Bourbaki

Exercise 6, § 3

Algebra, V, p. 147

It is set in small type as supplementary.

Let C be an uncountable algebraically closed field and P=C[(Xi)i∈I]P = C[(X_i)_{i \in I}] a polynomial algebra over C. Given two families (fα)α∈A,(gβ)β∈B(f_\alpha)_{\alpha \in A}, (g_\beta)_{\beta \in B} of elements of P suppose that: a) B and I are countable. b) For every finite subset A′A' of A and every finite subset B′B' of B there exists (xi)i∈I∈CI(x_i)_{i \in I} \in C^I such that

fα(xi)=0andgβ(xi)≠0for allα∈A′,β∈B′. f_\alpha(x_i) = 0 \quad \text{and} \quad g_\beta(x_i) \neq 0 \quad \text{for all} \quad \alpha \in A', \beta \in B'.

Show that then there exists (xi)i∈I∈CI(x_i)_{i \in I} \in C^I such that

fα(xi)=0andgβ(xi)≠0for allα∈A,β∈B. f_\alpha(x_i) = 0 \quad \text{and} \quad g_\beta(x_i) \neq 0 \quad \text{for all} \quad \alpha \in A, \beta \in B .

(Let (Yβ)β∈B(Y_\beta)_{\beta \in B} be a family of indeterminates indexed by B and let Q be the polynomial algebra C[(Xi)i∈I,(Yβ)β∈B]\mathbf{C}[(X_i)_{i \in I}, (Y_\beta)_{\beta \in B}]. Let A be the quotient of Q by the ideal generated by the fαf_\alpha and the 1−gαYα1 - g_\alpha Y_\alpha. Show that A≠0A \neq 0 thanks to a) and b) and apply Ex. 5 to a maximal ideal of A.) \*

Solution

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