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Show that if $A,B$ are modules over a pid $R$ and if $a\in A$, $b\in B$ are not torsion elements then $a\otimes b\neq 0$ in $A\otimes_R B$ and is not a torsion element.

My attempt: (i) Since $a,b$ are torsion elements, the map $r\mapsto ra,r\mapsto rb$ are isomorphisms and $Ra\cong R,Rb\cong R$. Let $u:Ra\otimes_R Rb \rightarrow R\otimes_R R$, such that $u(ra\otimes r'b)=r\otimes r'$. If $u(ra\otimes r'b)=r\otimes r'=0$, then $rr'=0$ and $r=0$ or $r'=0$ since $R$ is a domain. Thus $u$ is an isomorphism and we have $Ra\otimes_R Rb \cong R\otimes_R R\cong R$. If $a\otimes b=0$, then $Ra\otimes_R Rb=0=R$ which is a contradiction.

(ii) We know that for $r\neq 0$, $ra$ is clearly not a torsion element since $R$ is a domain. Using (i), we have $ra\otimes b\neq 0$. So if $r(a\otimes b)=ra\otimes b=0$, we must get $r=0$, which means $a\otimes b$ is not torsion.

However, I don't use the fact that $R$ is a principal ring. Does something go wrong in my proof?

Bernard
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GTM 73
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  • You said that if $a\otimes b=0$, then $Ra\otimes_R Rb=0$. This is the part which is not clear. $a\otimes b$ is an element of $A\otimes_R B$ by definition, but $Ra\otimes_R Rb$ is not a submodule of $A\otimes_R B$ in general (in other words, the map $Ra\otimes_R Rb\to A\otimes_R B$ is not injective without further assumptions). So we might as very well have $a\otimes b\neq 0$ as an element of $Ra\otimes_R Rb$ but $a\otimes b=0$ as an element of $A\otimes_R B$. – Roland May 14 '21 at 15:51
  • @Roland: My another attempt: I find that the case $A,B$ are finitely generated has been proved: https://math.stackexchange.com/questions/267819/when-are-elements-in-a-tensor-product-equal-to-0 For the general case, assuming $a\otimes b=0$ in $A\otimes B$, we may find finitely generated submodules $A_0\subseteq A, B_0\subseteq B$ such that $a\otimes b=0$ in $A_0\otimes B_0$. So either $a$ or $b$ will be torsion, a contradication. – GTM 73 May 16 '21 at 04:12
  • Yes this is correct, except you didn't justify why we may find finitely generated submodules $A_0,B_0$. Maybe you already have proven this, I don't know, but this is indeed the way to go. – Roland May 17 '21 at 16:18
  • @Roland: This is exercise ⅲ.7.4 in Hilton&Stammbach's book or can be found in chapter 2 (Corollary 2.13) of Atiyah's commutative algebra book. So I just cite this result. – GTM 73 May 18 '21 at 15:27

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