# Proof of Zariski lemma in Wikipedia

That wikipedia page has a proof of Zariski's lemma, which states that if $$A$$ is a Jacobson ring (in particular when $$A$$ is a field) and $$B$$ is a finitely-generated algebra of $$A$$, which is a field, then $$B$$ is a finite extension of $$A$$ (as a vector space).

In the proof, the following statement is proved:

(*) Let $$B\supseteq A$$ be integral domains such that $$B$$ is finitely generated as $$A$$-algebra. Then there exists a nonzero $$a$$ in $$A$$ such that every ring homomorphism $$\phi :A\to K$$, $$K$$ an algebraically closed field, with $$\phi (a)\neq 0$$ extends to $$\widetilde {\phi }:B\to K$$.

The proof of this statement begins with

If $$B$$ contains an element that is transcendental over $$A$$, then it contains a polynomial ring over $$A$$ to which $$φ$$ extends (without a requirement on $$a$$) and so we can assume $$B$$ is algebraic over $$A$$ (by Zorn's lemma, say).

I don't understand why is it possible to assume that. Using Zorn's lemma we can find a ring $$R$$ such that $$A\subseteq R\subseteq B$$, every element in $$R\setminus A$$ is trancendental over $$A$$ and $$B$$ is algebraic over $$R$$. However, if we prove the existence of $$a\in R$$ which satisfies the statement, how can we know that there is an element in $$A$$ which satisfies it?

• You mean a Jacobson ring, not a jacobian ring.
– KCd
Jun 20 at 13:12
• Thanks, I edited it. Jun 20 at 13:14
• A different proof of Zariski’s lemma (when $A$ is a field) is Theorem 2.11 in kconrad.math.uconn.edu/blurbs/ringtheory/maxideal-polyring.pdf. It uses induction on the number of generators of $B$ as an $A$-algebra.
– KCd
Jun 20 at 13:24
• Could it be that if $a \in R \setminus A$, then any ring morphism $\phi: A \to K$ extends to $\psi: R \to K$ in such a way that $\psi(a) \neq 0$? If that is true, given $\phi$, we can first extend it in a suitable manner to $R$, and then $(*)$ says you can extend it to $B \to K$. Jun 20 at 15:28

First, if $$a\in A$$ is any element so that there exists a homomorphism $$\phi:A\to K$$ with $$K$$ an algebraically closed field and $$\phi(a)\neq 0$$, then with $$R$$ as defined in your post, $$\phi$$ extends to a map $$\widetilde{\phi}:R\to K$$ which necessarily has $$\widetilde{\phi}(a)\neq 0$$, because $$\widetilde{\phi}$$ extends $$\phi$$ and $$\phi(a)\neq 0$$. (We get this extension by applying Zorn's lemma to the poset of pairs $$(T,\phi_T)$$ where $$A\subset T$$ is a transcendental ring extension and $$\phi_T:T\to K$$ extends $$\phi$$ with $$(T,\phi_T)\leq (T',\phi_{T'})$$ when the latter extends the former, and noting that any $$\phi_T$$ must have $$\phi_T(a)\neq 0$$ since it extends $$\phi$$.)

After that, one proves that if $$A\subset B$$ is an algebraic (even integral) extension and $$\phi:A\to K$$ is a homomorphism to an algebraically closed field, then $$\phi$$ extends to $$\widetilde{\phi}:B\to K$$. Again, if $$\phi(a)\neq 0$$, then we have $$\widetilde{\phi}(a)\neq 0$$ by the same reasoning as above.

Combining the two statements, we see that for any $$a\in A$$ such that there exists a homomorphism $$\phi:A\to K$$ to an algebraically closed field $$K$$ with $$\phi(a)\neq 0$$, and any extension of integral domains $$A\subset B$$, we can extend $$\phi$$ to a homomorphism $$\widetilde{\phi}:B\to K$$ with $$\widetilde{\phi}(a)\neq 0$$. This gives us the desired statement marked with a (*) in your post by taking any nonzero element $$a\in A$$.

(Let me point out that for most integral domains, we actually find that there are lots of $$a$$ which work in the statement - the reason we say "there exists" is because there's an integral domain where there is precisely one option for $$a$$, and that happens when $$A\cong \Bbb F_2$$. This sort of statement is actually something one runs in to a fair amount with "there exists" proofs in algebra.)

• It doesn't answer my question, how do we know that there is some $a\in A$? How do we "combine the two statements"? Jun 24 at 8:43
• Starting with $\phi:A\to K$ which has $\phi(a)\neq0$, we first extend $\phi$ to $R\to K$, and then we extend $\phi$ to $B\to K$. This preserves the property that $\phi(a)\neq0$, so we've shown that every $\phi:A\to K$ with $\phi(a)\neq0$ extends to $B$. Now note that for any nonzero element $a\in A$ we can produce a $\phi$ and $K$ with $\phi(a)\neq0$ - take the embedding of $A$ in to the algebraic closure of it's fraction field. So you can take any $a\neq0$, and since $A$ is an integral domain there is a nonzero element. Jun 24 at 12:51
• OK, now I understand it. But since R is not necessarily a field, how do you extend $\phi$ from R to B? 2 days ago
• Why would you need $R$ to be a field? An algebraic ring extension $R\subset B$ should at least mean "every element in $B$ satisfies a polynomial with coefficients in $R$", and that gives you enough to use the argument on the wikipedia page you link. 2 days ago
• The argument on wikipedia already assumes that the homomorphism is not zero in a specific element, how can you know that? 2 days ago