# Why is $A/\mathfrak p$ a field?

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 $$\mathcal O_{X,x}$$.

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 $$k$$.

• What is ${\cal O}_{X,x}$ and what is ${\cal O}_X(V)$? – uniquesolution Apr 29 '19 at 5:57
• @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 6:04
• Your question is answered by the accepted answer in this question. – KReiser Apr 29 '19 at 6:19
• @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 6:30
• 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 6:33

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 $$A/p\to k(x)$$.
• 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 7:57
• 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 8:02
• 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