# 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.

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.

• The scalar $x$ behaves very differently in $\mathbb R[x]/(x)$ then it does in $\mathbb R[x]/(x-1)$. If they were isomorphic as $\mathbb R[x]$ modules, this scalar should behave the same way. Feb 22 '19 at 2:08

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

• Could you expand this answer please? Not for the OP, but other viewers like me. Feb 22 '19 at 18:05
• If $\varphi$ is an $R[x]$-module homomorphism between the two modules in question, where does $x \cdot 1$ go? There are two possible answers to this question. Feb 22 '19 at 22:04
• Oh, Silly me! So, $\phi(x + \langle x \rangle) = \phi(\langle x \rangle)= \langle x+1 \rangle$ and at the same $\phi(x + \langle x \rangle)= x \varphi(1+\langle x \rangle) = x\cdot (1 + \langle x-1 \rangle)$ because $x$ is a scalar. Right? (+1) Feb 23 '19 at 3:29
• Are $\mathbb R$ and $\mathbb R[x]/(x)$ isomorphic as $\mathbb R[x]$ modules? Nov 25 '19 at 2:45
• @AlJebr: what is the action of $x$ on $\mathbb{R}$? If it doesn't act as $0$, then the answer is no. Nov 25 '19 at 3:19

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