I am not sure about my answer. Since x^2+1 is reducible with respect to Z2 but X^2-1 is irreducible with respect to Z2 so I made this finite field {0,1,i,1+i} nd said it as the splitting field of above polynomial. apologize for typing mistakes.
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???? $\Bbb Z_2$ is a field. – Martín-Blas Pérez Pinilla Dec 24 '15 at 12:30
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sorry for mistakes in typing. Read again my question. – As Ma Dec 24 '15 at 12:35
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???? In $\Bbb Z_2$, $1=-1$ so $x^2+1 = x^2-1$. – Martín-Blas Pérez Pinilla Dec 24 '15 at 12:38
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1$x^2-1 = (x+1)(x-1)$, so never is irreducible. – Martín-Blas Pérez Pinilla Dec 24 '15 at 12:40
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@Martín-Blas Pérez Pinilla, actually I had studied that if x^2-1 has no zeros with respect to Z2 then it is reducible in case of Z2 – As Ma Dec 24 '15 at 12:44
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I mean irreducible – As Ma Dec 24 '15 at 12:46
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1Well, $x=1$ is definitely a zero of $x^2+1$ in $\Bbb{Z}_2$, because $$1^2+1=2=0.$$ – Jyrki Lahtonen Dec 24 '15 at 12:47
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@AsMa, $x^2-1=x ^2+1$ has zeros in $\Bbb Z_2$. – Martín-Blas Pérez Pinilla Dec 24 '15 at 12:53
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ok, I get it thanks – As Ma Dec 24 '15 at 12:53
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No. The splitting field of $x^2+1$ over $\mathbf Z_2$ is $\mathbf Z_2$ itself since $x^2+1=(x+1)^2$ has $1$ as a double root in $\mathbf Z_2$.
Bernard
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The roots of above polynomial are i and -i and Z2 doen't contain these roots. – As Ma Dec 24 '15 at 12:37
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1@AsMa, stop thinking that you are in $\Bbb R$ or $\Bbb C$. – Martín-Blas Pérez Pinilla Dec 24 '15 at 12:39
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@Martín-Blas Pérez Pinilla, I got your above point but I am still confused that how Z2 is itself its splitting field can u please explain it little more.Thanks – As Ma Dec 24 '15 at 12:57
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As we know that splitting field of a polynomial is the smallest field extension of a field in which the polynomial has linear factors. – As Ma Dec 24 '15 at 13:00
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nothing, but what are the roots of x^2+1 ? My question is that Z2 doesn't contain i and -i it only contains 0 and 1 which are not the roots of x^2+1. – As Ma Dec 24 '15 at 13:06
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well, just like x^2+1 has splitting field C over R because C is the extension of R and it contain its roots i and -i . – As Ma Dec 24 '15 at 13:12
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1@AsMa If you define $i$ as a root of $x^2+1$, then $\Bbb Z2$ does contain $i$ and $-i$ - except in this case $1=i=-i$. – Wojowu Dec 24 '15 at 13:27
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