Fulton 2.10

Posted on November 15, 2017

2.48.* Verify that for any R-module homomorphism \varphi:M\to M&39;, \ker(\varphi) and \text{im}(\varphi) are submodules of M and M&39; respectively. Show that   0\longrightarrow \ker(\varphi)\longrightarrow M\overset{\varphi}{\longrightarrow}   \text{im}(\varphi)\longrightarrow 0 is exact.

The map \phi:M\to im(\phi) is surjective by definition and similarly \ker\phi\hookrightarrow M is the inclusion map and therefore injective.

2.49.* (a) Let N be a submodule of M, \pi:M\to M/N the natural homomorphism. Suppose \varphi:M\to M&39; is a homomorphism of R-modules, and \varphi(N)=0. Show that there is a unique homomorphism \overline{\varphi}: M/N\to M&39; such that \overline{\varphi}\circ\pi=\varphi.

See Lemma 1.

(b) If N and P are submodules of a module M, with P\subset N, then there are natural homomorphisms from M/P onto M/N and from N/P into M/P. Show that the resulting sequence   0\longrightarrow N/P\longrightarrow M/P\longrightarrow M/N   \longrightarrow 0 is exact (“Second Noether Isomorphism Theorem”).

The map N/P\hookrightarrow M/P is the natural inclusion map. The map \varphi:M/P\to M/N exists by part (a). Again by part (a) \ker\varphi=\ker(\varphi\circ\pi) where \pi:M\to M/P is the natural homomorphism. Thus the sequence is exact.

(c) Let U\subset W\subset V be vector spaces, with V/U finite dimensional. Then \dim V/U=\dim V/W+\dim W/U.

Observe that      0\longrightarrow W/U\longrightarrow V/U\longrightarrow V/W   \longrightarrow 0 is exact by part (b). The result then follows by Proposition 7.

(d) If J\subset I are ideals in a ring R, there is a natural exact sequence of R-modules:   0\longrightarrow I/J\longrightarrow R/J\longrightarrow R/I\longrightarrow 0.

Note that ideals of a ring R are clearly R-submodules of R. Therefore the result follows from part (b).

(e) If \mathcal{O} is a local ring with maximal ideal \mathfrak{m}, there is a natural exact sequence of \mathcal{O}-modules   0\longrightarrow\mathfrak{m}^n/\mathfrak{m}^{n+1}\longrightarrow\mathcal{O}/\mathfrak{m}^{n+1}   \longrightarrow\mathcal{O}/\mathfrak{m}^n\longrightarrow 0.

Observe that \mathfrak{m}^{n+1}\subset\mathfrak{m}^n\subset\mathcal{O}. Thus the result follows by part (d).

2.50.* Let R be a DVR satisfying the conditions of Problem 2.30. Then \mathfrak{m}^n/\mathfrak{m}^{n+1} is an R-module, and so also a k-module, since k\subset R.

(a) Show that \dim_k(\mathfrak{m}^n/\mathfrak{m}^{n+1})=1 for all n\ge 0.

By (b) R/\mathfrak{m}^n has dimension n over k. Moreover, by 2.49(e), the following sequence is exact      0\longrightarrow\mathfrak{m}^n/\mathfrak{m}^{n+1}\longrightarrow R/\mathfrak{m}^{n+1}   \longrightarrow R/\mathfrak{m}^n\longrightarrow 0. Thus, by Proposition 7,      \dim_k(\mathfrak{m}^n/\mathfrak{m}^{n+1})=\dim_k(R/\mathfrak{m}^{n+1})-\dim_k(R/\mathfrak{m}^n)         =n+1-n=1.

(b) Show that \dim_k(R/\mathfrak{m}^n)=n for all n>0.

By Problem 2.30 R=\{\lambda_0+\ldots+\lambda_{n-1}t^{n-1}+z^nt^n\}, where t is the uniformizing parameter. Therefore R/\mathfrak{m}^n has spanning linearly indpendent set \{t^d \, | \, 0\le d\le n-1\}.

(c) Let z\in R. Show that \text{ord}(z)=n if (z)=\mathfrak{m}^n, and hence that \text{ord}(z)=\dim_k(R/(z)).

If (z)=\mathfrak{m}^n then z=ut^n and \text{ord}(z)=n.

2.51. Let 0\longrightarrow V_1\longrightarrow\ldots\longrightarrow V_n \longrightarrow 0 be an exact sequence of finite-dimensional vector spaces. Show that \sum(-1)^i\dim(V_i)=0.

For n=2 the statment is trivial. Proposition 7 proves the statement for such sequences with n=3,4. Suppose n>4 and the statement holds for all such exact sequences of length less than n (and at least 2). For 1\le k\le n-1, let \varphi_k:V_k\to V_{k+1} be the homomorphism from the exact sequence. Define W=\ker\phi_{n-1}=\text{im}\phi_{n-2}. Then      0\longrightarrow V_1\longrightarrow\ldots\longrightarrow     V_{n-2}\longrightarrow W\longrightarrow 0 \ \text{and} \  0\longrightarrow     V_{n-1}\longrightarrow V_n\longrightarrow W\longrightarrow 0 are exact. The second sequence gives us that \dim W=\dim V_{n-1}-\dim V_n. Therefore, by the first sequence,      0=\sum_{i=1}^{n-2}(-1)^i\dim V_i + (-1)^{n-1}(W)     =\sum_{i=1}^n(-1)^i\dim V_i.

2.52.* Let N,P be submodules of a module M. Show that the subgroup N+P=\{n+p \, | \, n\in N, p\in P\} is a submodule of M. Show that there is a natural R-module isomomrphism of N/(N\cap P) onto (N+P)/P (“First Noether Isomorphism Theorem”).

Observe that for n,m\in N, n-m\in P if and only if n-m\in N\cap P. Thus the map \varphi:N/(N\cap P)\to (N+P)/P defined by n+(N\cap P)\mapsto n+P is a well defined injective R-module homomorphism. Moreover, if x\in N+P then x=n+p for some n\in N, p\in P. So x+P=n+P=\varphi(n+(N\cap P)). So \varphi is surjective.

2.53.* Let V be a vector space, W a subspace, T:V\to V a one-to-one linear map such that T(W)\subset W, and assume V/W and W/T(W) are finite dimensional.

(a) Show that T induces an isomorphism of V/W with T(V)/T(W).

Let \pi:V\to V/W and \pi_T:T(V)\to T(W) be the natural homomorphisms. Observe that \ker \pi_T\circ T=W=\ker\pi. Thus by Lemma 1, \pi_T\circ T induces an isomorphism V/W\to T(V)/T(W).

(b) Construct an isomorphism between T(V)/(W\cap T(V)) and (W+T(V))/W, and an isomorphism between W/(W\cap T(V)) and (W+T(V))/T(V).

Note that T(V),W are both submodules of V. Therefore both results follow from Problem 2.52.

(c) Use Problem 2.49(c) to show that \dim V/(W+T(V))=\dim (W\cap T(V))/T(W).

Observe that V/W is isomorphic to T(V)/T(W) since T is an isomorphism. Therefore \dim V/W=\dim T(V)/T(W)=n. Note that W\subset W+T(V)\subset V and V/W is finite dimensional. Therefore by Problem 2.49(c),      n=\dim V/W = \dim V/(W+T(V))+\dim (W+T(V))/W. Similarly, T(W)\subset W\cap T(V)\subset T(V) and thus      n=\dim T(V)/T(W)=\dim T(V)/(W\cap T(V))+\dim (W\cap T(V))/T(W). By part (b), \dim T(V)/(W\cap T(V))=\dim (W+T(V))/W, yielding the desired equality \dim V/(W+T(V))=\dim (W\cap T(V))/T(W).

(d) Conclude finally that \dim V/T(V)=\dim W/T(W).

Observe that (V/T(W))/(W/T(W))\cong T/W by the second isomorphism theorem for modules. Since W/T(W) and T/W are finitely dimensional, Lemma 2.1 implies that V/T(W) is finitely dimensional. Since T(V)\subset W+T(V)\subset V, Problem 2.49(c) shows      \dim V/T(V)=\dim V/(W+T(V))+\dim (W+T(V))/T(V). Moreover, since W/T(W) is finite dimensional and T(W)\subset W\cap T(V)\subset W,      \dim W/T(W)=\dim W/(W\cap T(V))+\dim (W\cap T(V))/T(W). By part (b),      \dim W/(W\cap T(V))=\dim (W+T(V))/T(V). By part (c),      \dim (W\cap T(V))/T(W)=\dim V/(W+T(V)). Therefore      \dim V/T(V)=\dim W/T(W).