2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Qiaochu Yuan Oct 13 '15 at 17:10
  • $\begingroup$ Ah quite right. I didn't notice the obvious connection between our answers. I had already begun writing when you answered :/ $\endgroup$ – Cameron Williams Oct 13 '15 at 17:12
3
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – Vincent Oct 13 '15 at 19:20
  • 1
    $\begingroup$ @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. $\endgroup$ – Qiaochu Yuan Oct 13 '15 at 22:23
  • $\begingroup$ Very interesting, thank you! $\endgroup$ – Vincent Oct 14 '15 at 5:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.