Showing that $\frac{\mathbb{R}[x]}{\langle x \rangle}$ and $\frac{\mathbb{R}[x]}{\langle x-1 \rangle}$ are not isomorphic as $\mathbb{R}[x]$ modules.

Question

I'm trying to solve an exercise which asks me to prove that $\frac{\mathbb{R}[x]}{\langle x \rangle}$ and $\frac{\mathbb{R}[x]}{\langle x-1 \rangle}$ are isomorphic as rings, but not as $\mathbb{R}[x]$ modules.

It's easy to show they're isomorphic as rings - you can either do it directly or use the first isomorphism theorem for rings to show they're both isomorphic to $\mathbb{R}$.

I'm not sure how to approach showing that they're not isomorphic as modules though - it seems natural to try to prove it by contradiction, but if I assume there is an isomorphism I'm struggling to see where the contradiction comes from.

Your help would be much appreciated.

Answer 1

Hint: what is $x \cdot 1$ in each of these?

Answer 2

If $φ$ is an $R[x]$-module homomorphism from $R[x]/⟨x⟩$ to $R[x]/⟨x−1⟩$

By definition $φ(x⋅1)=x⋅φ(1)$.

Scalar multiplication is defined as $x=0$ in $R[x]/⟨x⟩$ and $x=1$ in $R[x]/⟨x-1⟩$

So $LHS = φ(0*1)=φ(0), RHS = 1*φ(1)=φ(1)$