How do I show $x^2 + 1$ is reducible or irreducible in $\mathbb{Z}_7$? Is there a standardized procedure to check ?
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2For a quadratic or cubic polynomial, irreducibility is equivalent to having no roots in the base field. No “$i$” in $\Bbb Z/7\Bbb Z$? Irreducible! – Lubin Oct 16 '15 at 04:30
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1It is irreducible for any prime p = 3 (mod 4). – Shailesh Oct 16 '15 at 04:31
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Since $a^6\equiv 1\pmod 7$ for any $a$ not divisible by $7$, then if $-1$ is a square modulo $7$ then $(-1)^3\equiv 1\pmod 7$. We can see it is not.
More generally, the answer is to use various forms of quadratic reciprocity to determine if $x^2+n$ is irreducible modulo $p$.
Thomas Andrews
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6In this particular case, much easier to test the seven elements to see whether any of them satisfies $f(x)=0$. – Lubin Oct 16 '15 at 04:32
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Thank you very much. It makes sense. but would this method work for any other polynomial. in example let say we have x^3+x+1 in Z2 how would I use the modulo method – Samir Oct 16 '15 at 04:35
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1The general question is non-trivial. If $p$ is prime and $f(x)$ is a polynomial of degree $n$ then $f(x)$ is prime if $f(x)\mid x^{p^n}-x$ and relatively prime with other $x^{p^d}-x$, with $d\mid n$. – Thomas Andrews Oct 16 '15 at 04:44
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In the case of $x^3+x+1$ over the field with two elements, nothing makes it zero, so the polynomial is irreducible. The method fails for polynomials of degree $\ge4$, though, so don’t be misled. – Lubin Oct 16 '15 at 04:47
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Alternatively, as mentioned in comments, the first way I would think:
We will look at $x^2+1$ over $\Bbb F_7$. It is irreducible if it has no roots in the field $\Bbb F_7$. Now we can see that a root will have $x^2=-1\equiv 6\pmod 7$
$1^2=1$, $2^2=4$, $3^2\equiv 2\pmod 7$, $4^2\equiv 2\pmod 7$, $5^2 \equiv 4 \pmod 7$, $6^2\equiv 1\pmod 7$, $7^2 \equiv 0 \pmod 7$. Hence $x^2$ cannot equal $6\pmod 7$, and no root is present in $\Bbb F_7$.
