Unique unital $R$-algebra homomorphism $\varphi: R[x] \to A$ where $\varphi(x) = \alpha$ holds.

Question

In general, if $A$ is a unital algebra over a commutative unital ring $R$ and $\alpha\in A$ is an element, then there is a unique unital $R$-algebra homomorphism $\varphi : R[x] \to A$ such that $\varphi(x) = \alpha$ holds.

Why is this true?

Answer 1

Define the map $\varphi : R[x] \to A$ via $\varphi (x) = \alpha$. We want this map to be an $R$-algebra homomorphism, so we have to force the following conditions on $\varphi$: $\varphi(x \cdot x) = \varphi(x) \cdot \varphi(x)$ and $\varphi(r x) = r \varphi(x)$.

Incidentally, this already defines $\varphi$ for each element of $R[x]$, as each element in $R[x]$ is a linear combination of monomials in $x$ with coefficients in $R$. This is just a restatement of the fact that $R[x]$ is generated by $x$ as an $R$-algebra. Since it has only $1$ generator, we (can) only have to specify $1$ element as the rest is defined via the unitality of the homomorphism, and the homomorphism attribute $\varphi(r x) = r \varphi(x)$.

But once the image of $x$ is specified, the homomorphism is uniquely determined, as discussed above.