Algebra 2 Home Exercises 3 1. Let A be a Noetherian ring. Prove that an A-module M is a Noetherian A-module iff M is of finite type. Hint. The left arrow was proved in the lectures. To prove the right arrow assume that M isn't of finite type, then there are infinitely many elements mi in M such that mi+1 doesn't belong to Am1 + · · · + Ami. Show that the infinite sequence of A-submodules Am1; Am1 + Am2; Am1 + Am2 + Am3; : : : is strictly increasing. 1 formal power series A A X a Xn a A 2. Define the set of over a ring by [[ ]] = fPn=0 n : n 2 g. Define 1 1 1 n n n X anX + X bnX = X(an + bn)X ; n=0 n=0 n=0 1 1 1 n n n n (X anX )(X bnX ) = X cnX ; where cn = X akbn−k: n=0 n=0 n=0 k=0 Then A[[X]] is a commutative ring. Prove that A[[X]] is a Noetherian ring if A is a Noetherian ring. Hint: Modify cleverly the proof of Hilbert's theorem and consider the sequence of ideals n Jn = fa 2 A : there is f(X) = aX + terms of higher order in Jg: Note: higher order, not smaller order. Instead of polynomials which generate the ideal in the proof (m) (m) m of Hilbert's theorem consider power series fj (X) = aj X + terms of higher order . 3. Let A be a Noetherian ring and let f: A ! A be a ring homomorphism. Prove that f is an isomorphism iff f is surjective. [Remark: without the assumption of Noetherianity this is false]. (j) Hint. Assume that f is surjective and denote by Ij the kernel of f = f ◦ · · · ◦ f: A ! A. Show that Ij form an increasing sequence of ideals of A and therefore Ij = Ij+1 for some j. Deduce that f(f (j)(a)) = 0 implies f (j)(a) = 0 and use the surjectivity of f to complete the proof. Algebra 2 Solutions of Home Exercises 3 1. Assume that M isn't of finite type. Then there are infinitely many elements mi in M such that mi+1 doesn't belong to Am1 + · · · + Ami. Then the infinite sequence of A-submodules Am1; Am1 + Am2; Am1 + Am2 + Am3; : : : is strictly increasing, hence M is not a Noetherian A-module. 2. Let J be a non-zero ideal of A[[X]]. For n ≥ 0 define n Jn = fa 2 A : there is f(X) = aX + terms of higher order in Jg: Then Jn is an ideal of A and J1 ⊂ J2 ⊂ : : : . Since A is Noetherian, we deduce that there is n such that Jn = Jn+1 = : : : . For m ≤ n the ideal Jm as an ideal of the Noetherian ring A is finitely (m) (m) (m) m generated, let cj for 1 ≤ j ≤ nm be its generators. Pick up fj (X) = cj X + · · · 2 J. m Let f = amX + · · · 2 J. We shall pass from f to another series in J which starts with a larger power of X. (m) (m) (m) (m) (m) If m ≤ n, then write am = Pj aj cj with aj 2 A. Then f − Pj aj fj (X) = m+1 m+2 (m) (n) (m) bm+1X + bm+2X + · · · 2 J. If m > n, then write am = Pj aj cj with aj 2 A. Then (m) m−n (n) m+1 m+2 f − Pj aj X fj (X) = bm+1X + bm+2X + · · · 2 J. Continue this way (infinitely many times, keeping in mind that we work with power series that's (m) fine). Then f is a linear combination with coefficients in A[[X]] of fj (X), 0 ≤ m ≤ n, 1 ≤ j ≤ nm. Thus, J is finitely generated. (j) 3. The sequence of the ideals Ij stabilizes for sufficiently large j > j0. Since f is surjective, f (j) is surjective for all j. Now, if f(b) = 0 then for a j > j0 find a such that b = f (a). Then (j+1) f(b) = f (a) = 0, so a 2 Ij+1 = Ij , and so b = 0. Thus f is injective..