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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?

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  • $\begingroup$ You mean a Jacobson ring, not a jacobian ring. $\endgroup$
    – KCd
    Jun 20 at 13:12
  • $\begingroup$ Thanks, I edited it. $\endgroup$
    – uri gluck
    Jun 20 at 13:14
  • $\begingroup$ 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. $\endgroup$
    – KCd
    Jun 20 at 13:24
  • $\begingroup$ 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 $B \to K$. $\endgroup$ Jun 20 at 15:28

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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.)

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  • $\begingroup$ It doesn't answer my question, how do we know that there is some $a\in A$? How do we "combine the two statements"? $\endgroup$
    – uri gluck
    Jun 24 at 8:43
  • $\begingroup$ 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. $\endgroup$
    – KReiser
    Jun 24 at 12:51
  • $\begingroup$ OK, now I understand it. But since R is not necessarily a field, how do you extend $\phi$ from R to B? $\endgroup$
    – uri gluck
    2 days ago
  • $\begingroup$ 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. $\endgroup$
    – KReiser
    2 days ago
  • $\begingroup$ The argument on wikipedia already assumes that the homomorphism is not zero in a specific element, how can you know that? $\endgroup$
    – uri gluck
    2 days ago

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