Let $g \in \mathbb{R}[t]$ be a normed irreducible polynomial of degree 2, meaning that $g(t) = (t - \lambda)(t - \overline{\lambda}$) for a $\lambda = a + b i$, with $a, b \in \mathbb{R}$, $b ≠ 0$.
I want to show that for any $m \in \mathbb{N}$, the transformation
$$\phi: \mathbb{R}[t]/(g^m) \to \mathbb{C}[t]/(t-\lambda)^m, f + (g^m) \mapsto f + (t - \lambda)^m$$
is a well-defined $\mathbb{R}[t]$-module isomorphism. (Where $(g^m)$ is the ideal, generated by $g^m$.)
Now I know that I basically have to check that $\phi$ is well-defined (usually by choosing $f_1 + (g^m) = f_2 + (g^m)$ and show that $\phi$ sends them to the same equivalence class), that I have to check the criterias for a module homormorphism, and that I finally need to check injectivity und surjectivity.
But I'm already having trouble getting started. If
$$f_1 + (g^m) = f_2 + (g^m) <=> f_1 + (t-\lambda)^m(t-\overline{\lambda})^m = f_2 + (t-\lambda)^m(t-\overline{\lambda})^m$$
can we then simply divide through $(t-\overline{\lambda})^m$ to show that they are in the same equivalence class in $\mathbb{C}[t]/(t-\lambda)^m$? Seems a bit to easy. It still confuses me that in the first set, we divide the real polynomials by an irreducible polynomial, but in the second set, we consider them as polynomials over $\mathbb{C}$ that are "devided" by $(t-\lambda)^m$.
Rephrased, to show that the function is well-defined, you have to show that whenever $g^m|(f_1-f_2)$, you also have $(t-\lambda)^m|(f_1-f_2)$. But this is clear since $(t-\lambda)^m|g^m$.