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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.

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    $\begingroup$ 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. $\endgroup$ Feb 22 '19 at 2:08
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Hint: what is $x \cdot 1$ in each of these?

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  • $\begingroup$ Could you expand this answer please? Not for the OP, but other viewers like me. $\endgroup$ Feb 22 '19 at 18:05
  • $\begingroup$ 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. $\endgroup$
    – RghtHndSd
    Feb 22 '19 at 22:04
  • $\begingroup$ 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) $\endgroup$ Feb 23 '19 at 3:29
  • $\begingroup$ Are $\mathbb R$ and $\mathbb R[x]/(x)$ isomorphic as $\mathbb R[x]$ modules? $\endgroup$ Nov 25 '19 at 2:45
  • $\begingroup$ @AlJebr: what is the action of $x$ on $\mathbb{R}$? If it doesn't act as $0$, then the answer is no. $\endgroup$
    – RghtHndSd
    Nov 25 '19 at 3:19
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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 $LHS = φ(0*1)=φ(0), RHS = 1*φ(1)=φ(1)$

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