Show that there is no isomorphism

Question

I want to show that $\mathbb R[x]/(x-a)^2$ is not isomorphic to $\mathbb R\times \mathbb R$. If we take $a = b = x-a$ in $\mathbb R[x]/(x-a)^2$, then $ab=0$ and $f(ab)=0$. If there is isomorphism than $f(ab)=f(a)f(b)$, but $0\ne(x-a)^2$. Is this a right solution?

Answer 1

You are onto a good idea: $\Bbb R[x]/(x-a)^2$ has nilpotent elenments while $\Bbb R^2$ doesn't, and thus they cannot be isomorphic. However, your phrasing could be better.

First off, the letter $a$ is already in use as a real number in the definition of $\Bbb R[x]/(x-a)^2$. You shouldn't also use it to signify an element of $\Bbb R[x]/(x-a)^2$. So please pick another letter, like $u$.

Also, there is no need for both $a$ and $b$. They are equal, so why have two different names for it?

Finally, you're using the letter $f$ without introducing it.

Taking your proof and making these three changes, we get

Take a homomorphism $f:\Bbb R[x]/(x-a)^2\to \Bbb R^2$. If we take $u= x-a$ in $\mathbb R[x]/(x-a)^2$, then $u^2=0$ and $f(u^2)=0$. If there is isomorphism than $f(u^2)=f(u)f(u)$.

which is much better. ("if there is isomorphism" can be swapped with "because $f$ is a homomorphism" and once you've done that I see nothing mathematically wrong with what we have.)

To finish off the proof properly, note that since $f(u)\in \Bbb R^2$, there is only one element which fulfills $(f(u))^2 = 0$, which is $(0,0)$. Since $u\neq 0$ but $f(u) = (0,0)$, $f$ cannot be injective and therefore not an isomorphism.

Answer 2

In $\mathbb R[x]/(x-a)^2$, there is a nonzero element $u$ such that $u^2=0$. In $\mathbb R\times \mathbb R$ there is no such element.