Bourbaki

Lemma 1

Tag 0D7Y, permanent. Cite it as [Bourbaki, Tag 0D7Y].

Commutative Algebra, chapter X, § 6. Extension des scalaires dans les algèbres régulières

Let AA be a noetherian ring, the union of an increasing filtered family (Aα)α∈I(A_\alpha)_{\alpha \in I} of noetherian subrings.

a) If the rings AαA_\alpha are regular and if AA is a flat AαA_\alpha-module for every α∈I\alpha \in I, the ring AA is regular.

b) If the rings AαA_\alpha are normal (§1§ 1, No. 8), AA is normal.

a) Let mm be a maximal ideal of AA; for each α∈I\alpha \in I, let us denote by mαm_\alpha the ideal m∩Aαm \cap A_\alpha of AαA_\alpha. Since AA is noetherian, mm is of finite type and there exists an element α\alpha of II such that m=Amαm = A m_\alpha, so that the AA-module (Aα/mα)⊗AαA(A_\alpha / m_\alpha) \otimes_{A_\alpha} A is isomorphic to A/mA / m. Since AA is flat over AαA_\alpha, we have dpA(A/m)≤dpAα(Aα/mα)\mathrm{dp}_A(A/m) \leq \mathrm{dp}_{A_\alpha}(A_\alpha / m_\alpha) (A, X, p. 141, Lemma 2). Since the rings AαA_\alpha are regular, it follows that AA is regular (§4§ 4, No. 2, Prop. 4).

b) Since the AαA_\alpha are reduced, AA is reduced. Let aa and bb be elements of AA such that bb is not a zero divisor and that the element a/ba/b of the total ring of fractions of AA is integral over AA. There exists a monic polynomial P∈A[X]P \in A[X] such that P(a/b)=0P(a/b) = 0. Let α\alpha be an element of II such that the ring AαA_\alpha contains aa, bb and the coefficients of PP. Since AαA_\alpha is normal, there exists c∈Aαc \in A_\alpha such that a=bca = bc. Thus a/b=ca/b = c belongs to AA and AA is normal.

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

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