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I found a result in ring theory, which is as follows:

Let $f(x) \in F[x]$. Suppose $f = f_1 ^{a_1}\cdots f_m ^{a_m}$ is a factorization of $f$ into irreducibles.

Suppose $f$ is square free, then $\mathbb{Q}[x]/(f)$ has no nilpotent not equal to $0$.

I want to know that using the above result can I say about the quotient ring $\mathbb{Q}[x]/(x^2)$. Also, how to prove $\mathbb{Q}[x]/(x^2)$ is not isomorphic to $\mathbb{Q}[x]/(x^2-1)$.

I know about the ring $\mathbb{Q}[x]/(x^2-1)$. Since $x^2 - 1 = (x+1)(x-1)$ is a factorization into irreducibles, then by CRT we have $$\mathbb{Q}[x]/(x^2-1) \cong (\mathbb{Q}[x]/(x+1)) \times (\mathbb{Q}[x]/(x-1)) \cong \mathbb{Q} \times \mathbb{Q}.$$

But how to deal with $\mathbb{Q}[x]/(x^2)$? Does the above theorem tell anything?

Please help me. Thanks.

user26857
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  • Compare with posts like this one. The first one cannot be a product of fields, having a nonzero nilpotent element. In particular, it is not isomorphic to $\Bbb Q\times \Bbb Q$. – Dietrich Burde Oct 20 '21 at 14:55
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    $\mathbb{Q[x]}/(x^2)$ has a nilpotenet element it(can you find it), while $\mathbb{Q}^2$ doesn't have any nilpotents – Amr Oct 20 '21 at 15:02
  • The situation is similar to the difference between $\Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z$ and $\Bbb Z/4\Bbb Z$. – WhatsUp Oct 20 '21 at 18:58

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