Bourbaki

Exercise 10, § 3

Commutative Algebra, V, 373

Bourbaki marks this one as harder than the rest.

Let kk be a field, LL an infinite set and AA the polynomial ring k[Xλ]λ∈Lk[X_\lambda]_{\lambda \in L}. If k0k_0 is the prime field contained in kk, let bb denote the cardinal of a transcendence basis of kk over k0k_0 and let us write c=Card(L)c = \mathrm{Card}(L).

(a) For every ideal aa of AA, show that there exists a system of generators SS of aa such that Card(S)≤c\mathrm{Card}(S) \leq c. (Consider the intersections of aa and the subrings k[Xλ]λ∈Jk[X_\lambda]_{\lambda \in J} for the finite subsets JJ of LL.)

(b) Suppose that c<bc < b. Show that, if mm is a maximal ideal of AA, A/mA/m is an algebraic extension of kk. (Let KK be the subfield of kk generated over k0k_0 by the coefficients of the polynomials forming a system of generators of mm of cardinal ≤c\leq c; let m0=m∩K[Xλ]λ∈Lm_0 = m \cap K[X_\lambda]_{\lambda \in L}, so that m=m0⊗Kkm = m_0 \otimes_K k and

A/m=(K[Xλ]λ∈L/m0)⊗Kk. A/m = (K[X_\lambda]_{\lambda \in L}/m_0) \otimes_K k.
Note finally that there exists a KK-isomorphism of the field of fractions of K[Xλ]λ∈L/m0K[X_\lambda]_{\lambda \in L}/m_0 onto an algebraic closure of kk.) Deduce that AA is then a Jacobson ring (use Exercise 8).

(c) Suppose that b<cb < c; then Card(k)≤c\mathrm{Card}(k) \leq c, so that there exists a bijection λ↦ξλ\lambda \mapsto \xi_\lambda of a subset L′L' of LL onto kk, L′L' having a non-empty complement in LL. Let λ0∈L−L′\lambda_0 \in L - L'; show that there exists a kk-homomorphism uu of AA onto a subfield of the field of rational functions k(Y)k(Y) such that u(Xλ0)=Yu(X_{\lambda_0}) = Y,

u(Xλ)=1/(Y−ξλ) u(X_\lambda) = 1/(Y - \xi_\lambda)
for λ∈L′\lambda \in L', u(Xλ)=0u(X_\lambda) = 0 for λ∈L−(L′∪{λ0})\lambda \in L - (L' \cup \{\lambda_0\}); if mm is the kernel of uu, A/mA/m is therefore not an algebraic extension of kk.

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

Solution

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Transcribed from the printed volume. Not translated.