Bourbaki

Exercise 2, § 4

Commutative Algebra, III, 255

Bourbaki marks this one as harder than the rest.

Let A be a local ring, m its maximal ideal, k=A/mk = A/m its residue field and f:A→kf : A \to k the canonical homomorphism. Let P∈A[X]P \in A[X] be a monic polynomial of degree n. We write B=A[X]/P.A[X]B = A[X]/P.A[X] and denote by xx the canonical image of X in B.

(a) Let Q, Q′Q' be two strongly relatively prime monic polynomials in A[X]A[X] such that P=QQ′P = QQ'. Show that the ring B is the direct sum of the ideals B.Q(x)B.Q(x) and B.Q′(x)B.Q'(x).

(b) Conversely, let B=b⊕b′B = b \oplus b' be a decomposition of B as a direct sum of two ideals. Show that there exist polynomials Q, Q′Q' in A[X]A[X] satisfying the hypotheses of (a) and such that b=B.Q(x),b′=B.Q′(x)b = B.Q(x), b' = B.Q'(x). (Show first that b/mBb/mB and b′/mBb'/mB are generated by the images in B/mBB/mB of monic polynomials Q0∈f~(b),Q0′∈f~(b′)Q_0 \in \tilde{f}(b), Q'_0 \in \tilde{f}(b')) such that

f~(P)=f~(Q0)f~(Q0′). \tilde{f}(P) = \tilde{f}(Q_0) \tilde{f}(Q'_0).
Let r=deg⁡(Q0),s=deg⁡(Q0′)r = \deg(Q_0), s = \deg(Q'_0). Show that b is a free A-module with basis Q0(x),xQ0(x),…,xs−1Q0(x)Q_0(x), xQ_0(x), \ldots, x^{s-1}Q_0(x) (apply Chapter II, § 3, no. 2, Proposition 5); we may then write xsQ0(x)=a0Q0(x)+a1xQ0(x)+⋯+as−1xs−1Q0(x)x^s Q_0(x) = a_0 Q_0(x) + a_1 x Q_0(x) + \cdots + a_{s-1} x^{s-1} Q_0(x), where ai∈Aa_i \in A (0≤i≤s−10 \leq i \leq s-1); show that, if we write
Q′(X)=Xs−(as+a1X+⋯+as−1Xs−1), Q'(X) = X^s - (a_s + a_1 X + \cdots + a_{s-1} X^{s-1}),
then f~(P)=f~(Q0)f~(Q′)\tilde{f}(P) = \tilde{f}(Q_0) \tilde{f}(Q') and Q′∈f~(b′)Q' \in \tilde{f}(b'). Similarly define Q starting with Q′Q' and Q0Q_0 and show that Q and Q′Q' solve the problem, using Proposition 2 of no. 1.)

Solution

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Transcribed from the printed volume. Not translated.