## Tensor product of division algebras (2)

Posted: October 29, 2011 in Division Rings, Noncommutative Ring Theory Notes
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Here you can see part (1). We are now going to prove a more interesting result than the one we proved in part (1). But we need to get prepared first. The following important result is known as Zariski’s lemma.

Lemma. (Zariski, 1946) Let $k$ be a field and let $A=k[a_1, \ldots , a_n]$ be a finitely generated commutative algebra. If $A$ is a field, then $A$ is algebraic over $k$  and thus $\dim_k A < \infty.$

Proof. The proof is by induction over $n.$ If $n=1,$ then $A=k[a_1]$ and since $A$ is a field, $a_1^{-1} \in A.$ Thus $a_1^{-1}=\sum_{i=0}^m \gamma_i a_1^i$ for some integer $m \geq 0$ and $\gamma_i \in k.$ Then $\sum_{i=0}^m \gamma_i a_1^{i+1}=0$ and so $a_1,$ and hence $A,$ is algebraic over $k.$ Now suppose that $n \geq 2.$ If all $a_i$ are algebraic over $k,$ then $A=k[a_1, \ldots , a_n]$ is algebraic over $k$ and we are done. So we may assume that $a_1$ is transcendental over $k.$ Since $A$ is a field, $K =k(a_1) \subseteq A$ and thus $A=K[a_2, \ldots , a_n].$ By the induction hypothesis, $A$ is algebraic over $K.$ So every $a_i$ satisfies some monic polynomial $f_i(x) \in K[x]$ of degree $m_i.$ Let $v_i \in k[a_1]$ be the product of the denominators of the coefficients of $f_i(x)$ and put $v=\prod_{i=1}^n v_i.$ Let $m$ be the maximum of $m_i.$ Then multiplying $f(a_i)=0$ through by $v^m$ shows that each $va_i$ is integral over $k[a_1].$ Note that since $a_1$ is transcendental over $k,$ $k[a_1] \cong k[x],$ the polynomial algebra over $x.$ Thus I can choose an irreducible polynomial $p(a_1) \in k[a_1]$ such that

$\gcd(p(a_1), v)=1. \ \ \ \ \ \ \ \ \ (*)$

Now $(p(a_1))^{-1} \in A=k[a_1, \ldots , a_n],$ because $A$ is a field. Thus for a large enough integer $r,$ we have $v^r(p(a_1))^{-1} \in k[va_1, \ldots , va_n]$ and hence $v^r(p(a_1))^{-1}$ is integral over $k[a_1].$ But $k[a_1]$ is a UFD and we know that every UFD is integerally closed (in its field of fraction). Therefore $v^r(p(a_1))^{-1} \in k[a_1],$ which is absurd because then $p(a_1) \mid v,$ contradicting $(*). \ \Box$

Corollary. Let $k$ be a field and let $A=k[a_1, \ldots , a_n]$ be a finitely generated commutative algebra.

1) If $\mathfrak{m}$ is a maximal ideal of $A,$ then $\dim_k A/\mathfrak{m} < \infty.$

2) If $A$ is a field and $k$ is algebraically closed, then $A=k.$

3) If $A$ is a domain, $k$ is algebraically closed and $b_1, \ldots , b_m \in A \setminus \{0\},$ then there exists a ring homomorphism $\varphi : A \longrightarrow k$ such that $\varphi(b_i) \neq 0$ for all $i=1, \ldots , m.$

Proof. 1) Let $\overline{a_i}=a_i + \mathfrak{m}, \ i=1, \ldots , n.$ Then $A/\mathfrak{m} = k[\overline{a_1}, \ldots , \overline{a_n}]$ and we are done by the lemma.

2) By the lemma, $A$ is algebraic over $k$ and thus, since $k$ is algebraically closed, $A=k.$

3) Let $S$ be the set of all monomials in $b_1, \ldots, b_m.$ Clearly $S$ is multiplicatively closed and $0 \notin S$ because $A$ is a domain. Consider $B=S^{-1}A,$ the localization of $A$ at $S.$ Clearly $B=k[a_1, \ldots , a_n, b_1^{-1}, \ldots , b_m^{-1}]$ and so $B$ is finitely generated. Let $\mathfrak{m}$ be a maximal ideal of $B.$ Note that $b_i \notin \mathfrak{m},$ for all $i,$ because each $b_i$ is a unit in $B.$ Now, we have $B/\mathfrak{m} \cong k,$ by 2). Let $f: B/\mathfrak{m} \longrightarrow k$ be a ring isomorphism. We also have the natural ring homomorphism $g: B \longrightarrow B/\mathfrak{m}$ and an inejctive ring homomorphism $h: A \longrightarrow B=S^{-1}A$ defined by $f(a)=a/1.$ Let $\varphi = fgh.$ Then $\varphi : A \longrightarrow k$ and $\varphi(b_i) \neq 0$ for all $i,$ because $f$ is an isomorphism and $b_i \notin \mathfrak{m}$ for all $i. \ \Box$

See part (3) here.