer pairing $$\begin{aligned} B: G\times \mathrm{kum}(L/K) &\to \mu_n\\ (\sigma,\alpha K^{\times}) &\mapsto \sigma \alpha/\alpha \end{aligned} $$ Kummer theory said that $B$ is a non-degenerate bilinear form. It defines an isomorphism $$\begin{aligned} \mathrm{kum}(L/K) &\to \mathrm{Hom}(G,\mu_n)\\ h=\alpha K^{\times} &\mapsto B_h:\sigma \mapsto \sigma \alpha/\alpha \end{aligned}$$ Note that $\mathrm{Hom}(G,\mu_n)=\mathrm{Hom}(G,\mathbb{C}^{*})\cong G$, by Pontryagin duality. Since the isomorphism $\mathrm{Hom}(G,\mathbb{C}^{*})\cong G$ is non-canonical, it suggests that one is easy to compute and one is not. In this case, the duality $\mathrm{kum}(L/K)$ is easier to deal with.

One might notice that $B_x$ and $B_y$ generate $\mathrm{Hom}(G,\mu_n)$, since we have $B_x(\sigma)=i, B_y(\sigma)=j$ and $i,j$ determine $\sigma$ uniquely. In consequence, $xK^{\times}$ and $yK^{\times}$ generate $\mathrm{kum}(L/K)$. Let $r=xK^{\times},s=yK^{\times}$, we have $r^as^b=r^cs^d=1$. Moreover, one can prove that $x^py^q\in K$ if and only if $x^py^q$ can be written in the form $(x^ay^b)^u(x^cy^d)^v$. In other words, $$\mathrm{kum}(L/K)=\langle r,s\mid rs=sr,r^as^b=r^cs^d=1\rangle\cong \mathbb{Z}^2/\langle (a,b),(c,d)\rangle$$ Here I prove that if $x^py^q\in K$ then it must be of this form. By adding a multiply of $n$, we can assume that $p,q,a,b,c,d>0$. There are $F,G\in K[X,Y],G\neq 0$ such that $G(x^ay^b,x^cy^c)x^py^q=F(x^ay^b,x^cy^d)$. Compare the leading terms (the terms with highest total degree and there can be many such terms), one obtain $$c(x^ay^b)^u(x^cy^d)^vx^py^q=c(x^ay^b)^{u'}(x^cy^d)^{v'}\Leftrightarrow x^py^q=(x^ay^b)^{u'-u}(x^cy^d)^{v'-v}$$

Last step can be done using Smith normal form for matrix over PID. Consider the matrix $$A=\begin{pmatrix}a &c\\b &d\end{pmatrix}$$ with Smith normal form $$A'=\begin{pmatrix}u &0\\0 &v\end{pmatrix}$$ Here $uv=\pm \det A=\pm n$. Then $$\mathrm{kum}(L/K)\cong \mathbb{Z}^2/\langle (a,b),(c,d)\rangle=\mathbb{Z}^2/A\mathbb{Z}^2\cong \mathbb{Z}^2/A'\mathbb{Z}^2\cong \mathbb{Z}/u\mathbb{Z}\times \mathbb{Z}/v\mathbb{Z}$$ Then $\vert G\vert=\vert \mathrm{kum}(L/K)\vert=n$ obviously.

For a treatment of Kummer extension, see Field and Galois theory by Patrick Morandi or Lectures of Abstract Algebra - III. Theory of Fields and Galois Theory by N. Jacobson. See this answer for more detail about the use of Smith normal form.

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