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The following lemma is from Qing Liu's "Algebraic Geometry and Arithmetic Curves" p. 61. I don't understand why $A/\mathfrak p$ is a field (line 5-6 of the proof). In the proof, $k(x)$ is a residue field $\mathcal O_{X,x}/\mathfrak m_x$, where $\mathfrak m_x$ is a maximal ideal of $\mathcal O_{X,x}$.

Lemma 4.3

I know that if $A_{\mathfrak p}$ is a finitely generated algebra over $k$, then by corollary 1.12, $k(x)\cong A_\mathfrak p/(\mathfrak p A_{\mathfrak p})$ is a finite extension of $k$. After that, I think one can somehow use the following theorem (from darij grinberg's answer in this thread) to show that $A/\mathfrak p$ is a field.

Theorem 1. Let $K$ be a field. Let $R$ and $L$ be two $K$-algebras such that $L$ is a finite-dimensional $K$-vector space and $R$ is an integral domain. Let $g:R\to L$ be an injective $K$-linear map. Then, $R$ is a field.

It's just that I couldn't prove that there is an injective $k$-linear map $A/\mathfrak p\to k(x)$.

In short, there are two missing steps in my attempt:
1. I couldn't prove that $A_\mathfrak p$ is a finitely generated algebra over $k$.
2. I couldn't prove that there is an injective $k$-linear map $A/\mathfrak p\to k(x)$.

The proof of lemma 4.3 uses some theorems from the book. I'll write them down here.

Part of Remark 1.3. Let $\mathfrak p\in\operatorname{Spec}A$. Then the singleton $\{\mathfrak p\}$ is closed for the Zariski topology if and only if $\mathfrak p$ is a maximal ideal of $A$.

Corollary 1.12. Let $A$ be a finitely generated algebra over a field $k$. Let $\mathfrak m$ be a maximal ideal of $A$. Then $A/\mathfrak m$ is a finite algebraic extension of $k$.

zxcv
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  • What is ${\cal O}_{X,x}$ and what is ${\cal O}_X(V)$? – uniquesolution Apr 29 '19 at 05:57
  • @uniquesolution $\mathcal O_X$ is the sheaf of rings given on $X$, and $\mathcal O_{X,x}$ is the stalk of $\mathcal O_X$ at $x$. – zxcv Apr 29 '19 at 06:04
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    Your question is answered by the accepted answer in this question. – KReiser Apr 29 '19 at 06:19
  • @KReiser No, I tried to use that answer but I couldn't prove the hypothesis of the theorem in that answer. – zxcv Apr 29 '19 at 06:30
  • What's the problem? You know $k\subset A/p \subset k(x)$, $k(x)$ is a finite extension of $k$, and $A/p$ is an integral domain since $p$ is a prime ideal. These are all the hypotheses. – KReiser Apr 29 '19 at 06:33
  • The assumption is that $k(x) = A_p / p A_p$ is a field and a finite extension of $k$, so $A/p$ contains $k$ and is an integral domain and a finite $\le [k(x):k]$ dimensional $k$-vector space, any $a \in A/p$ is the root of some polynomial $\sum_{j=0}^d c_j t^j \in k[t]$, which implies $a^{-1} = -\sum_{j=1}^d c_j a^{j-1} \in A/p$, and hence $A/p$ is a field. Example : $p$ is the point $(0,0)$ in $X = { (u,v) \in \overline{k}^2, u^3=v^2}, A = O_X(X)=k[u,v]/(u^3-v^2)[u/v], A_p = {\frac{f}{g},f,g \in A, g(0,0) \ne 0}$ – reuns Apr 29 '19 at 06:54

1 Answers1

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Your specific problems should be resolved by the answer to your previous question, which shows that $k\subset A/p\subset k(x)$, and the fact mentioned in the text that $k(x)$ is a finite extension of $k$.

Problem 1: $k(x)$ is a finite-dimensional vector space over $k$. Since $A/p$ is a sub $k$-algebra of $k(x)$, it is also a sub $k$ vector space of $k(x)$ and thus finite dimensional as a $k$ vector space, which implies it has a finite basis and this basis can be taken to be a finite generating set.

Problem 2: The inclusion given in the answer to your previous question gives an injective $k$-linear map $A/p\to k(x)$.

KReiser
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  • Here's a thing that I'm stuck with. I see that I have injective ring homomorphisms $f:k\to A/\mathfrak p$, $g:A/\mathfrak p\to k(x)$, and $h:k\to k(x)$. Also, $k(x)$ is a finite-dimensional vector space over $k$, with respect to $h$. But I don't know if $g$ is $k$-linear. If $h=g\circ f$, then $g$ is definitely $k$-linear, but I can't prove that $h=g\circ f$. – zxcv Apr 29 '19 at 07:57
  • Oh, actually, I was not seeing that $A/\mathfrak p$ is also finite-dimensional. That solves my second problem. Thanks! – zxcv Apr 29 '19 at 08:02
  • But I don't understand 1. Keeping the notation of my first comment, $k(x)$ is finite-dimensional vector space over $k$, with respect to $h$. But shouldn't it be finite-dimensional with rsspect to $g\circ f$, in order to say that $A/\mathfrak p$ is finite-dimensional? – zxcv Apr 29 '19 at 10:35
  • What are you talking about? Your previous question gives inclusions $k\subset A/p \subset k(x)$, all of these are $k$-algberas, and all of these inclusions are $k$-linear. $f$ and $g$ are exactly the inclusions in this sequence! – KReiser Apr 29 '19 at 17:16
  • Strictly speaking, $k\subseteq A/\mathfrak p\subseteq k(x)$ is not true, but rather, there are injective ring homomorphisms $f:k\to A/\mathfrak p$ and $g:A/\mathfrak p\to k(x)$, right? So we should say that $g(f(k))\subseteq g(A/\mathfrak p)\subseteq k(x)$. Also, corollary 1.12 gives an injective ring homomorphism $h:k\to k(x)$ such that $k(x)$ is a finite extension of $h(k)$. But I don't know if $k(x)$ is a finite extension of $g(f(k))$. So isn't it not enough to claim that the inclusion map $\psi:g(A/\mathfrak p)\to k(x)$ is $g(f(k))$-linear? – zxcv Apr 29 '19 at 22:15
  • There's no good reason not to identify a distinguished (in the case of $k\to A/p$) or canonical (in the case of $A/p\to k(x)$) inclusion with it's image. And you absolutely do know that the two maps are the same - the maps $f$ and $g$ are $k$-linear which means that the map $g\circ f$ must also be $k$ linear, and this map must equal $h$ since both are $k$-linear and take the same value on $1$. – KReiser Apr 29 '19 at 22:43
  • But if we consider $k(x)$ as a vector space over $k$, with respect to $g\circ f$, then the scalar multiplication is given by $\alpha\cdot y=(g\circ f)(\alpha)y$, where $\alpha\in k$ and $y\in k(x)$. On the other hand, if we consider $k(x)$ as a vector space over $k$, with respect to $h$, then the scalar multiplication is given by $\alpha\cdot y=h(\alpha)y$. So claiming that $g\circ f$ is $k$-linear is merely claiming that $(g\circ f)(\alpha\beta)=(g\circ f)(\alpha)(g\circ f)(\beta)$, and claiming that $h$ is $k$-linear means that $h(\alpha\beta)=h(\alpha)h(\beta)$, where $\alpha,\beta\in k$. – zxcv Apr 29 '19 at 23:35
  • I don't know how these (plus the fact that $(g\circ f)(1)=h(1)=1$) imply that $g\circ f = h$. – zxcv Apr 29 '19 at 23:36
  • Perhaps this previous comment was slightly imprecise. Recall that every $k$-algebra $R$ comes equipped with a unit map $i_R:k\to R$, and one of the conditions on a morphism $f:R_1\to R_2$ of $k$-algebras is that $f\circ i_{R_1} = i_{R_2}$. This means that the copies of $k$ go to the copies of $k$ in the natural way, which you may also see by checking that restricting the constant global functions on your variety to the closed subscheme defined by $p$ and then evaluating them at points inside that subscheme is exactly the same as evaluating them at those points (without restriction). – KReiser Apr 30 '19 at 00:16
  • Let us continue this discussion in chat. – zxcv Apr 30 '19 at 02:46