# Direct sum of modules

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 5 '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 7 '11 at 9:28

## 3 Answers

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).

-

Is $N \simeq \mathbb{Z}/2\mathbb{Z}$, via $(x,x) \mapsto x$ – Juan S Jun 3 '11 at 23:33
@Qwirk: yes, exactly. – Jack Schmidt Jun 3 '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 3 '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 3 '11 at 23:47
ahh, got it, thank you! – Juan S Jun 4 '11 at 0:51

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 $M_1, M_2$

-

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 $R=\mathbb C$).

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 24, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8696092963218689, "perplexity_flag": "head"}