If $S$ be a ring with identity and $f: M_{2}(\mathbb{R})\longrightarrow S$ is a nonzero ring homomorphism, which cases is true?

Question

Let $S$ be a ring with identity and $f: M_{2}(\mathbb{R})\longrightarrow S$ is a nonzero ring homomorphism,then which cases is true?

$S$ is left Artinian
$S$ is left Noetherian
$S$ is simple ring
$S$ has infinite number nilpotent elements

I think that the key observation is that $f$ must be injective. — Alex Youcis, Mar 06 '14 at 07:28
$\text{Mat}_2(\mathbb{R})$ is simple--it has no non-trivial two-sided (proper) ideals. Why? — Alex Youcis, Mar 06 '14 at 07:36
$f$ is only nonzero ring homomorphism. what you want? do you want that we have $f(1_{R})=1_{S}$? if i consider $R=M_{n}(\mathbb{R})$. — Ali Qurbani, Mar 06 '14 at 08:26

score 1 · Accepted Answer · answered Mar 06 '14 at 13:20

Take any nonzero ring homomorphism $g:M_2(\Bbb R)\to T$ for some nonzero ring $T$.

Then $S=\prod_{i=1}^\infty T$ obviously isn't Noetherian on either side, and it obviously isn't simple.

Define $f:M_2(\Bbb R)\to S$ by $f(x)=(g(x),g(x),\ldots,g(x),\ldots)$, and you have a counterexample to the first three.

There's hope for the fourth statement, though. As Alex pointed out in the comments, $M_2(\Bbb R)$ is simple, and therefore any nonzero ring homomorphism leaving it is injective. The images of nilpotent elements in $M_2(\Bbb R)$ will therefore be nilpotents in $S$. So, if you can show that $M_2(\Bbb R)$ has infinitely many nilpotents $S$ will also have at least that many. (Try it!)

ok, i find my answer in http://math.stackexchange.com/questions/271566/has-s-infinitely-nilpotent-element — Ali Qurbani, Mar 06 '14 at 15:18

If $S$ be a ring with identity and $f: M_{2}(\mathbb{R})\longrightarrow S$ is a nonzero ring homomorphism, which cases is true?

1 Answers1