Bourbaki marks this one as harder than the rest.
Let k be a field, L an infinite set and A the polynomial ring k[Xλ]λ∈L. If k0 is the prime field contained in k, let b denote the cardinal of a transcendence basis of k over k0 and let us write c=Card(L).
(a) For every ideal a of A, show that there exists a system of generators S of a such that Card(S)≤c. (Consider the intersections of a and the subrings k[Xλ]λ∈J for the finite subsets J of L.)
(b) Suppose that c<b. Show 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 k0 by the coefficients of the polynomials forming a system of generators of m of cardinal ≤c; let m0=m∩K[Xλ]λ∈L, so that m=m0⊗Kk and
A/m=(K[Xλ]λ∈L/m0)⊗Kk. Note finally that there exists a
K-isomorphism of the field of fractions of
K[Xλ]λ∈L/m0 onto an algebraic closure of
k.) Deduce that
A is then a Jacobson ring (use Exercise 8).
(c) Suppose that b<c; then Card(k)≤c, so that there exists a bijection λ↦ξλ of a subset L′ of L onto k, L′ having a non-empty complement in L. Let λ0∈L−L′; show that there exists a k-homomorphism u of A onto a subfield of the field of rational functions k(Y) such that u(Xλ0)=Y,
u(Xλ)=1/(Y−ξλ) for
λ∈L′,
u(Xλ)=0 for
λ∈L−(L′∪{λ0}); 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), show 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 Card(B)=b.)