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Let $R$ be a ring and let $a \in R$ satisfy $a^2 = a$. Show that there is a subset $S$ of $R$, closed under addition and multiplication, such that $S$ is a ring with the induced operations and $a$ is the multiplicative identity in $S$.

Attempt:

The set $S = \{xa | x \in R\}$ seems to satisfy the above requirements in case of $R$ being a commutative ring, but I am not sure how to approach it if $R$ is a non commutative ring.

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    $\begingroup$ Try $S=\{axa: x\in R\}$. $\endgroup$ – A. Salguero-Alarcón Apr 13 '17 at 18:30
  • $\begingroup$ @DietrichBurde this question is different here the assumption is for one particular $a$ we have $a^2 = a$, unlike in the question in your comment ... where this property is assumed for all $a \in R$ $\endgroup$ – spaceman_spiff Apr 13 '17 at 18:58
  • $\begingroup$ Ah, I see. I thought, that $a$ was arbitrary, i.e., that any $a$ satisfies $a^2=a$. $\endgroup$ – Dietrich Burde Apr 13 '17 at 19:01
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I figured I would write another proof using the corner ring.

Let $S=aRa=\{ara\mid r\in R\},$ then $S$ is a subring of $R$ with unity $a\in S.$

Proof: First, it is clear that for any of the binary operations on $S$ inherited from $R$ will certainly be associative, and in the case that they are commutative in $R$ then they will be in $S$ as well. The distributive property also is inherited.

Since $a\in R,$ and $a^2=a,$ then the element $aaa=a\in S,$ furthermore, $0\in S$ since $a0a=0.$

Now, consider any $axa, aya\in S,$ then $axa+aya=a(xa+ya)=a(x+y)a,$ so $S$ is closed under addition. Now, consider $axaaya=axaya,$ if we write $r=xay,$ then it is clear that $axaya=ara\in S,$ and thus $S$ is closed under addition and multiplication.

We need to show that $a\in S$ is a suitable unity. Since $a^2=a,$ we're off to a good start, we just need to show that $ax=x,$ for any $x\in S.$ We saw already that any sum or product of elements of $S$ can be written as follows $x=ara$ for some $r\in R,$ so $ax=aara=ara=x,$ and a similar argument works for $xa.$ Thus $ax=x=xa$ for every $x\in S,$ so $a$ is a suitable unit, thus the corner ring is a subring of $R.\Box$

Since $a^2=a,$ then $a$ is a zero divisor. I had worried that this would mess up the corner ring, but it works out that $a(1-a)a=(a-a^2)a=0a=0.$ So $1-a$ gets "sent to zero" in $S$. Oddly enough, my approach yielded a subring of $S,$ since $a\in S.$

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Consider the subset $S=\{na\mid n\in \mathbb{Z}\},$ where $na=a+\cdots+a$ $n-$times. Certainly, this subset is closed under addition: $ka+ma=\underset{\text{$k+m$ times}}{a+\cdots+a}=(k+m)a\in S,$ for any $k,m\in\mathbb{Z}.$ Furthermore, if you consider $ka\cdot ma=ka\cdot(a+\cdots+a),$ the distributive property assures us that this is equal to $k (a^2+\cdots+a^2)=k(a+\cdots +a)=k(ma)=(km)a,$ and thus is closed under multiplication. Now try to justify that $a$ is the unit of this subset.

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  • $\begingroup$ i did not get what you mean by $S = \{a | n \in Z\}$ $\endgroup$ – spaceman_spiff Apr 13 '17 at 18:46
  • $\begingroup$ I fixed the typo, please see again. $\endgroup$ – Chickenmancer Apr 13 '17 at 18:53
  • $\begingroup$ seems fine to me but why the down vote ?? $\endgroup$ – spaceman_spiff Apr 13 '17 at 18:56
  • $\begingroup$ There isn't anything wrong with this idea. $xRx$ comes to mind first, but this one works just as well. Maybe it was the typo. $\endgroup$ – rschwieb Apr 13 '17 at 19:32
  • $\begingroup$ @rschwieb, what do you mean by xRx? I'm not familiar with this notation. $\endgroup$ – Chickenmancer Apr 14 '17 at 13:42

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