$\mathfrak{u} \cdot R =R$ for every left ideal $\mathfrak{u}$ in a simple ring $R$

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 $J_R$?

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

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.