All homomorphisms from a simple ring to a non-zero ring are injective

Question

Let $R$ be a simple ring and $T$ be a non-zero ring. Let $f\colon R \rightarrow T$ be a ring homomorphism. Show $f$ is injective.

Proof: $\ker f \lhd R$, so $\ker f=R$ or $\ker f=\{0\}$. If $\ker f=R$ then $R/R \cong \{0\}$, but $T \neq \{0\}$, which is a contradiction. I can't see why does the last part means contradiction.

Thanks for any assistance!

Answer 1

The assertion is true only if you assume

1. rings have unity and
2. ring homomorphisms map the unity to the unity.

With these conventions, a ring homomorphism $f\colon R\to T$, where $T$ is a nonzero ring, cannot have $\ker f=R$, because $f(1_R)=1_T$ and $1_T\ne0_T$ because $T$ is a nonzero ring.

Answer 2

You're right to feel it's a little strange, because this does not make a conclusive argument:

If $\ker(f)=R$ then $R/R \cong \{0\}$, but $T \neq \{0\}$...

The reader is left hanging: so what if $T\neq 0$? What relationship does that have to $R/R$?

Well what this sentence wants to say is that $R/\ker(f)\cong\mathrm{Im}(f)$ is a subring of $T$, and somehow that subring is nonzero. While $T$ was assumed to be nonzero, that may not (depending on your assumptions: keep reading) directly imply that $\mathrm{Im}(f)$ is nonzero.

The real issue is that we need something to ensure $\mathrm{Im}(f)\neq \{0\}$, in order to conclude that $\ker(f)\neq R$. This can be done in a variety of ways.

For one thing (as already mentioned) you could check your definition of rings and homomorphisms to see whether or not they mention multiplicative identity and preservation of this identity by homomorphisms. This setup is very common, and it ensures that no homomorphism between nonzero rings is the zero map.

Alternatively, you could simply require $f$ to be something other than the zero homomorphism. That immediately says "$\ker(f)\neq R$," and therefore you could immediately eliminate that case and automatically conclude $\ker(f)=\{0\}$.