Show that $\text{ker}(\alpha) \subseteq \text{ker}(\beta\alpha)$. If $\beta$ is one-to-one, show that $\text{ker}(\beta\alpha) =\text{ker}(\alpha) $

Question

Let $\alpha : R \rightarrow S$ and $\beta : S \rightarrow T$ be ring homomorphisms. Show that $\text{ker}(\alpha) \subseteq \text{ker}(\beta\alpha)$. If $\beta$ is one-to-one, show that $\text{ker}(\beta\alpha) =\text{ker}(\alpha) $

What I have done: Let $r\in \text{ker}(\alpha)$, then $\alpha(r)=0$. But $0\in \text{ker}(\beta) $. Thus $r\in \text{ker}(\beta\alpha)$, and hence $\text{ker}(\alpha) \subseteq \text{ker}(\beta\alpha)$. Also, if $\beta$ is one-to-one, we know that $\text{ker}(\beta) =\{0\}$, and hence $\text{ker}(\beta\alpha) =\text{ker}(\alpha) $, as desired.

This problem is easy to understand, but it was hard for me to write the proof. Is what I did okay? More than anything I have doubts in the first part, I think a little more clarity is needed. Can you help me please?

Answer 1

Your proof of $\ker(\alpha)\subseteq\ker(\beta\alpha)$ is perfect.

As for the latter part, it would be better if you can write more details, e.g., since $\ker(\beta)=\{0\}$, for every $x\in\ker(\beta\alpha)$, we have $\alpha(x)=0$ whence $x\in\ker(\alpha)$. This gives the converse inclusion.