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Let $R$ be a commutative Ring with identity and $x\in R$ with $x=x^{2}$. Show that $\langle x \rangle$ is a Ring.

For me there are two possibilities coming to my mind, namely that x could be 1 or 0, however I think that this question demands an unambigous answer.

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  • $\begingroup$ No, $x$ does not need to be $1$ or $0$. There are many rings where $x^2 = x$ for all ring elements. $\endgroup$ – T. Bongers Mar 19 '18 at 15:59
  • $\begingroup$ Am ideal is a ring so you are done. $\endgroup$ – Sorfosh Mar 19 '18 at 16:01
  • $\begingroup$ @Sorfosh, An ideal need not be a ring. In particular, the only ideal of $R$ that is a ring is $R$ itself -- an ideal need not contain the multiplicative identity. $\endgroup$ – Joe Mar 19 '18 at 16:03
  • $\begingroup$ @joe by definition of a a two sided ideal, it is a subring. Since it passes the subring test from definition $\endgroup$ – Sorfosh Mar 19 '18 at 16:08
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    $\begingroup$ @Joe It depends on whether or not you require a ring to have an identity. Even with this requirement you can still have proper ideals be rings. For example, the ideal $(2) \subseteq \mathbf Z/6\mathbf{Z}$ has $4$ as its identity. $\endgroup$ – Trevor Gunn Mar 19 '18 at 16:08
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EDIT: The post was edited so that $R$ is a commutative ring.

Note (as I failed to in the comments) that $\langle x \rangle$ being a ring need not make it a subring of $R$. Now as an ideal, $\langle x \rangle$ inherits every ring axiom from $R$ except for having identity. Now I claim that $x$ takes on the role of multiplicative identity in $\langle x \rangle$

Take $a \in \langle x \rangle$. Since $\langle x \rangle$ is an ideal, we can take $b \in R$ such that $a = bx$. Now since $x$ is idempotent and R is commutative we have: $$xa = ax = (bx)x = bx^2 = bx = a$$

Thus $x$ satisfies the definition of a multiplicative identity and $\langle x \rangle$ is a ring with operations inherited from $R$, additive identity $0 \in R$, and multiplicative identity $x$.

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  • $\begingroup$ "Since $\langle x \rangle$ is a two sided ideal, we can take $b, b' \in R$ such that $a = bx = xb'$." This is not what a general element of a two sided ideal looks like. Rather the elements of $RxR$ are finite sums $\sum a_ixb_i$. $\endgroup$ – Trevor Gunn Mar 19 '18 at 16:32
  • $\begingroup$ I was assuming that ideal meant 2-sided ideal. Wouldn't it be true in that case? $\endgroup$ – Joe Mar 19 '18 at 17:30
  • $\begingroup$ @Joe If I give you the element $axb$ of $RxR$, can you write it as $rx$ or $xr'$? Can you combine $axb + a'xb'$ into a single term? This is why general elements of $RxR$ are of the form $\sum a_ixb_i$. $\endgroup$ – Trevor Gunn Mar 19 '18 at 18:21
  • $\begingroup$ @TrevorGunn yes you are right. The OP edited the post so that $R$ is commutative and I have fixed my answer. $\endgroup$ – Joe Mar 19 '18 at 18:57
  • $\begingroup$ @Joe Trevor did a better job of explaining than I did, so I deleted my comment :/ $\endgroup$ – rschwieb Mar 19 '18 at 19:36
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I'm going to use $e$ instead of $x$ because it is more common.

The general situation is that $eRe$ is a ring with identity $e$, which you can easily check. It is not necessarily an ideal, however.

If you don't require that a subring has identity, then this is all trivial since ideals are "subrings" in that case.

In general it is false that $\langle e \rangle$ is a ring with identity. For example, in the ring of linear transformations of a countably infinite dimensional vector space, you can pick an idempotent element given by a transformation which projects onto a single $1$-d subspace. The ideal generated by $e$ is the only nontrivial ideal of the endomorphism ring, and it does not have an identity.

If you were secretly assuming $R$ is commutative (or, more interestingly, when $e$ is assumed to be central), then yes, $eRe=\langle e \rangle = eR$ and it is, in fact, a subring with identity $e$.

There, you have your unambiguous answer. Now you can go ahead and verify axioms.

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