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I'm having trouble with this homework problem (from Algebra by Hungerford).

If $S$ is a ring (possibly without identity) with no proper left ideals, then either $S^2=0$ or $S$ is a division ring.

We've just proven in the previous part that if $S$ is a ring with identity and with no proper left ideals, then $S$ is a division ring. So the strategy is to suppose $S^2 \neq 0$ and prove that $S$ has an identity. The problem also has a hint:

Hint: Show that $\{a \in S \mid Sa = 0\}$ is an ideal. If $cd \neq 0$, show that $\{r \in S \mid rd = 0\} = 0$. Find $e \in S$ such that $ed = d$ and show that $e$ is a two-sided identity.

Here's what I have so far, with some boring details removed:

Note that $S_0 = \{a \in S \mid Sa = 0\}$ is an ideal, because [proof it's an ideal].

Since $S$ has no proper left ideals, we must have $S_0 = 0$ or $S_0 = S$. This means that either $S^2 = 0$ (if $S_0 = S$) or for every non-zero $a \in S$, there is some $b \in S$ with $ba \neq 0$ (if $S_0 = 0$). From now on suppose $S_0 = 0$, and we will show that $S$ is a division ring.

Note also that $A(d) = \{r \in S \mid rd = 0\}$ is a left ideal, because [proof it's a left ideal].

Now if $cd \neq 0$ for some $c, d \in S$, then $A(d) \neq S$ because $c \notin A(d)$; but $S$ has no proper left ideals, so $A(d) = 0$. Then $rd = sd$ if and only if $r = s$, for otherwise $(r-s)d = 0$ contradicts $A(d) = 0$.

Consider the left ideal $I_a = \{ra \mid r \in S\}$. [proof that it's a left ideal]. Since $S$ has no proper left ideals, we must have $I_a = 0$ or $I_a = S$ for all $a \in S$.

If $cd \neq 0$ then $I_d = S$ (so $I_d = S$ for all non-zero $d \in S$, since we showed that for all $d$ there is a $c$ with $cd \neq 0$), and therefore $d \in I_d$ so there is some $e \in S$ with $ed = d$. As we observed before, $rd = sd$ iff $r = s$, and $(re)d = r(ed) = rd$, so $re = r$ for each $r \in S$.

In summary, for every $a \in S$ we have some $e \in S$ (specific to this $a$) such that $e a = a$ and $r e = r$ for all $r \in S$. I feel like it should be very easy to show that there is a two-sided identity (e.g. by showing that all of these $e$ are in fact the same), but I can't seem to get there.

Thanks for any help!

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The last paragraph is not clear. Could you rewrite it please ? – Amr Nov 26 '12 at 4:30
@Amr the last paragraph is not terribly important, since it's just a summary. Unless you're talking about the last paragraph of my attempted proof. – smackcrane Nov 26 '12 at 4:37
up vote 1 down vote accepted

It turns out that I missed a key implication in an earlier part of the proof: since $cd \neq 0$ implies $A(d) = 0$, and since we know for every $d$ there is such a $c$, we must have that $A(d) = 0$ for all $d$, and so $S$ has no zero divisors. This lets us use cancellation!

So the ending goes like this. Consider the left ideal $I_d = \{rd \mid r \in S\}$. We must have $I_d = 0$ or $I_d = S$, but we know that for $d \neq 0$ there is some $c$ such that $cd \neq 0$ so $I_d = S$. Thus $d \in I_d$, so there is some $e \in S$ with $ed = d$.

Now we use cancellation to say $ed = d \Rightarrow red = rd \Rightarrow re = r$ for all $r \in S$, and $e$ is a right identity; also $re = r \Rightarrow rex = rx \Rightarrow ex = x$ for all $x \in S$, and we see that $e$ is a two-sided identity. So we're done!

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Technically you're not quite done. You've shown it's an integral domain. One more line is needed to show it's a division algebra: if $u \ne 0$ then $S u = S$, so $v u = 1$ for some $v$. – TBrendle Jun 24 '15 at 2:40

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