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Let $\varphi : \mathbb{Z}[x,y] \to \mathbb{C}$ be a ring homomorphism. Then $\ker{\varphi}\subset \mathbb{Z}[x,y]$ is prime.

I'm assuming $\mathbb{Z}[x,y]$ a ring?

The argument I have is that $\mathbb{Z}[x,y]/\ker{\varphi}$ is an integral domain. I'm looking back at my notes and I see I've written that for $\operatorname{Im}(\varphi)\subset \mathbb{C}$, we can't say the image is a field because $\mathbb{C}$ is not a finite field. I'm confused because I thought any 'structured' subset of $\mathbb{C}$ could be a subfield.

Bernard
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cheeseboardqueen
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    What do you mean by "'structured' subset"? Is the ring $\mathbb Z\subset \mathbb C$ structured? To say the obvious, it's not a field, of course. – peter a g May 05 '21 at 13:48
  • To elaborate on peter a g's comment, it is not true that a subring of a field is necessarily a field, and the image of a ring homomorphism is a subring of the codomain – Noah Solomon May 05 '21 at 13:49
  • For the conclusion that $\ker \varphi$ is prime you don't need the fact that $\text{im} \varphi$ is not a field. You just need that $\text{im} \varphi$ is an integral domain, which it is, because it is a subring of a field. If you want to argue that $\ker \varphi$ is not maximal, then of course you need that $\text{im} \varphi$ is not a field. – Magdiragdag May 05 '21 at 15:48

1 Answers1

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The point is that if the image is a field, then it either contains a finite field or $\mathbb Q$. If it contains a finite field then it can't be a subfield of $\mathbb C$ because they have different characteristics, so that's one possibility ruled out. If it contains $\mathbb Q$, then you also get a contradiction, because no field containing $\mathbb Q$ is a finitely generated $\mathbb{Z}$-algebra, but $\mathbb{Z}[x,y]/\operatorname{ker}\varphi$ (which is isomorphic to $\operatorname{im}\varphi$ by one of the isomorphism theorems) is a finitely generated $\mathbb{Z}$-algebra by definition.

A reference for the last fact is Fields finitely generated as $\mathbb Z$-algebras are finite?