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We know that in vector spaces such a complementary substructure exists. But in this proof we took advantage of the fact that every vector space has a basis, ie. the subvectorspace has a basis and this basis then can be extended to a full basis of the whole vector space and then the basis vectors which are only in the extension create a basis for the desired complement.

In an $R$-module $M$, where $R$ is a principal ideal domain, we cannot use this argument with a basis, is it still possible to find a submodule $F$ such that every element of $M$ can be expressed as $x=f+t$ with $f\in F$ and $t\in T$ and that every such description is unique? If it is not true for the general case, how can it be true if $T$ is the torsion submodule of $M$?

Bernard
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New2Math
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  • The terminology for this is "semisimple" https://en.wikipedia.org/wiki/Semisimple_module – Trevor Gunn Oct 02 '19 at 13:23
  • @freakish There exists a finite System $(a_i)_n$ of Elements ins $M$ such that $\sum_i Ra_i=M$. The torision submodule $T$ of a module $M$ is the set ${a\in M:\exists\alpha\in R-{0}:\alpha a=0}$. It is indeed a submodule if you know that $R$ is a principal ideal domain – New2Math Oct 02 '19 at 14:26
  • @freakish how can you prove that $T$ is a direct summand? – New2Math Oct 02 '19 at 14:28

2 Answers2

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It works for the torsion module primarily because a finitely generated module $M$ over a P.I.D. $R$, which is a noetherian ring, has a finite presentation, i.e. there exists an exact sequence $$F_1\longrightarrow F_0\longrightarrow M\longrightarrow 0,$$ where $F_0$ and $F_1$ are finitely generated free modules. The map $F_1\longrightarrow F_0$ is represented by a matrix, and in suitable bases for $F_0$ and $F_1$, this matrix has a Smith normal form: $$\begin{pmatrix} d_1&0&0&\dots&0&\dots&0\\ 0&d_2&0&\dots&0&\dots&0\ \\ 0&0&d_3&\dots&0 &\dots&0\\ \vdots&&&\ddots&&&\vdots \\ 0&0&0&&d_r&\dots&0\\ 0&0&0&\dots&0&\dots&0 \\ \vdots&&&&&\ddots&\vdots\\ 0&0&0&\dots&0&\dots&0 \end{pmatrix}$$ where $\;d_i\mid d_{i+1}$ for each $1\le i<r$. There results that the module $M$ is isomorphic to $$\underbrace{R/d_1R\times R/d_2R\times\dots\times R/d_rR}_{\text{torsion submodule}}\times R^{\text{rk}(F_0)-r}.$$

Bernard
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  • Why can you say that the left part of $\underbrace{R/d_1R\times R/d_2R\times\dots\times R/d_rR}_{\text{torsion submodule}}\times R^{\text{rk}(F_0)-r}.$ corresponds to the torision submodule? – New2Math Oct 02 '19 at 14:52
  • The other factor is free, and the quotient is isomorphic to $M$. – Bernard Oct 02 '19 at 14:53
  • With the Quotient you mean $R/d_1R\times R/d_2R\times\dots\times R/d_rR\times R^{\text{rk}(F_0)-r}.$?. I understand that this set is isomorphic to $M$. I also see that if M is free,the torision submodule is $0$. But I don't understand how this implies the existence of a $F$ such that $M=F\oplus T$. I think I also could prove that $R/d_1R\times R/d_2R\times\dots\times R/d_rR$'s torision submodule is it itself. To be more precise why if $\psi:M\rightarrow A\oplus B$ isomorph such that $B$ is torioson module of $A\oplus B$, why $\phi^{-1}(B)=T$ and $\phi^{-1}(A)=T$ as desired? – New2Math Oct 02 '19 at 15:10
  • What you denote $F$ (the free part) is $R^{\text{rk}F_0-r}= and $T$ (the torsion part) is the contents of the underbrace. As to your last question, it is easy to check that a homomorphism maps torsion elements to torsion elements (including $0$ in $T$), and to deduce from this fact that an isomorphim maps the torsion submodule of its domain onto the torsion submodule of its codomain. – Bernard Oct 02 '19 at 15:32
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I presume that $R$ is always a PID here.

is it still possible to find a submodule $F$ such that every element of $M$ can be expressed as $x=f+t$ with $f\in F$ and $t\in T$ and that every such description is unique?

No. Consider $R=M=\mathbb{Z}$ and $T=2\mathbb{Z}$ and note that $\mathbb{Z}$ is not decomposable as a group, let alone as a module (well, over $\mathbb{Z}$ it is the same thing).

In fact rings with this property (i.e. every submodule is a direct summand) are precisely semisimple rings and these are all well known by the Artin-Wedderburn theorem. As a consequence an integral domain is semisimple if and only if it is a field. And so we are back at vector spaces.

If it is not true for the general case, how can it be true if $T$ is the torsion submodule of $M$?

It is true. Since $M$ is finitely generated then so is $M/T$. Also $M/T$ is torsion free, which I leave as an exercise. Finally finitely generated torsion free modules over PID are free, in particular projective. Therefore

$$0\to T\to M\to M/T\to 0$$

splits which shows that $T$ is a direct summand of $M$.

freakish
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