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Given a rng $(R,+,\cdot)$ and $e \in R$ with $\forall a \in R: e \cdot a = a$, I would like to prove $a \cdot e = a$. What I have so far is the following:

Assume there exists $r \in R$ with $\forall a \in R: a \cdot r = a$. Plug in $e$ for $a$ to get $e \cdot r = e$. But by the given conditions, $e \cdot r = r$. Thus, $r=e$ and consequently $e \cdot r = a$.

Now the problem I have with this is that I have to assume there exists a right identity element. Any tips on how I could solve this problem? Is the thing I'm trying to prove even true? I made this exercise for myself, but my feeling tells me it's true.

Vincent
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2 Answers2

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Your argument shows that if both a left and a right identity exists, they must be equal (and hence must be an identity, and hence must be unique). But it's not true that the existence of a left identity implies the existence of a right identity. For example, let $S$ be the semigroup consisting of two elements $\{ a, b\}$ with multiplication table

$$a^2 = ba = a, b^2 = ab = b.$$

That is, every word in the generators evaluates to its rightmost letter. Then $a$ and $b$ are both left identities, but neither of them are right identities. The "semigroup rng" $\mathbb{Z}[S]$ similarly has two left identities, and hence it can't have any right identities (because that contradicts uniqueness).

Qiaochu Yuan
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  • Thank you very much for your answer. I'm afraid I do not completely understand it, since I don't know what the notation $\mathbb{Z}[S]$ means. But it's my first semester studying mathematics, and it started yesterday, so that's probably why! – Vincent Oct 13 '15 at 19:20
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    @Vincent: $S$ is a semigroup, namely a set equipped with an associative binary operation $S \times S \to S$ (described above), and $\mathbb{Z}[S]$ is the free abelian group on $S$ with multiplication given by extending the multiplication on $S$ by linearity (which is a rng). It works out to the same thing as in Cameron's answer, but with $\mathbb{Z}$ coefficients and transposed. – Qiaochu Yuan Oct 13 '15 at 22:23
  • Very interesting, thank you! – Vincent Oct 14 '15 at 05:18
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This is not true in general. Take the following example

$$R = \left\{\left(\begin{array}{cc} x & x \\ y & y\end{array}\right):a,b\in\mathbb{R}\right\}$$

with usual matrix addition and multiplication. This is a ring. (I'll leave you to show this.) However the right identities are those elements such that $x+y=1$ but these are not left identities. To see this, consider the matrix $\left(\begin{array}{cc} 0 & 0 \\ 1 & 1\end{array}\right)$.

To address the spirit of the question: consider $(a\cdot e)\cdot e$. Then by associativity, this is the same as $a\cdot(e\cdot e) = a\cdot e$. So for elements of the form $a\cdot e$, $e$ acts as a right identity. However the sticking point is that the image of $R$ under right multiplication by $e$ may not be all of $R$. If it were, then indeed $e$ would be a right identity.

Cameron L. Williams
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    Note that this is $\mathbb{R}[S^{op}]$ where $S$ is the semigroup in my answer; the two obvious generators have the property that every word in them evaluates to its leftmost letter. This is opposite from what the OP wanted though. – Qiaochu Yuan Oct 13 '15 at 17:10
  • Ah quite right. I didn't notice the obvious connection between our answers. I had already begun writing when you answered :/ – Cameron L. Williams Oct 13 '15 at 17:12