I was thinking about an explicit isomorphism between $\mathbb{R} [X]/(f)\cong \mathbb{R} [\epsilon] $ where $f(X) = (X-z)^2$ for $z\in\mathbb{R} $.
I found an explicit isomorphism $\mathbb{R} [X]/(f)\cong \mathbb{C} $ where $f(X)$ has two complex roots, and I would like to prove $\mathbb{R} [X]/(f)\cong \mathbb{R} [\epsilon] $ the same way.
I know the isomorphism can be proven using the isomorphism theorem, however, I would like to find an explicit isomorphism. But I can't seem to find the correct mapping which gives an isomorphism. I know that every $g\in \mathbb{R} [X]/(f)$ can be written as $a+bx$ using the polynomial division theorem. And that in $\mathbb{R} [X]/(f)$ the following holds: $[X^2]=[2zX-z^2]$.
So the question is what is an explicit mapping $\phi:\mathbb{R} [X]/(f)\rightarrow\mathbb{R} [\epsilon] $ which gives an isomorphism.