3
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $$\overline{\frac{g_0}{h_0} + \frac{g_1}{h_1} y + \dots + \frac{g_{n - 1}}{h_{n - 1}} y^{n - 1}} = 0$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.