## Krull-Schmidt Theorem

Here, we will prove that the process of decomposing $M = M_1 \oplus \ldots \oplus M_k$ is unique, given that M is noetherian and artinian.

Again, R is a ring, possibly non-commutative.

Definition. A decomposition of an R-module M is an expression $M = M_1 \oplus \ldots \oplus M_k$ for non-zero modules $M_1, \ldots, M_k.$

An R-module M is said to be indecomposable if no such expression exists for k≥2 (or equivalently, for k=2).

A simple module is clearly indecomposable, but the converse is not true.

For example, let $R = \begin{pmatrix} * & * \\ 0 & *\end{pmatrix}$ be the ring of upper-triangular 2 × 2 real matrices. Consider the module M of column vectors R2, where R acts by matrix multiplication. Then M is not simple since it has a submodule spanned by (1, 0). But M is indecomposable since any submodule N ⊂ M which contains an (xy) for y≠0 must be the whole M.

We have a criterion for a module to be indecomposable:

Proposition. Let M be an artinian and noetherian R-module. Then M is indecomposable if and only if $\text{End}_R(M)$ is a local ring.

Proof

⇐ : if MM1 ⊕ M2 is a decomposition of M, let π1, π2  : M → M be the projection onto M1M2 respectively. Then π1, π2 are not units but π12 is.

⇒ : recall that for an artinian and noetherian M, a module map fM → M is bijective iff it is surjective or injective. Now suppose fg : M → M are module maps such that f+g is a unit and g is not. We need to show that f is a unit.

Write (f+g)h = 1 for some h : M → M, so fh = 1-gh. Now k := gh is not an isomorphism since g is not surjective. Since M is artinian and noetherian, eventually $\text{ker} (k^n)= \text{ker}(k^{n+1}) = \ldots, \qquad \text{im}(k^n)= \text{im}(k^{n+1}) \ldots$

for some large n. We claim that $M = \text{ker}(k^n)\oplus \text{im}(k^n).$

• To show $M = \text{ker}(k^n)+ \text{im}(k^n)$: for any xM, we have $k^n(x) \in \text{im}(k^n) = \text{im}(k^{2n})$ so we can write $k^n(x) = k^{2n}(y).$ But then $x = (x - k^n(y)) + k^n(y)$; clearly the second term lies in $\text{im}(k^n)$ and the first term gives $k^n(x - k^n(y)) = k^n(x) - k^{2n}(y) = 0$ so it lies in $\text{ker}(k^n).$
• To show $\text{ker}(k^n) \cap \text{im}(k^n) = 0$: if $y = k^n(x)$ lies in $\text{ker}(k^n)$ then we have $k^n(y) = k^{2n}(x) = 0$ which gives $x \in \text{ker}(k^{2n}) = \text{ker}(k^n)$ and so $y =k^n(x) = 0.$

Since M is indecomposable either $\text{ker}(k^n) = 0$ or $\text{im}(k^n) = 0.$ The former case says k is injective and hence an isomorphism, so we have the latter case kn=0. Thus fh = 1-k is invertible (since 1 + nilpotent = unit), so f is right-invertible. By symmetry, it is also left-invertible. ♦ ## The Krull-Schmidt Theorem

Before stating and proving our main theorem, here is a useful criterion for splitting a module as a direct sum:

Splitting Lemma. If $f: M \to N, g:N\to M$ satisfy $gf = 1_M$ then N splits as $\text{im}(f)\oplus \text{ker}(g).$ Furthermore f is injective, thus M is a direct summand of N.

Note

To remember this, imagine NMP with fM → MP and gMP → M given by the inclusion and projection maps.

Proof

f is injective since gf is. For the first statement:

• M = im(f) + ker(g) : any xM can be written as x = fg(x) + (xfg(x)); the first term lies in im(f); the second term gives g(x – fg(x)) = g(x) – gfg(x) = 0.
• im(f) ∩ ker(g) = 0: if f(y) lies in ker(g), then gf(y) = 0 so y = 0. ♦

Finally the main theorem we would like to prove:

Krull-Schmidt Theorem. Let M be an artinian and noetherian module. If we decompose: $M \cong U_1 \oplus U_2 \oplus \ldots \oplus U_k, \qquad M\cong V_1 \oplus V_2 \oplus \ldots \oplus V_l$

then k=l and there is a permutation σ of {1, 2, …, k} such that $U_i \cong V_{\sigma(i)}$ for all i=1,…,k.

Proof

We shall prove that U1 ≅ Vj for some j, such that $M \cong V_j \oplus U_2 \oplus \ldots \oplus U_k,$ after which we can apply induction on k.

Let $p: M\to M$ be the projection onto U1 and $q_i : M\to M$ be the projection onto Vi.  This gives $\sum_i q_i = 1_M$ and thus $p = p\sum_i q_i = \sum_i (pq_i).$ Since im(p) ⊆ U1 this restricts to a map $\alpha_i = p(q_i|_{U_1}) : U_1 \to U_1.$

Since U1 is indecomposable, End(U1) is local; and since $\sum_j \alpha_j = 1_{U_1}$, one of the αj is invertible, say α1. Let $r:U_1 \to U_1$ be the inverse of $\alpha_1 : U_1 \to U_1$, so $r(p|_{V_1})(q_1|_{U_1}) = 1_{U_1},$ where $q_1|_{U_1} : U_1 \to V_1, p|_{V_1} : V_1 \to U_1, r: U_1 \to U_1.$ (*)

From the splitting lemma, Uis a direct summand of V1; since V1 is indecomposable we must have U1V1. To prove that $M \cong V_1 \oplus (U_2 \oplus \ldots \oplus U_k)$ we write (*) as $rp (q_1|_{U_1}) = 1_{U_1}$ where $q_1|_{U_1} : U_1 \to M, p : M\to U_1, r : U_1\to U_1.$

Again, splitting lemma tells us $M \cong \text{ker}(rp) \oplus \text{im}(q_1|_{U_1}).$ The second term is isomorphic to V1 and the first is just ker(p), which is $U_2 \oplus \ldots \oplus U_k$ as desired. ♦

Note

Because M is artinian and noetherian it is always possible to decompose M as a direct sum of finitely many indecomposable modules. Simply keep decomposing the terms until we’re left with indecomposable modules: since M is of finite length, the process must terminate. Krull-Schmidt theorem says that the resulting terms are unique to isomorphism and permutation.

Example

Let R be the ring of upper-triangular 2 × 2 matrices with real entries. Here are two different ways of decomposing R as a direct sum of indecomposable left ideals: $R = \left\{\begin{pmatrix} * & * \\ 0 & * \end{pmatrix}\right\} = \left\{ \begin{pmatrix} * & 0 \\ 0 & 0 \end{pmatrix}\right\} \oplus \left\{ \begin{pmatrix} 0 & * \\ 0 & *\end{pmatrix}\right\} = \left\{ \begin{pmatrix} a & a \\ 0 & 0\end{pmatrix}\right\} \oplus \left\{ \begin{pmatrix} 0 & * \\ 0 & * \end{pmatrix} \right\}.$

Regardless of the decomposition we pick, the second term must be there (prove this!).

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