The usual is stated something like this:
Kronecker's Theorem: if $p(x) \in \mathbb{F}[x]$ is a monic $n$-th order irreducible polynomial where $\mathbb{F}$ is a field then $\mathbb{F}[x]/(p(x))$ forms a degree $n$ extension field of $\mathbb{F}$. Moreover, if $\theta = x \ mod(p(x))$ then $\mathbb{F}[x]/(p(x))$ has basis $1, \theta, \dots , \theta^{n-1}$.
For example, $\mathbb{R}[x]/(x^2+1) \approxeq \mathbb{C}$. Indeed, if you survey any of the standard elementary abstract algebra texts you can find many theorems about such quotients. For example, see Theorem 4 of $\S 13.1$ of Dummit and Foote on page 513.
I'm studying certain questions involving polynomials over an algebra. Well, let's narrow it a bit. In this question I mean for $\mathcal{A}$ to denotes an $n$-dimensional, associative, commutative, unital algebra over the real numbers. I am hopeful the generalization of Kronecker's Theorem below holds true:
Kronecker's Theorem for Algebras: If $p(x) \in \mathcal{A}[x]$ is a monic $n$-th order polynomial where $\mathcal{A}$ is an algebra then $\mathcal{A}[x]/(p(x))$ forms a free $\mathcal{A}$-module of rank $n$. If we denote $\theta = x \ mod(p(x))$ we find basis $1, \theta, \dots , \theta^{n-1}$ for $\mathcal{A}[x]/(p(x))$.
I've read the proof of the usual theorem in Dummit and Foote and it appears it adapts to prove "Kronecker's Theorem for Algebras" as detailed above.
Question: is my assertion correct? Is "Kronecker's Theorem for Algebras" true? Or, can you provide evidence to the contrary?
I thank the MSE in advance for your insights! (I put the extension field tag because this question is closely related to that arena of thought)