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I have a question concerning the post

Closed points of $Spec(A)$ are dense

The last paragraph says:

I know we have $A_f/pA_f=(A/p)_f$, then if p is a prime ideal of $A_f$, then the left side is a field and is a finite extension of k. Then how should I get the conclusion that $A/p$ is also a field?

I don't understand why "the left side is a field and is a finite extension of $k$" and the answer below "$(A/\mathfrak p)_f$ is finite-dimensional".

Maybe he is using "Any integral domain $R$ which is a finite-dimensional $k$-algebra must be a field". I know $A_f/pA_f$ is an integral domain and a finitely generated algebra. But why is it a finite-dimensional $k$-algebra?

No One
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1 Answers1

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There appears to simply be an error in the statement: $p$ should be assumed to be a maximal ideal of $A_f$, not just a prime ideal. Then $A_f/pA_f$ is a field which is a finitely generated $k$-algebra, so it is a finite extension of $k$ by the Nullstellensatz.

(If $p$ is merely prime, then it is certainly not true in general that $A_f/pA_f$ is a field, or finite dimensional over $k$. For instance, if $A=k[x]$ and $f=1$, you could have $p=0$ and then $A_f/pA_f=k[x]$.)

Eric Wofsey
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