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Let $C$ be the projective plane curve of degree $d$ over an algebraically closed field $k$ with affine equation $$ y^d = \prod_{i=1}^{d-1}(x-\alpha_i) $$ where $a_i\in k$ are distinct. Writing $f(x,y) = y^d-\prod_{i=1}^{d-1}(x-\alpha_i)$ I want to show $$ k(C) \simeq k(x)[y]/\left(f(x,y)\right) $$ where $k(C)$ is the field of fractions of $k[X,Y]/(f(x,y))$ and $k(x)$ is the field of fractions of polynomials in one variable $x$.

I have been staring at this for a while now without making much progress, so any help is appreciated!

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  • $\begingroup$ Maybe we should first try to show $k(x)[y] / (f)$ is indeed a field by showing $y^d - \prod_{i = 1}^{d - 1}(x - \alpha^i)$ is irreducible over the field $k(x)$. I have no idea how to do it though... $\endgroup$ – Alex Vong Feb 25 '17 at 12:44
  • $\begingroup$ I think I know how to prove $y^d - \prod_{i = 1}^{d - 1}(x - \alpha^i)$ being irreducible now. Suppose not, say $\frac{g}{h} \in k(x)$ is a root with deg $g = m$, deg $h = n$. Then $(\frac{g}{h})^d - \prod_{i = 1}^{d - 1}(x - \alpha^i) = 0 \implies g^d = h^d \prod_{i = 1}^{d - 1}(x - \alpha^i)$. Now takes degree on both sides, we have $md = nd + d - 1 \implies (n - m + 1)d = 1 \implies d = 1$ contradicting the unstated assumption $d >= 2$ in the question. $\endgroup$ – Alex Vong Feb 25 '17 at 13:20
  • $\begingroup$ So we really know $k(x)[y] / (f)$ is a field now. Now we want to show it is the smallest field containing $k[x, y] / (f)$, i.e. namely it posses the universal property that any embedding from $k[x, y] / (f)$ to a field must factors through $k(x)[y] / (f)$. Now this looks harder than the last question... $\endgroup$ – Alex Vong Feb 25 '17 at 13:27
  • $\begingroup$ I was wrong, I have only shown $f$ had no root but not irreducible. $\endgroup$ – Alex Vong Feb 25 '17 at 13:57
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I think I know how to answer this question now! Our plan is first to show $K(x)[y] / <f>$ is a field and then to show $K(x)[y] / <f> \cong$ Frac$(K[x, y] / <f>)$ by using the universal property of field of fractions. From now on, I will denote the ideal generated by $f$ using $<f>$ since I will be using a lot of parenthesis.

To show $K(x)[y] / <f>$ is a field, it suffices to show $f \in K(x)[y]$ is irreducible. We will use the generalized Eisenstein's criterion. Note $x - \alpha_j \in K[x]$ being irreducible $\implies <x - \alpha_j>$ is a prime ideal in $K[x]$. Now $-\prod_{i = 1}^{d - 1}(x - \alpha_i) \in <x - \alpha_j>$, $0 \in <x - \alpha_j>$, $1 \notin <x - \alpha_j>$ and $-\prod_{i = 1}^{d - 1}(x - \alpha_i) \notin <x - \alpha_j>^2 = <(x - \alpha_j)^2>$. Here we assume $\alpha_i$ are distinct, which I guess is an unstated assumption. So by the generalized Eisenstein's criterion, $f \in K[x][y]$ is irreducible. Since $K[x]$ is a UFD, $f \in K(x)[y]$ is also irreducible. Hence $K(x)[y] / <f>$ is a field.

We are left to show that $K(x)[y] / <f>$ is the smallest field containing $K[x, y] / <f>$ up to isomorphism, i.e. it posses the universal property that for any embedding $\sigma: K[x, y] / <f> \hookrightarrow \Omega$, where $\Omega$ is a field, we can find some embeddings $\iota: K[x, y] / <f> \hookrightarrow K(x)[y] / <f>$ and $j:K(x)[y] / <f> \hookrightarrow \Omega$ such that $\sigma = j \circ \iota$. Before proceeding, we first observe $K[x, y] \cong K[x][y]$, so we can replace all instances of $K[x, y]$ with $K[x][y]$. Also, we will write the coset $a + I$ as $\overline{a}$.

Given embedding $\sigma: K[x][y] / <f> \hookrightarrow \Omega$ by $$\overline{g_0(x) + g_1(x) y + \dots + g_{n - 1}(x) y^{n - 1}} \mapsto \sigma(\overline{g_0(x) + g_1(x) y + \dots + g_{n - 1}(x) y^{n - 1}})$$ with $g_i(x) \in K[x]$.

We define $\iota: K[x][y] / <f> \hookrightarrow K(x)[y] / <f>$ by $$\overline{g_0(x) + g_1(x) y + \dots + g_{n - 1}(x) y^{n - 1}} \mapsto \overline{\frac{g_0(x)}{1} + \frac{g_1(x)}{1} y + \dots + \frac{g_{n - 1}(x)}{1} y^{n - 1}}$$ with $g_i(x) \in K[x]$

and $j: K(x)[y] / <f> \hookrightarrow \Omega$ by $$\overline{\frac{g_0(x)}{h_0(x)} + \frac{g_1(x)}{h_1(x)} y + \dots + \frac{g_{n - 1}(x)}{h_{n - 1}(x)} y^{n - 1}} \mapsto \frac{\sigma(\overline{g_0(x)})}{\sigma(\overline{h_0(x)})} + \frac{\sigma(\overline{g_1(x))}}{\sigma(\overline{h_1(x)})} \sigma(\overline{y}) + \dots + \frac{\sigma(\overline{g_{n - 1}(x)})}{\sigma(\overline{h_{n - 1}(x)})} \sigma(\overline{y})^{n - 1}$$ with $g_i(x), h_i(x) \in K[x]$.

Of course, it is then an exercise to check that $\iota, j$ are well-defined embeddings such that $j \circ \iota = \sigma$ :) Finally, since $K(x)[y] / <f>$ satisfies the universal property of Frac$(K[x][y] / <f>)$ and universal objects are isomorphic, we have $K(x)[y] / <f> \cong$ Frac$(K[x][y] / <f>) \cong$ Frac$(K[x, y] / <f>)$.

EDIT-1: We will check $j$ is injective. Suppose $$\frac{\sigma(\overline{g_0(x)})}{\sigma(\overline{h_0(x)})} + \frac{\sigma(\overline{g_1(x))}}{\sigma(\overline{h_1(x)})} \sigma(\overline{y}) + \dots + \frac{\sigma(\overline{g_{n - 1}(x)})}{\sigma(\overline{h_{n - 1}(x)})} \sigma(\overline{y})^{n - 1} = 0$$ From now on we write $g_i(x), h_i(x)$ as $g_i, h_i$. Then $$\frac{\sigma(\overline{g_0 h_1 \dots h_{n - 1} + h_0 g_1 h_2 \dots h_{n - 1} y + \dots + h_0 \dots h_{n - 2} g_{n - 1} y^{n - 1}})}{\sigma(\overline{h_0 \dots h_n})} = 0$$ We have $$\sigma(\overline{g_0 h_1 \dots h_{n - 1} + h_0 g_1 h_2 \dots h_{n - 1} y + \dots + h_0 \dots h_{n - 2} g_{n - 1} y^{n - 1}}) = 0$$ Since $\sigma$ is injective, $$\overline{g_0 h_1 \dots h_{n - 1} + h_0 g_1 h_2 \dots h_{n - 1} y + \dots + h_0 \dots h_{n - 2} g_{n - 1} y^{n - 1}} = 0$$ Since $\overline{h_i}$ are nonzero and $K(x)[y] / <f>$ is a field (it is what we've just proved!), $\overline{h_1 \dots h_{n - 1}}$ is nonzero and hence a unit in $K(x)[y] / <f>$. Thus we can divide the whole thing by $\overline{h_1 \dots h_{n - 1}}$ and obtain $$\overline{\frac{g_0}{h_0} + \frac{g_1}{h_1} y + \dots + \frac{g_{n - 1}}{h_{n - 1}} y^{n - 1}} = 0$$

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  • $\begingroup$ Yea, you are right: $\alpha_i\neq\alpha_j$ for all $i\neq j$ $\endgroup$ – user114158 Feb 25 '17 at 18:56
  • $\begingroup$ I can't quite follow your proof of irreducibility - it seems to me you prove it is irreducible in $k[x][y]$, not $k(x)[y]$. Am I right, or could you maybe elaborate on that? $\endgroup$ – user114158 Feb 26 '17 at 10:36
  • $\begingroup$ Yes, I only prove $f$ is irreducible in $k[x][y]$ but since $k[x]$ is a UFD, by Gauss's lemma, $f$ is also irreducible in $k(x)[y]$. See en.wikipedia.org/wiki/Eisenstein's_criterion#Generalization for details. I will edit the answer to include this. $\endgroup$ – Alex Vong Feb 26 '17 at 11:01
  • $\begingroup$ Ah, of course - thanks! I'll accept your answer, when I've gone through all of your arguments. $\endgroup$ – user114158 Feb 26 '17 at 11:02
  • $\begingroup$ I do not quite see how to show $j$ is an embedding, in particular why it is injective. Could you give hint? $\endgroup$ – user114158 Feb 26 '17 at 11:46

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