Direct sum of modules

Question

Let $R$ be a PID,$M$ an $R$-module and $M$ is the direct sum of $M_1, \dots, M_k$ where $M_i \leq M$ for $1 \leq i \leq k$.

Now let $N$ be a submodule of $M$ .

Is it true that $N$ is the direct sum of $N_1,\dots,N_t$ where $N_i \lt N$ and $N_{k-t+1} \le M_i$?

I have tried to format it. Please correct if there were mistakes. — Aryabhata, Jan 05 '11 at 21:04
There are no mistakes.Thank you for your help.Next time I' ll try to do it myself — t.k, Jan 07 '11 at 09:28

score 5 · Answer 1 · answered Jan 05 '11 at 21:13

5

Take $R=\mathbb{Z}$, $M=\mathbb{Z} \oplus \mathbb{Z}= M_1 \oplus M_2$ and $N=\mathbb{Z}\cdot (1,1)$.

$N\cap M_1 =N\cap M_2 = 0$ so it can't be a direct sum of submodules of $M_1, M_2$

answered Jan 05 '11 at 21:13

Ofir

8,015

score 5 · Answer 2 · answered Jan 05 '11 at 21:14

5

If I understand the question correctly then $$ \mathbb C\cdot(1,1)\subset \mathbb C\oplus\mathbb C $$ is a counterexample (here if have chosen $R=\mathbb C$).

answered Jan 05 '11 at 21:14

Rasmus

18,404

score 3 · Accepted Answer · answered Jan 05 '11 at 21:14

3

For a general module M over a PID R, its submodules need not even be isomorphic to direct sums of submodules of its proper direct summands. For instance, the Z-module Q¹⁰ is a direct sum of proper submodules, but it has directly indecomposable submodules of rank 10 (so they cannot be written as any non-trivial direct sum). Since the torsion-free rank of a submodule can only decrease, this means M = Q⊕Q⁹ is a counterexample.

If we try to do what the question asks, and get a direct sum decomposition from submodules of the direct summands, then things go wrong even for M=Z/2Z ⊕ Z/2Z and the submodule N={(x,x):x in Z/2Z }. N is directly indecomposable, and is not a submodule of the specified direct summands of M (though it is itself a summand in a different decomposition).

answered Jan 05 '11 at 21:14

Jack Schmidt

55,589

Is $N \simeq \mathbb{Z}/2\mathbb{Z}$, via $(x,x) \mapsto x$ – Juan S Jun 03 '11 at 23:33
@Qwirk: yes, exactly. – Jack Schmidt Jun 03 '11 at 23:35
and so if $M = M_1 \oplus M_2 = \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$, then $N \cap M_1 = N \cap M_2 = \mathbb{Z}/2\mathbb{Z}$? (I am trying to reconcile all the answer's here - specifically Prometheus' comment that $N \cap M_1 = 0)$ – Juan S Jun 03 '11 at 23:43
@Qwirk, the problem is ≅-isomorphism is not the same as set equality. M1 = { (0,0), (1,0) }, M2 = { (0,0), (0,1) } and N = { (0,0), (1,1) }, so M1∩N = M2∩N = { (0,0) }. Rasmus's and Prometheus's examples are the same as this one, using C or Z instead of my Z/2Z. – Jack Schmidt Jun 03 '11 at 23:47
ahh, got it, thank you! – Juan S Jun 04 '11 at 00:51

Direct sum of modules

3 Answers3