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For rings $R_1, \: R_2$ let $\phi : R_1 \rightarrow R_2$ be a homomorphism.

It's easy to show that $\ker \left (\phi \right)$ is an ideal for $R_1$

Now if we assume $R_1$ is a field (or a division ring), we know that the only ideals of $R_1$ are $\left \{ 0 \right \}$ and $R_1$.

Clearly if $\ker \left ( \phi \right ) = R_1$, then $\phi$ is the trivial homomorphism.

For all homomorphisms we know that:

$\ker \left ( \phi \right ) = \left \{ 0 \right \} \iff \phi$ is injective

From this can we conclude that if $R_1$ is a division ring, any nontrivial homomorphism $\phi : R_1 \rightarrow R_2$ is injective?

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  • $\begingroup$ I don't see why we couldn't, all the arguments are there. $\endgroup$ – Captain Lama May 18 '16 at 18:02
  • $\begingroup$ Yeah because the kernel is a two-sided ideal and the only two-sided ideals of a division ring are $0$ and the whole thing. So if it's nontrivial then the kernel must be $0$. $\endgroup$ – Alex Mathers May 18 '16 at 19:47
  • $\begingroup$ I believe we can actually lessen the requirement that $R_1$ is a division ring. I think it only needs to be a simple ring (no proper two sided ideals) $\endgroup$ – bthmas May 18 '16 at 19:54
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Yes, that is what we may conclude. A different proof (and I believe the standard proof) for the same result is this:

Let $x \in \ker(\phi)$. If $x \neq 0$, then $x^{-1}$ exists, and we have $$ \phi(1) = \phi(xx^{-1}) = \phi(x)\cdot \phi(x^{-1}) = 0\cdot \phi(x^{-1}) = 0 $$ which means that $\phi$ is trivial. Therefore, if there is even a single non-zero element in the kernel, then the homomorphism is trivial.

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Yes, either via the argument that you give, or more directly:

Suppose $a, b \in R_1$.

If $\phi(a) = \phi(b)$, then $\phi(a - b) = 0 = \phi(0)$.

If $a - b = 0$, then we are done, as $a = b$.

Otherwise, we may take an inverse to $a - b$ to get that $\phi(1) = \phi((a - b)(a - b)^{-1}) = \phi(0)\phi((a - b)^{-1}) = 0$.

Then for any $r \in R_1$, we have that $\phi(r) = \phi(r)\phi(1) = \phi(r)\cdot 0 = 0$, so $\phi$ is the zero homomorphism.

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