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Let $Z(R) = \{ a \in R : ax = xa,\text{ for all $x \in R$}\}$

Is $Z(R)$ an ideal of $R$?

Attempt: I already proved that $Z(R)$ is a subring of $R$. I would say yes, since if $x \in R$, then $xa$ is an element in $Z(R)$ and if $a\in Z(R)$ then we have $ax\in Z(G)$. So by definition it is an ideal for $R$.

Please can anyone please give some feedback? Thank you.

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    $\begingroup$ Have you tried to do this in one example? $\endgroup$ – Mariano Suárez-Álvarez May 13 '14 at 5:28
  • $\begingroup$ No. I was thinking in using my proof that Z(R) is a subring of R. $\endgroup$ – user3425 May 13 '14 at 5:34
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    $\begingroup$ Well... do an example then! $\endgroup$ – Mariano Suárez-Álvarez May 13 '14 at 5:43
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    $\begingroup$ "Is $Z(R)$ and ideal of $R$? I would say yes, since if $x\in R$ then $xa$ is in $Z(R)$..." That is not how a proof goes: that is simply asserting that what you desire to prove is true. $\endgroup$ – rschwieb May 13 '14 at 19:04
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To show $I = Z(R)$ is an ideal, you need to show two things: (1) $I$ is an additive subgroup of $R$. (2) $I$ absorbs $R$ on both sides. You showed (2) already. The quickest way to (1) is showing that $I$ is closed under the binary map $f(x,y) = x - y$. Then from that you can prove that you indeed have an additive group.

Let $x, y \in I$. Then for any $a \in R$, $a(x-y) = ax - ay = xa - ya = (x - y) a$ from the ring axioms, so $I$ is closed under subtraction. QED

$a \in R , x \in Z(R) \implies (ax)b = a(xb) = abx \neq b(ax)$ necc. So you were wrong about the first part. It is an additive subgroup though!

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  • $\begingroup$ Thanks for this great post. In the third paragraph, b is an arbitrary element of R? Right?. $\endgroup$ – Geoffrey Critzer Jun 24 '15 at 10:26
  • $\begingroup$ @GeoffreyCritzer yes $\endgroup$ – Shine On You Crazy Diamond Oct 28 '18 at 20:20
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Hint: If $I$ is an ideal in a ring $R$ and $1\in I$, then $I=R$. Moreover, one may show that $1\in Z(R)$. Hence $Z(R)$ is an ideal if and only if $Z(R)=R$. Can you use this to find a counterexample?

The hint above tells us that $Z(R)$ is an ideal of $R$ if and only if $R$ is a commutative ring. So letting $R$ be your favorite noncommutative ring (say the ring of $n\times n$ matrices over $\Bbb C$) gives a counterexample.

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  • $\begingroup$ I agree with the first two sentences. How does this imply the third sentence? Also, I cannot "use this to find a counterexample" ( I assume you mean to find a ring R so that the Z(R) is NOT and ideal of R). Could you please give such a counter example? $\endgroup$ – Geoffrey Critzer Jun 24 '15 at 10:44
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    $\begingroup$ If $I$ is an ideal in a ring $R$, then $1\in I$ if and only if $I=R$. But it is always true that $1\in Z(R)$. It follows that if $Z(R)$ is an ideal, then $Z(R)=R$ since $1\in Z(R)$. Of course, if $Z(R)=R$, then $Z(R)$ is an ideal of $R$. Hence $Z(R)$ is an ideal of $R$ if and only if $Z(R)=R$. $\endgroup$ – Brian Fitzpatrick Jun 24 '15 at 18:55
  • $\begingroup$ But doesn't this assume that R has a multiplicative identity,1? What if R doesn't have such an element?. $\endgroup$ – Geoffrey Critzer Jun 24 '15 at 20:25
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    $\begingroup$ @GeoffreyCritzer My rings always have identities but this is a source of controversy. $\endgroup$ – Brian Fitzpatrick Jun 24 '15 at 21:20

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