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Let $R$ be a ring and $T$ be the set of torsion elements of $R$.

  1. Prove that the only torsion element of $R/T$ is $\{0\}$. I can show that $T$ is an ideal of $R$. Or in fact, $T$ is a normal subgroup of $R$ if we treat $R$ simply as an additive group. If I take this latter view, then I guess I can use the first isomorphism theorem to prove this statement. Define $\psi: R\rightarrow\{0, \infty\}$ by sending element of $R$ to 0 if its order is finite and $\infty$ otherwise. This is an onto homomorphism if we assume that $0+\infty=\infty$. Therefore, $R/T\cong\{0,\infty\}$ since the kernel of $\psi$ is $T$. Then I can conclude that $R/T$ is torsion free since $0$ is the only element of $\{0, \infty\}$ which is of finite order. Is this argument right, please? Thank you!
  2. Give an example of a ring $R$ for which $T\neq \{0\}$ and $T\neq R$. I tried $2\times 2$ matrices with entries in $\mathbb{Z}/4\mathbb{Z}$. It seems to work. I am wondering whether there is a general rule for this, please? Thank you!

BTW, a torsion element is defined as an element with finite order.

LaTeXFan
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1 Answers1

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For the proof, check here: Prove that $M/Tor(M) $ is torsion-free.

For an example of a ring with some elements but not all , torsion, take $\mathbb Z\oplus \mathbb Z_n$ , where the only torsion elements are $\{ (0,z_n): z_n \in \mathbb Z_n\}$

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user99680
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  • Thank you for your help. Just wondering what $\mathbb{Z}(+)\mathbb{Z}_n$ is please. – LaTeXFan Nov 17 '13 at 09:02
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    @XuS : No problem; it is the direct sum of $\mathbb Z$ with $\mathbb Z_n$ , i.e., a ring with elements $(a,b)$ ; $a$ in $\mathbb Z$, $b$ in $\mathbb Z_n$, and sum is done component-wise, i.e., $(a,b)+(a',b')=(a+a', (b+b')mod(n))$ , i.e., the sum in the 1st component is the usual sum of integers, and the sum in the second case is sum mod(n). Consider then, say, $(0,1)$, where $1$ is the class of $1(mod(n))$. Then $n(0,1)=(0,1)+(0,1)+..(0,1)=(0,n)=(0,0)$. – user99680 Nov 17 '13 at 09:08