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Where $R$ is a principal ideal and $x$ is a basiselement in the $R$ module $M=R^d$ where $d\in\mathbb{N}$ and $\alpha$ is a non-unit and not zero and a element in $R$. With $Rx$ defining the subideal in $M$ for which we have $Rx=\{a\in M ; a=rx , r\in R\}$ and similarily $R\alpha x=\{a\in M ; a=r\alpha x , r\in R\}$ and $\alpha R=\{b\in R:b=r\alpha\}$ an ideal in $R$. I am looking for an explicit isomorphism.

Note: A basis of a $R$ module $M$ is a family of elements $(a_i)$ that constructs the ring, i.e $\sum Ra_i=M$ such that every such sum has a unique representation. One can prove that if a module has a finite basis then every other basis must have the same cardinality.

For context I refer to this question:

The torisionmodule of a $R$ module (where $R$ is PID) is isomorphic to a direct sum of ideals $\bigoplus_{j=1}^sR/\alpha_j R$

New2Math
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2 Answers2

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No.

For example let $R = \mathbb Z^2$ and $x=(0,0)$ and $\alpha = (0,1)$.

Then $Rx = R \alpha x = Rx/ R \alpha x = \{0\}$.

but $\alpha R = \{0\} \times \mathbb Z$ and so $R/\alpha R = \mathbb Z \times \{0\}$.

Edit: If you want $x$ to be a basis element of $R$ then take $R = \mathbb Z^3$ and $x=(1,0,0)$ and $\alpha = (0,0,1)$. Then we have $\alpha x=(0,0,0)$ and $$Rx = \mathbb Z \times \{0\} \times \{0\} \qquad R\alpha = \{0\} \times \{0\}\times \mathbb Z$$ $$R\alpha x = \{0\} \times \{0\}\times \{0\} $$and $$\frac{Rx}{R \alpha x} =Rx = \mathbb Z \times \{0\} \times \{0\} \qquad \frac{R }{R x} = \{0\}\times \mathbb Z \times \mathbb Z$$

Daron
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  • but what if we know that $x$ is an element of a Basis of the $R$ module? – New2Math Oct 02 '19 at 10:39
  • Take $R= \mathbb Z^3$ and $x=(1,0,0)$ and $\alpha =(0,1,1)$. – Daron Oct 03 '19 at 19:05
  • $Rx=\mathbb{Z}\times{0}\times{0}$, $R\alpha x={0}$. $Rx/R\alpha x=\mathbb{Z}\times{0}\times{0}$. $\alpha R={0}\times\mathbb{Z}\times\mathbb{Z}$. But then $\mathbb{Z}^3 / \alpha R=\mathbb{Z}\times{0}\times{0} $ If this was a counterexample it does not work because the sets are isomorphic. My question was about that it always works in this situation. It would be odd if it doesn't work because in the book I was reading it was assumed without further comment – New2Math Oct 05 '19 at 17:57
  • See the edit. I think one of us doesn't understand the question or what the book is assuming. For example where does the module over $R$ come into play? – Daron Oct 05 '19 at 18:44
  • Daron I have looked at the statement again. Before I make my comment I want to share our definition of basis with you: A basis of a $R$ module $M$ is a family of elements $(a_i)$ that constructs the ring, i.e $\sum Ra_i=M$ such that every such sum has a unique representation. One can prove that if a module has a finite basis then every other basis must have the same cardinality. In this case $(1,1,1)$ is a basis element of the $\mathbb{Z}^3$ module $\mathbb{Z}^3$, therefore every other basis must have the samecardinality and hence your claim that $x=(1,0,0)$ is a basiselement is wrong (cont) – New2Math Oct 10 '19 at 18:56
  • (cont) because $(1,0,0)$ does not construct $\mathbb{Z}^3$. I want to add that I am interested in the case where $x$ is a basiselement of $R^d$, your example is covered in this situation $d=1$. I have made a seperate question which is the context of my question: https://math.stackexchange.com/questions/3388330/the-torisionmodule-of-a-r-module-where-r-is-pid-is-isomorphic-to-a-direct – New2Math Oct 10 '19 at 19:01
  • Nope. The family with single element $(1,1,1)$ constructs only ${(1,1,1),(2,2,2),(3,3,3),\ldots } \ne \mathbb Z^3$. For example it does not construct $(1,2,3)$. – Daron Oct 11 '19 at 10:53
  • No it constructs $(1,2,3)$ too because $(1,2,3)\cdot (1,1,1)= (1,2,3)$. When I say $(1,1,1)$ is a basis I mean that $\mathbb{Z}^3(1,1,1)=\mathbb{Z}^3$ and every representation is unique – New2Math Oct 11 '19 at 13:08
  • Okay that is a rather strange definition of a basis for $\mathbb Z^3$. Usually we mean pretty much the same thing as a basis for $\mathbb R^3$ but with $\mathbb R $ replaced by $\mathbb Z$. I guess you're talking about a basis for $\mathbb Z^3$ when it is treated as a module over itself. – Daron Oct 11 '19 at 22:38
  • But in your notation $R=\mathbb Z$ and $M=\mathbb Z^3$. Thus $A \subset M$ being a basis means $R A = M$ which is $ \mathbb Z A = \mathbb Z^3$ rather than $ \mathbb Z^3 A = \mathbb Z^3$. – Daron Oct 11 '19 at 22:42
  • Yes you are right if you choose $R=\mathbb{Z}$ then $A$ being a basis of $M$ means $RA=M$, or in particular $\mathbb{Z}A=\mathbb{Z}^3$. But in the counterexample you proposed you said $R=\mathbb{Z}^3$. If you had choosen $R=\mathbb{Z}$ your counterexample wouldn't work, i.e. the residueclasses constructed as proposed are indeed isomorphic – New2Math Oct 12 '19 at 14:47
  • This is wrong: if $R=\mathbb{Z}^3$, then $(1,0,0)$ cannot be an element of a basis of $R=R^1$. – egreg Oct 13 '19 at 10:04
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Since $x$ is a member of a basis of a free module, you know that the map $\mu_x\colon R\to Rx$, $\mu_x(r)=rx$ is an isomorphism (only that $x$ has no torsion, that is, zero annihilator, would be sufficient).

Consider the canonical map $\pi\colon Rx\to Rx/R\alpha x$; then the kernel of $\pi\circ\mu_x$ is $$ \ker(\pi\circ\mu_x)=\{r\in R:rx\in R\alpha x\} $$ Clearly $R\alpha$ is contained in $\ker(\pi\circ\mu_x)$. Suppose $rx\in R\alpha x$, that is, $rx=s\alpha x$, for some $s\in R$.

Since $x$ has no torsion, this implies $r=s\alpha$, so $r\in R\alpha$.

egreg
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