Exercise 10, § 3

Let $k$ be a field, $L$ an infinite set, $A$ the polynomial ring $k[X_\lambda]_{\lambda \in L}$. If $k_0$ is the prime field contained in $k$, denote by $b$ the cardinal of a transcendence basis of $k$ over $k_0$, and put $c = \mathrm{Card}(L)$.

a) For every ideal $a$ of $A$, now prove that there exists a system of generators $S$ of $a$ such that $\mathrm{Card}(S) \leq c$. (Consider the intersections of $a$ and the subrings $k[X_\lambda]_{\lambda \in J}$ for the finite subsets $J$ of $L.$

b) Suppose that $c < b$. Now prove that if $m$ is a maximal ideal of $A$, $A/m$ is an algebraic extension of $k$. (Let $K$ be the subfield of $k$ generated over $k_0$ by the coefficients of the polynomials forming a system of generators of $m$, of cardinal $\leq c$; let $m_0 = m \cap K[X_\lambda]_{\lambda \in L}$, so that one has $m = m_0 \otimes_K k$, and $A/m = (K[X_\lambda]_{\lambda \in L}/m_0) \otimes_K k$. Finally remark that there exists a $K$-isomorphism of the field of fractions of $K[X_\lambda]_{\lambda \in L}/m_0$ into an algebraic closure of $k.$) Deduce that $A$ is then a Jacobson ring (use exerc. 8).

c) Suppose that $b < c$; then $\mathrm{Card}(k) \leq c$, so that there exists a bijection $\lambda \to \xi_\lambda$ of a subset $L'$ of $L$ onto $k, L'$ having a nonempty complement in $L$. Let $\lambda_0 \in L - L'$; now prove that there exists a $k$-homomorphism $u$ of $A$ onto a subfield of the rational fraction field $k(Y)$ such that $u(X_{\lambda_0}) = Y, u(X_\lambda) = 1/(Y - \xi_\lambda)$ for $\lambda \in L'$, $u(X_\lambda) = 0$ for $\lambda \in L - (L' \cup \{\lambda_0\})$; if $m$ is the kernel of $u$, $A/m$ is therefore not an algebraic extension of $k$.

d) Under the same hypotheses as in c), now prove that $A$ is not a Jacobson ring. (Note that there exist $k$-algebras $B$ which are not Jacobson rings (for example local rings) such that $\mathrm{Card}(B) = b.$)