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I'm trying to get my head around modules, and there's a problem that's bothering me regarding scalar multiplication from the left vs from the right.

In many books/articles I've read, the author may refer to a left $R$-module $M$, and then continue to talk of terms such as $xr$, where $x\in M$, $r\in R$. What is right scalar multiplication in a left module? Does it even make sense to talk about it?

I've been trying to show that over a commutative ring, right scalar multiplication and left scalar multiplication are the same thing, but I've not managed it so far. I've googled it to no avail, and I can't see why the distinction between left and right modules disappears when the base ring is commutative.

Thanks for any replies!

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    $\begingroup$ Other than that, $xr$ for $x ∈ M$, $r ∈ R$ makes no sense if there’s no way to interpret $M$ as a right $R$-module. $\endgroup$ – k.stm Apr 28 '14 at 22:08
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    $\begingroup$ One has to be a little careful with the statement "$R = R^{op}$ if and only if $R$ is commutative." What's true is that the "identity" map from $R$ to $R^{op}$ is a ring isomorphism if and only if $R$ is commutative. However, various noncommutative rings admit isomorphisms $R \cong R^{op}$, e.g. the quaternions. Any choice of such an isomorphism (making $R$ a *-ring) allows you to identify left modules with right modules, but the identification depends on the choice. $\endgroup$ – Qiaochu Yuan Apr 28 '14 at 22:47
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    $\begingroup$ Oh ok, so there are noncommutative rings such that there is an isomorphism between $R$ and $R^{op$, it's just that each element wont be mapped to itself? I think I get that. Thanks for pointing it out! $\endgroup$ – James Machin Apr 28 '14 at 23:18
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    $\begingroup$ @JamesMachin Just take Qiaochu's example of the quaternions and look at quaternion conjugation: $a+bi+cj+dk\mapsto a-bi-cj-dk$. The conjugation map $q\mapsto \overline{q}$ satisfies $\overline{ab}=\overline{b}\overline{a}$ as a map from $\Bbb H \to \Bbb H$, but if you view it as a map from $\Bbb H\to \Bbb H^{op}$, then it satisfies $\overline{ab}=\overline{a}\overline{b}$. It also satisfies $\overline{a+b}=\overline{a}+\overline{b}$, and it is then a ring homomorphism. All of the quaternions with zero "imaginary" part are mapped to themselves, but the rest are not mapped to themselves. $\endgroup$ – rschwieb Apr 29 '14 at 13:00
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    $\begingroup$ @JamesMachin (cont.) The upshot is that $\Bbb H\cong \Bbb H^{op}$ as rings. $\endgroup$ – rschwieb Apr 29 '14 at 13:03
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Over a commutative ring there is no difference between a left and right module. In general the difference between a left and right module is that in a left module $$a\cdot(b\cdot m)=(ab)\cdot m$$ while in a right module $$(m\cdot a)\cdot b=m\cdot(ab)$$

So if a ring $R$ is non commutative there is a difference between a left and right action. Given a left action $a\cdot m$ if one tries to define $m\cdot a=a\cdot m$ this does not in general give a right action, because $$(m\cdot a)\cdot b=b\cdot(m\cdot a) = b\cdot(a\cdot m)=(ba)\cdot m$$ while $$m\cdot (ab)= (ab)\cdot m$$ However, if $R$ is commutative then there is no problem. Every left action defines a right action which is "the same" and vice versa. In this case it is a matter of convenience whether one chooses to write $a\cdot m$ or $m\cdot a$.

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