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Le $\Bbb{K}$ be a field. I want to show wether $\frac{\Bbb{K}[X]}{\langle X^2 \rangle}$ and $\frac{\Bbb{K}[X]}{\langle X^2 + 1\rangle}$ are isomorphic.

First thing to note is that there is an nilpotent element $[X]$ in the first quotient. But in the second one I think it will depend on the roots of $X^2+1$ because for example in $\Bbb{F}_2$, $x^2+1=(X-1)^2$ so $[X-1]$ is nilpotent. So in the case where it can't be factorized there is no nilpotent element (how to prove that ?). Otherwise I don't see any morphism we can exhibit.

Any hint ?

acd3456
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    It depends on $K$. For example, for the complex numbers both polynomials are reducible. For real numbers, the rings are of course not isomorphic. The second quotient ring then is a field, the first one not. For $2=0$ see this post. – Dietrich Burde Dec 02 '20 at 09:36
  • Hint: if $K[X]/(X^2+1)$ has a nilpotent element, it means that $X^2+1$ has a repeated irreducible factor in its decomposition, in particular that $X^2+1$ isn’t coprime to its derivative. Can you work out in which $K$ this happens? – Aphelli Dec 02 '20 at 09:42
  • @DietrichBurde Ok I didn't found that this question was already answered. Sorry. – acd3456 Dec 02 '20 at 09:44

1 Answers1

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I will change your $\Bbb K$ to $K$, for simplicity. All isomorphism below are isomorphisms of rings (or $K$-algebras, actually).

The structure of $K[X]/(X^2 + 1)$ depends on the number of different roots of the polynomial $X^2 + 1$ in $K$. There can be three situations:

  • There are two different roots $r_1, r_2$. We then have $$K[X]/(X^2 + 1) \simeq K[X]/(X - r_1) \times K[X]/(X - r_2)$$ by chinese remainder theorem, and each factor $K[X]/(X - r_i)$ is canonically isomorphic to $K$. Hence $K[X]/(X^2 + 1) \simeq K^2$ in this case.

  • There is one double root $r$. It is easy to see that this happens iff the characteristic of $K$ is equal to $2$, and in that case we have $X^2 + 1 = (X + 1)^2$, hence $$K[X]/(X^2 + 1) \simeq K[X]/(X + 1)^2 \simeq K[X]/X^2$$ via evident isomorphisms.

  • There is no root, i.e. $-1$ is not a square in $K$. In this case, $K[X]/(X^2 + 1)$ is a field, which is the quadratic extension $K[\sqrt{-1}]$ of $K$.

We see that in the second case, we do have $K[X]/(X^2 + 1) \simeq K[X]/X^2$.

In the remaining two cases, $K[X]/(X^2 + 1)$ is a product of fields, hence does not have any nilpotent element, while $X$ is a nilpotent element in $K[X]/X^2$. Thus they are not isomorphic in these two cases.

To summerize, we have $K[X]/(X^2 + 1) \simeq K[X]/X^2$ iff the characteristic of $K$ is equal to $2$.

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