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Are there any field $F$ such that $F[x]/(x^2)\cong F[x]/(x^2-1)$ are isomorphic?

My answer was No. And I thought I had proof but I assume it fails and I have no idea why.

I have also seen this answer but it does not answer my question.

Say if $F[x]/(x^2)\cong F[x]/(x^2-1)$ then $F[x]/(x^2, x^2)\cong F[x]/(x^2-1,x^2)$ and note that $1\in(x^2-1,x^2)\implies F[x]/(x^2-1,x^2)=$ the zero ring.

But $F[x]/(x^2,x^2)$ is not a zero ring as it has element $x$ in it unless char $F=1$.

However, taking $F=\Bbb Z/2\Bbb Z$ works as both sets we get are of the same elements.

Raheel
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  • @Kandinskij respectfully, please read my question and explanation. I want someone to clarify what is wrong with my solution :) – Raheel Sep 03 '24 at 07:16
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    This is wrong: $F[x]/(x^2, x^2)\cong F[x]/(x^2-1,x^2)$, because the image of $x^2$ under the (assumed) isomorphism between $F[x]/(x^2)$ and $F[x]/(x^2-1)$ does not have to be (and will not be) $x^2$. – Magdiragdag Sep 03 '24 at 07:20
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    If you seek to pose a solution-verifification question then please read the tag info to learn what is required (and please also be sure to proved enough details to make it clear what methods you are applying). I recommend that you first read the linked proofs - since doing so may help you debug your intended argument. – Bill Dubuque Sep 03 '24 at 07:28

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