Bourbaki

Lemma 1

Tag 04RE, permanent. Cite it as [Bourbaki, Tag 04RE].

Commutative Algebra, chapter V, § 3. Finitely generated algebras over a field

Let AA be a finitely generated algebra over a field kk, (ai)1≤i≤n(a_i)_{1 \leq i \leq n} a system of generators of this algebra and cc the ideal of algebraic relations between the aia_i with coefficients in kk (Algebra, Chapter IV, § 2, no. 1). For every extension field LL of kk, the mapping f↦(f(ai))1≤i≤nf \mapsto (f(a_i))_{1 \leq i \leq n} is a bijection of the set of k-homomorphisms from AA to LL onto the set of zeros of rr in LnL^n.

There exists a unique k-algebra homomorphism hh of k[X1,…,Xn]k[X_1, \ldots, X_n] onto AA such that h(Xi)=aih(X_i) = a_i, for 1≤i≤n1 \leq i \leq n and by definition rr is the kernel of hh. The mapping f ↦\mapsto his a bijection of the set of k-homomorphisms from AA to LL onto the set of k-homomorphisms from k[X1,…,Xn]k[X_1, \ldots, X_n] to LL which are zero on c. For every polynomial P∈k[X1,…,Xn]P \in k[X_1, \ldots, X_n] and every element x=(x1,…,xi)∈Lnx = (x_1, \ldots, x_i) \in L^n we write hx(P)=P(x)h_x(P) = P(x); then the mapping x↦hxx \mapsto h_x is a bijection of LnL^n onto the set of k-homomorphisms from k[X1,…,Xn]k[X_1, \ldots, X_n] to LL (such a homomorphism being determined by its values at the XiX_i (1≤i≤n1 \leq i \leq n) ); to say that hxh_x is zero on c means that xx is a zero of c in L′′L'', whence the lemma.

If Proposition 1 is applied to the algebra A=k[X1,…,Xn]/rA = k[X_1, \ldots, X_n]/r, where c is an ideal of k[X1,…,Xn]k[X_1, \ldots, X_n] distinct from the whole ring, we obtain by Lemma 1 the following statement:

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Transcribed from the printed volume. Not translated.