Prove the following Lemma in the polynomial rings.

Question

Let $R$ be a ring. Then, the natural inclusion $R \to R[x]$ which just sends an element $r \in R$ to the constant polynomial $r$, is a ring homomorphism.

Attempts: Let $r \in R$ and define $g : R \to R[x]$ as $g(r) = f(x)$ where $f(x) = r$ for all $x \in R$.

Then, \begin{align*} g(r_1 + r_2) &= r_1 + r_2 \\ &= g(r_1) + g(r_2) \end{align*} and \begin{align*} g(r_1 r_2) &= r_1 r_2 \\ &= g(r_1) g(r_2) \end{align*}

Hence, proved.

Is above true?

Answer 1

Your proof looks fine. Depending on your definition of group homomorphism, you may also want to verify that $g(0)=0$ and $g(1)=1$.

A nice general fact is that if $R$ is a subring of a ring $S$, then the inclusion $R\ \hookrightarrow\ S$ is a group homomorphism.