2
$\begingroup$

I am reading Noncommutative Rings by Lam. In his proof of the Wedderburn-Artin theorem (§1.3.11) he seems to use the following:

If $\mathfrak{u}$ is a left ideal in a simple ring $R$, then $\mathfrak{u} \cdot R = R$.

I wonder why this is true. I mean, why can't $\mathfrak{u}\cdot R \subset J_R$ for some right ideal $J_R$?

$\endgroup$
  • 1
    $\begingroup$ Because $\mathfrak{u}R$ will be a left ideal (since $\mathfrak{u}$ is), and a right ideal (since $R$ is). Signed-Alex. $\endgroup$ – Alex Youcis Jul 21 '14 at 1:55
  • $\begingroup$ Oh, yes! I guess I was being really dumb. Thanks! $\endgroup$ – Alex Jul 21 '14 at 2:02
2
$\begingroup$

Here I assume $R$ has a unit, in accord with http://en.m.wikipedia.org/wiki/Ring_(mathematics); I also take it as understood that ${\frak u} \ne \{0\}$.

Note that ${\frak u} R$ is a two-sided ideal in $R$, for if $r \in R$, $r \frak u \subset \frak u$ since $\frak u$ is a left ideal; thus $r {\frak u} R \subset {\frak u} R$, showing ${\frak u} R$ is a left ideal. Likewise, ${\frak u} R$ is a right ideal, since $R r \subset R$ implying ${\frak u} R r \subset {\frak u} R$; so ${\frak u} R$ is two-sided. Now $R$, being a simple ring, has no two-sided ideals other than $\{0\}$ and $R$ itself. Note that since $\{0\} \ne {\frak u} = {\frak u} 1_R \subset {\frak u} R$, ${\frak u} R \ne \{0\}$; thus we must have ${\frak u} R = R$.

Note added Sunday 20 July 2014 9:19 PM PST: It strikes me as worth mentioning that for right ideals $\frak v$ we have $R {\frak v} = R$, by an argument which simply reverses the roles of left and right in the above. End of Note.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

$\endgroup$

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.