# Left Multiplication Ring Homomorphism

Assume we have a non-commutative unital ring $R$ and an element $r$ not in the center. Define a map $$\phi_r:R\rightarrow R$$ $$x\mapsto rx$$ Can this ever be a ring homomorphism?

If it can be then $r$ has to be idempotent, and the kernel (the set of $r$'s zero divisors with $0$) must be an ideal. I'm pretty sure that this can't be a homomorphism, but I can't prove it. Although I can't promise that I just haven't been clever enough to think of an example where this works. Some simple proof showing that $r$ has to be in the center is what I've striven for, but alas...

Edit: Suppose I should add that I'm adopting the definition of a ring homomorphism that isn't necessarily (multiplicative) identity invariant.

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You mean a ring without unity, right? –  Stefan Walter Jul 21 '13 at 7:14
@StefanWalter It has unity, but it doesn't have to be invariant wrt the homomorphism. Unless this is problematic? –  AsinglePANCAKE Jul 21 '13 at 7:16

Let $R$ be the ring of $2\times 2$ upper triangular matrices with coefficients in some field, and let $r=\begin{pmatrix}0&0\\0&1\end{pmatrix}$.

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It is certainly possible to concoct an artificial example where $x\mapsto rx$ is a homomorphism but $r$ does not commute with everything.

The elements shall be non-commutative polynomials in two variables, say $x$ and $y$, such that no term of the polynomial contains more than one occurrence of $x$. Addition is ordinary polynomial addition. For multiplication, multiply the polynomials as usual and then delete from each term the second occurrence of $x$, if there is one.

For example:

$\begin{array}{rcl}(2y^2xy + x + 2y)(xy+yx)&=&2y^2xyxy + x^2y + 2yxy + 2y^2xy^2x + xyx + 2y^2x \\&=& 2y^2xy^2 + xy + 2yxy + 2y^2xy^2 + xy + 2y^2x \\&=& 4y^2xy^2 + 2xy + 2yxy + 2y^2x\end{array}$

In other words, this is the free ring on $x$ and $y$ quotiented by the relations $xab=xaxb$ for all $a$ and $b$.

Clearly $x$ does not commute with $y$, so is not in the center, and yet $xab=(xa)(xb)$ for all $a$ and $b$.

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Strictly, you need to show that your universal example does not collapse. One way to do that is to map it into a concrete example like mine :-) –  Mariano Suárez-Alvarez Jul 21 '13 at 20:33
I agree, but I think the concrete description I gave (i.e. “non-commutative polynomials in two variables, say x and y, such that no term of the polynomial contains more than one occurrence of x”) suffices to show it does not collapse. I like your concrete example! –  Robin Houston Jul 21 '13 at 20:35
(Suppose you want an example of a ring in which $1$ is a commutator and there is a non-zero element whose square is zero: you could try to consider the algebra generated by $x$ and $y$ subject to $[x,y]=1$ and $x^2=0$. But that algebra is trivial.) –  Mariano Suárez-Alvarez Jul 21 '13 at 20:35
Well, I did not say that your description does not suffice (how could it not given that it is is a presentation of the algebra!) but only that this is in fact required. –  Mariano Suárez-Alvarez Jul 21 '13 at 20:39
A example in the same same vein , but finitely presented, is $k\langle x,y:xy=0,x^2=x\rangle$. –  Mariano Suárez-Alvarez Jul 21 '13 at 21:16

Let's assume that $x\mapsto ex$ is a ring homomorphism. Then of course the image $eR$ is a ring with identity $e$, and so (as you observed) it will be idempotent and central in the ring $eR$. Those are certainly necessary conditions, so it's natural to ask if they're sufficient also.

Suppose that $e^2=e$ and $e$ commutes with everything in $eR$. Then $x\mapsto ex$ is obviously an additive homomorphism from $R$ to $eR$, and additionally $(ex)(ey)=(exe)y=(eex)y=exy$, so it's a multiplicative homomorphism as well. Thus $eR$ is a ring with identity $e$, and the multiplication map is a ring homomorphism.

The map $x\mapsto ex$ from $R\to R$ using the element $e\in R$ is a ring homomorphism iff $e$ is an idempotent of $R$ commuting with each element of $eR$.
In particular, any idempotent that's in the center of $R$ will create the ring $eRe=eR$ inside of $R$. Mariano's example shows us a noncentral idempotent satisfying the condition highlighted above. If you look at the same idempotent in the full ring of $2\times 2$ matrices, you'll find it fails to commute with $eR$. (Just check its product with $\begin{bmatrix}0&0\\1&1\end{bmatrix}$ on both sides, for example.)