Tensor product of division algebras (2)

Posted: October 29, 2011 in Division Rings, Noncommutative Ring Theory Notes
Tags: , ,

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.


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