I know that if a ring has a multiplicative identity, then the multiplicative identity must be unique. Are there simple-to-describe examples of rings with two (or more) multiplicative right-identities?
2 Answers
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Take the semigroup ring of a semigroup with two or more multiplicative right identities. For example, the semigroup $$S = \langle a, b | ab = aa = a, ba = bb = b \rangle$$ works (it is the universal example, so if it fails then no example can work). Multiplication in this semigroup can be described as follows: every word evaluates to the first letter in it. The resulting semigroup ring is literally the free ring on two not necessarily identical right identities.
answered May 1, 2011 at 14:32
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2$\begingroup$ Not that it matters, but doesn't your example have two left identities, not two right identities? $\endgroup$ Commented May 1, 2011 at 15:38
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$\begingroup$ @RichardStanley, could it be a difference of terminology (like which ones are right cosets, where one has to decide whether the representative or the group action is on the right)? I'd call a right identity of $R$ an element $i$ that satisfies $x i = x$ for all $x \in R$, in which sense it seems that $a$ and $b$ are both right identities here. $\endgroup$– LSpiceCommented Apr 25, 2020 at 1:37
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I was browsing some "What is...?" threads, and bumped in here: Let $n$ be a positive integer, and consider the subring of the opposite of the ring of $n$-by-$n$ matrices over a commutative unital ring $\mathbb A = (A, +, \cdot)$ consisting of those matrices all of whose rows, except at most for the first, are zero; this has $|A|^{n-1}$ right identities, given by those matrices whose first row is any vector of $A^n$ with first element equal to $1_\mathbb{A}$.
answered Dec 23, 2015 at 22:13