If F is a field then is $F[x]/\left<x^2\right> = F \times F$ correct by CRT ? Can we write $F[x]/\left<x^2-1\right> = F \times F$ by CRT ? I think both can be written in the corresponding forms. If any mistake, please verify it.
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Special cases of this Lemma – Bill Dubuque Aug 08 '22 at 07:45
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The first guess you gave will always be wrong. The quotient by $x^2$ has nonzero nilpotents but $F\times F$ does not. It cannot be “correct by the CRT” because the CRT doesn’t apply. I think you need to go back and read it again.
If $F$ has characteristic $2$ the second guess also fails for the same reason.
For fields not of characteristic $2$ though the Chinese remainder theorem will take hold because $R=(2)=(x+1, x-1)$, the factors are coprime.
rschwieb
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Following your hint, could I say that in the first case , F[x]/<x²> has two elements (namely x +<x²> and 0+<x²> )that satisfies equation of the form t²=0 , whereas in F² there are only one element, which is 0 , satisfies the equation t²=0 . So F[x]/<x²>and F² are not isomorphic as field ? – Pritam Roy Aug 08 '22 at 13:12
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