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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 $F={\bf Z}_2$

HK Lee
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    $i$ 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 ${\bf Z}_2$. – Pedro Mar 06 '14 at 01:49
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    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. –  Mar 06 '14 at 01:53
  • @ Tamaroff : My additional question is : Which polynomial makes ${\bf Z}_2(i)$ to be a splitting field over ${\bf Z}_2$ ? – HK Lee Mar 06 '14 at 01:54
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    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. – anon Mar 06 '14 at 01:56
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    If you mean to be denoting ${a+bi:a,b\in\mathbb{Z}_2}$, then the correct notation would be $\mathbb{F}_4=\mathbb{Z}_2[i]$. This is a four element set though, $\mathbb{Z}_2[i]={0,1,i,1+i}$ – Stella Biderman Mar 06 '14 at 02:03
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    In response to your edit: the analogy breaks down because $x^2+1$ is irreducible over $\bf R$ but it splits completely over ${\bf F}_2$. As a result, you need to adjoin an element to $\bf R$ to get $x^2+1$ to factor, but you don't need to do anything to ${\bf F}_2$ to get it to factor. – anon Mar 06 '14 at 02:04
  • @anon : Thank you. I see difference, irreducibility. – HK Lee Mar 06 '14 at 02:06
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    @Stella That is not a field, if you mean to have $i^2=-1=1$. –  Mar 06 '14 at 02:10
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    @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. – Pedro Mar 06 '14 at 02:19
  • @ Pedro : Thank you for your comment. From Mike's answer, I know that $K={ 0,\ 1,\ \alpha,\ 1+\alpha}$ is a splitting. And irreducible polynomial $f(x)=x^2+x+1$ has factorization $(x-\alpha)(x-1-\alpha)$ in $K$. – HK Lee Mar 06 '14 at 02:45
  • Oh, yes, I, like Mike, was using i as a formal symbol because that's what the OP used. – Stella Biderman Mar 06 '14 at 06:21

1 Answers1

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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 $\Bbb Z_2$.