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On page 35, the proof of corollary 1.8:

If k is an algebraically closed field and A is a k-algebra, then A = A(X) for some algebraic set X iff A is reduced and finitely generated as a k-algebra.

In the proof, it says: "... Conversely, if A is a finitely generated k-algebra, then after choosing generators we may write A=k[x1,...,xn]/I for some ideal I. ..."

Can someone explain this to me explicitly why this assertion is valid? How to choose the generators and ideal I so that A=[x1,...,xn]/I?

user2690457
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1 Answers1

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I mean, this is almost by definition.

If $A(X)$ is finitely generated as a $k$-algebra, then $A(X)=k[\alpha_1,\ldots,\alpha_n]$ for some $\alpha_i\in A(X)$. Now, since $k[t_1,\ldots,t_n]$ is the "free commtutative $k$-algebra on $n$ generators" you know that there exists a unique $k$-algebra homomorphism $k[t_1,\ldots,t_n]\to k[\alpha_1,\ldots,\alpha_n]=A(X)$ such that $t_i\mapsto \alpha_i$. Note that this map is obviously surjective, and so if $I$ is its kernel, then

$$k[t_1,\ldots,t_n]/I\cong k[\alpha_1,\ldots,\alpha_n]=A(X)$$

So, $A(X)=A(V(I))$. Of course, you need to note in this last step that since $A(X)$ is reduced, that $I$ is necessarily radical.

Alex Youcis
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  • Thank you very much for the answer. It's really a matter of definition.I thought finitely generated k-algebra is generated the same way as k-module. But as a newbie in the area, it's not obvious for me to see why the k-algebra homomorphism is not necessarily injection, is it because we don't ask for generators to be algebraically independent? – user2690457 Oct 08 '13 at 21:41
  • @user2690457 Yes, roughly that is correct :) All the elements of our algebra are polynomials in the $\alpha_i$, but there's nothing stopping them from being related to one another! – Alex Youcis Oct 09 '13 at 07:37