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Still struggling with some algebra exercises:

Let $\phi : R \rightarrow S $ be a ring homomorphism.

i) show that $ker(\phi)$ is a two sided ideal of R

ii) Show that $\phi$ is injective iff $ker(\phi)$ = {0}

For i) I think I can see why, since kernel maps to zero then it will ker multiplied R will also map to zero from both sides, but I am not sure how to write this mathematically.

For ii) I'm not really sure on how to start

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1) For $r\in R$ and $x\in\ker\phi$, so $\phi(x)=0$, then $\phi(rx)=\phi(r)\phi(x)=0$, so $rx\in\ker\phi$. Similar reasoning applied to $xr$.

2) Only if part. For $x\in\ker\phi$, then $\phi(x)=0$. But then $\phi(0)=0$, so $\phi(x)=\phi(0)$. As $\phi$ is injective, then $x=0$.

If part. For $\phi(x)=\phi(y)$, then $\phi(x)-\phi(y)=0$, so $\phi(x-y)=0$, and hence $x-y\in\ker\phi=\{0\}$, we have $x-y=0$, then $x=y$.

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