Question on finitely generated algebras

Question

While reading Reid's "Introduction to Algebraic Geometry", I came across the following passage:

"A finitely generated $k$-algebra is a ring of the form $A = k[a_1,\cdots,a_n]$, so that $A$ is generated as a ring by $k$ and $a_1, \cdots , a_n$; clearly, every such ring is isomorphic to a quotient of the polynomial ring, $A = k[X_1, \cdots ,X_n]/I$"

My question is: why must such ring be isomorphic to a ring of the form $k[X_1, \cdots ,X_n]/I$?

The short answer is "because of the evaluation homomorphism", which has an ideal $I$ as its kernel. — hardmath, Aug 12 '15 at 18:22
Now your job is to write up the Answer with more detail, or find a previous Question that we can claim this duplicates! — hardmath, Aug 12 '15 at 18:25
Thank you for the link, I added an answer with the reference. — u1571372, Aug 12 '15 at 18:30

score 1 · Answer 1 · edited Apr 13 '17 at 12:21

1

As user hardmath pointed out in the comments, this easily follows from the evaluation homomorphism. For more details, see this.

edited Apr 13 '17 at 12:21

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answered Aug 12 '15 at 18:30

u1571372

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Question on finitely generated algebras

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