Fulton 2.10

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.

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 $\{t^d \, | \, 0\le d\le n-1\}.$

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.

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 $\dim V/T(V)=\dim W/T(W).$