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Note that $f(x)=x^2+1=(x+1)^2$ has distinct two roots $1,\ i$ so that it is separable polynomial.

Hence spliting field is $K={\bf Z}_2(i)=\{ 0,\ 1,\ i\}$

Over $K$, $f(x)$ has factorization $(x-1)(x-i)=x^2+x+ix+1\neq f(x)$

Why does such phenomenon happen ?

More detailed explanation :

Consider ${\bf R}$-case. For $f(x)=x^2+1 \in {\bf R}[x]$ has a root $\pm i$ so that we have an extension $$ K={\bf R}(i)={\bf C}={\bf R}[x]/(x^2+1)$$ and we have factorization $$ f(x)=(x+i)(x-i)$$ over $K$.

I want to do similar thing in case $F={\bf Z}_2$

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    $\begingroup$ $i$ is not an element of ${\bf Z}_2$! In fact, $X^2+1$ has $1$ as its only root. Note $f=(X+1)(X+1)$ splits over ${\bf Z}_2$. $\endgroup$
    – Pedro
    Mar 6, 2014 at 1:49
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    $\begingroup$ I don't understand what you mean by "$f$ has factorization..." As you note immediately afterwards that is NOT $f$. Of course this is ignoring what Mr. Tamaroff pointed out above. $\endgroup$
    – user98602
    Mar 6, 2014 at 1:53
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    $\begingroup$ What do you mean by $i$? There are many things wrong with your statements. The polynomial $(x+1)^2$ has $x=1$ as its only root, it doesn't have distinct roots. Since the root has multiplicity two, the polynomial is not separable. The splitting field of $x^2+1$ over ${\Bbb F}_2$ is ${\Bbb F}_2$ itself, since it contains the root $1$. The notation ${\bf Z}_2(i)$ doesn't seem to make sense, and there isn't even any ring with characteristic two and precisely three elements. $\endgroup$
    – anon
    Mar 6, 2014 at 1:56
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    $\begingroup$ @Stella That is not a field, if you mean to have $i^2=-1=1$. $\endgroup$
    – user98602
    Mar 6, 2014 at 2:10
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    $\begingroup$ @Mike It might be my fault, since I thought we were talking about $\{0,1,X,1+X\}$ with $i$ playing the role as $X$, not as $i^2=-1$, but $i^2=i+1$, and made a comment accordingly, which I then deleted when I spotted my mistake. $\endgroup$
    – Pedro
    Mar 6, 2014 at 2:19

1 Answer 1

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There is no $\Bbb Z_2(i)$; it's nonsense to adjoin a complex number. Colloquially, though, this is meant to mean the splitting field of $x^2+1$ over $\Bbb Z_2$ - but this is $ \Bbb Z_2$, as $x^2+1=x^2+2x+1=(x+1)^2 \in \Bbb Z_2[x]$. There is, however, a field of characteristic 2 with 4 elements: the splitting field of $x^2+x+1$ over $\Bbb Z_2$.

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  • $\begingroup$ Thank you : The field is $\{ 0,\ 1,\ x,\ 1+x\}$, and $x$ has an inverse $1+x$ with $x(x+1)=1$, which is stated in Pedro's comment. $\endgroup$
    – HK Lee
    Mar 6, 2014 at 2:21
  • $\begingroup$ Yep! $\ \ \ \ $ $\endgroup$
    – user98602
    Mar 6, 2014 at 2:25

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