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For a module homomorphism $f:M\to N$, the map $m+\ker f\mapsto f(m)$ is a module isomorphism $M/\ker f\to \mathrm{Im}f$. The formula is well-defined because two representatives differ by an element of the kernel of a module homomorphism exactly when their images agree; the same observation proves injectivity, and surjectivity onto the image follows by its definition.

A free module is a module with a basis of a module $(e_i{)}_{i\in I}$: every element has exactly one expression $m={\sum }_{i\in I}{r}_i{e}_i$ with $r_i\in R$ and only finitely many $r_i$ nonzero. Equivalently, the coefficient map ${⨁}_{i\in I}R\to M$ is a module isomorphism. Existence means generation; uniqueness means linear independence in a module. For finite $I$ of size $n$, this says $M\cong R^n$, a finite free module. See Neil Strickland's Rings, Modules and Linear Algebra, section 11.

The integer lattice ${\mathbb{Z}}^2$ in the opening diagram is a free module over $\mathbb{Z}$ with basis of a module $e_1=(1,0)$, $e_2=(0,1)$. For example, $(3,2)=3e_1+2e_2$, and comparing the two coordinates makes the coefficients unique. A different basis of a module is $u=(1,0)$, $v=(1,1)$, because $(a,b)=(a-b)u+bv$; in particular $(3,2)=u+2v$.

By contrast, $w_1=(2,0)$ and $w_2=(0,1)$ are linearly independent in a module, but generate only the submodule $2\mathbb{Z}\times \mathbb{Z}$. They miss $(1,0)$ because $2a=1$ has no integer solution. They would form a basis over the rational numbers. Even generation need not let one extract a basis of a module: the elements $2,3$ generate $\mathbb{Z}$, since $1=3-2$, but neither singleton generates it, and $3\cdot 2-2\cdot 3=0$ is a nontrivial linear relation between them. These are concrete failures of the usual vector space argument that divides by any nonzero coefficient.

The cyclic module $\mathbb{Z}/6\mathbb{Z}$ has no basis of a module over $\mathbb{Z}$: every element has the nontrivial relation $6m=0$, so no nonempty subset is linearly independent in a module, while the empty subset cannot generate this nonzero module. Specifying the scalar ring is essential: over $R=\mathbb{Z}/6\mathbb{Z}$, the same additive object is a free module with the one-element basis of a module $\{\overline{1}\}$, since each coefficient is itself a unique class in $R$.

A module isomorphism is a bijection of modules over the same ring preserving addition and scalar multiplication. Its inverse preserves these operations too. When a linear operator $T$ gives a vector space an $F[t]$-module structure by $t\cdot v=T(v)$, a module isomorphism is exactly an invertible linear map intertwining the two operators. This translates classification of modules into matrix similarity and rational canonical form.

Use the defining criterion that a flat module $E$ makes $u\otimes 1_E$ injective for every injective module homomorphism $u$. By right exactness of the tensor product, preserving injections is equivalent to preserving every short exact sequence.

First assume $E$ is flat over $A$, and fix a maximal ideal $\mathfrak{m}$. Let $u:U\hookrightarrow V$ be an injection of $A_{\mathfrak{m}}$-modules. Regard it as an injection of $A$-modules. There are natural module isomorphismsFor example, $u_0\otimes (e/s)$ maps to $s^{-1}{u}_0\otimes e$, with inverse $u_0\otimes e\mapsto u_0\otimes (e/1)$. These are well-defined because every $s\notin \mathfrak{m}$ already acts invertibly on $U$. Under these isomorphisms the tensor map is $u{\otimes }_A{1}_E$, which is injective by flatness. Thus $E_{\mathfrak{m}}$ is flat over $A_{\mathfrak{m}}$.

For the converse, assume every $E_{\mathfrak{m}}$ is flat over $A_{\mathfrak{m}}$ and take any injection $u:U\hookrightarrow V$ of $A$-modules. By part (c), each $u_{\mathfrak{m}}$ is injective. Since localization commutes with tensor products, the localization of $u{\otimes }_A{1}_E$ identifies withIt is injective by the assumed local flatness. Apply part (c) again, now to $u{\otimes }_A{1}_E$, to conclude that it is injective globally. Since $u$ was arbitrary, $E$ is flat. ThereforeThis proves flatness is local without any finite-generation hypothesis.

For any injection $u:U\hookrightarrow V$, the direct sum decomposition $F=P\oplus Q$ gives natural module isomorphismsUnder these, $u\otimes 1_F$ is the block-diagonal map $(u\otimes 1_P)\oplus (u\otimes 1_Q)$. It is injective because $F$ is a flat module. If an element lies in the kernel of $u\otimes 1_P$, insert zero in the other component; injectivity of the whole map forces that element to be zero. Hence $u\otimes 1_P$ is injective for every $u$, which proves $\boxed{P\text{ is flat}}$. This establishes that direct summands of flat modules are flat, without a finite-generation assumption.

The annihilating polynomial excludes any free $\mathbb{C}[X]$-summand. In a cyclic primary decomposition summand $\mathbb{C}[X]/(P^a)$, the same condition requires $P^a$ to divide $(X-2{)}^4$. Thus $P=X-2$, up to a scalar unit, and $1\le a\le 4$. The complex dimension of this summand is $a$.

Write $A_j=\mathbb{C}[X]/((X-2{)}^j)$. The exponents must form a partition of an integer $4$, giving exactlyThey are distinct module isomorphism types by uniqueness in the structure theorem for finitely generated modules over a principal ideal domain; equivalently they are the five possible Jordan block patterns for eigenvalue $2$ in dimension four.

The structure theorem for finitely generated modules over a principal ideal domain applies in particular to a Euclidean domain $R$: every finitely generated module has an isomorphismwith each $d_i$ nonzero and a nonunit. The rank of a free module $r$ and the invariant factors of a finitely generated module $d_i$ are unique up to multiplication by units; empty sums are allowed. This is the classification theorem being used, without a proof of that theorem.

The rational canonical form theorem states that for a linear operator $T$ on a finite-dimensional vector space over any field $F$, there are unique monic nonconstant polynomials $p_1\mid \cdots \mid p_k$ such that some basis givesHere, for $p(t)=t^d+a_{d-1}{t}^{d-1}+\cdots +a_0$, the companion matrix has ones on its subdiagonal and last column $(-a_0,\dots ,-a_{d-1}{)}^T$. The empty matrix covers a zero-dimensional space. Two linear operators have matrix similarity exactly when they have the same invariant factors of a linear operator.

For the proof, make $V$ an $F[t]$-module by $t\cdot v=T(v)$. It is a finitely generated module, since a vector-space basis also generates it as a module. It is a torsion module: the powers of $T$ have linear dependence in the finite-dimensional space ${\mathrm{End}}_F(V)$, so some nonzero polynomial annihilates every vector. Because $F[t]$ is a Euclidean domain, the classification theorem givesThere is no free summand, since a nonzero free module over $F[t]$ is not a torsion module. Choose monic generators of the ideals. On $F[t]/(p_i)$, the classes of $1,t,\dots ,t^{d_i-1}$ form an $F$-basis by polynomial division. Multiplication by $t$ is exactly $C(p_i)$ in that basis, proving existence. Uniqueness in the classification theorem proves uniqueness of the rational canonical form. Finally, an $F[t]$-module isomorphism is precisely an invertible $F$-linear map intertwining the operators, establishing the assertion about matrix similarity. In particular the characteristic polynomial is ${\prod }_i{p}_i$ and the minimal polynomial is $p_k$ for nonzero $V$.

Put $K=\ker \varphi$, $H=\ker \psi$, and $Q=\mathrm{Im}q$. The kernel of a module homomorphism $H$ is a submodule, so its image $L=q(H)$ is a submodule. For $x\in H$, $\varphi (q(x))=\psi (x)=0$, showing $L\subseteq K$. Thus the quotient module $K/L$ exists. Also $L\subseteq Q$.

Consider the restriction to $K$ of the natural module homomorphism $R^m\to R^m/Q$:Its kernel of a module homomorphism is $K\cap Q$. We have $L\subseteq K\cap Q$. Conversely, if $y\in K\cap Q$, choose $x\in R^n$ with $y=q(x)$. Thenso $x\in H$ and $y\in L$. Hence $\boxed{K\cap Q=L}$.

The map $\alpha$ is surjective. Given $v+Q\in R^m/Q$, use the surjectivity of $\psi$ to choose $x\in R^n$ with $\psi (x)=\varphi (v)$. Set $y=v-q(x)$. ThenThus $y\in K$ is a preimage of the arbitrary class $v+Q$.

By the first isomorphism theorem for modules, $\alpha$ induces the module isomorphismExplicitly, replacing $y$ by $y+\ell$ with $\ell \in L$ leaves its class modulo $Q$ unchanged, proving well-definedness. Its formula preserves addition and scalar multiplication. If its value is zero, $y\in K\cap Q=L$, so $y+L=0$, proving injectivity; surjectivity was proved above. These checks give all the requested properties of the induced map.

A linear operator $T:V\to V$ on a vector space over a field $k$ makes $V$ a module over the polynomial ring $k[t]$: let $t$ act as $T$, let constants act as the original scalars, and define $p(t)\cdot v=p(T)v$. The identity $(pq)(T)=p(T)q(T)$ for polynomials $p,q$ verifies the multiplication axiom. The submodules are precisely the invariant subspaces of $T$. This construction is described in Keith Conrad's Introductory Notes on Modules, section 6.

For a computed example, take $k=\mathbb{Q}$, $V=k^2$, andThen $T{e}_2=e_1$, $T{e}_1=0$, and $T^2=0$. ConsequentlyEvery $(a,b)$ equals $(b+at)\cdot e_2$. The module homomorphism $k[t]\to V$, $p\mapsto p(T)e_2$, is therefore surjective, and its kernel is the ideal $(t^2)$. It induces a module isomorphism $V\cong k[t]/(t^2)$. This is a cyclic module and a torsion module over $k[t]$, since $t^{2}V=0$. Over $k$ it is a free module with basis $(e_1,e_2)$ and has no nonzero torsion elements of a module. Over $\mathbb{Q}$ its nonzero elements also have infinite additive order of a group element. Changing scalars changes the questions one asks about the same addition law.