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Show that $x^2 + 1$ is irreducible over $\Bbb Z_3$ and reducible over $\Bbb Z_5$.

I can't figure any way to express $x^2 + 1$ as a product of two polynomials in either ring. Each product I try either ends up with a number being off by $1$ or $2$.

Anyone have any ideas?

Oliver G
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    More generally, $x^2+1$ is irreducible in $\mathbb Z_p$ (where $p$ is prime) if and only if $p$ is of the form $4k+3$, $k\in\mathbb Z$. See Quadratic Reciprocity. – user236182 Sep 12 '16 at 13:39
  • In $\mathbb{Z}_5$, $x^2+1= (x-3)(x-2)$ (or equivalently $(x+3)(x+2)$), for instance. – Nicholas Stull Sep 12 '16 at 13:41
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    How about you calculate the roots with the quadratic formula, instead of making some blind guess? –  Sep 12 '16 at 13:42
  • You can use the set of quadratic residues modulo ${3,5,...}$ to determine whether there is a factor; in particular, modulo $5$, the residues are $0,1,4,4,1$, showing that $2,3$ are roots. – abiessu Sep 12 '16 at 13:45
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    @G.Sassatelli What do you mean? I don't see how I would use this for $\Bbb Z_3$ – Oliver G Sep 12 '16 at 13:48
  • If $f(x)g(x)=x^2+1$, then (up to an invertible constant) $f(x)=x-\alpha_1$ and $g(x)=x-\alpha_2$, where $\alpha_1,\alpha_2$ are the roots of $x^2+1$ in $\Bbb Z_3$. These must be given by $2^{-1}\cdot(-0\pm\beta)$, where $\beta$ is any number such that $\beta^2=0^2-4\cdot 1=-1$. –  Sep 12 '16 at 19:31

3 Answers3

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You can check the roots. Over $\mathbb{Z}_3$ we have $$0^2 + 1 = 1 \qquad 1^2 + 1 = 2 \qquad 2^2 + 1 = 2$$ Hence $x^2 + 1$ is irreducible. Over $\mathbb{Z}_5$ we have $$2^2 + 1 = 0 \qquad 3^3 + 1 = 0$$ Thus $$x^2 + 1 = (x + 2)(x + 3)$$ is reducible.

TheGeekGreek
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  • But how would you write $x^2 + 1$ as a product of two polynomials in $\Bbb Z_3[x]$ with one of them being a unit? – Oliver G Sep 12 '16 at 14:02
  • $1 \cdot (x^2 + 1)$ or $2 \cdot (x^2 + 1)$. – TheGeekGreek Sep 12 '16 at 14:04
  • Ah, I see I was misunderstanding the definition of an irreducible polynomial. So reducibility for $\Bbb Z_5$ would imply that just there exists a product combination that is not a unit $\times$ un-reducible expression? – Oliver G Sep 12 '16 at 14:12
  • Roughly speaking, an irreducible polynomial is a non-constant polynomial, which cannot be factored as two non-constant polynomials. A non-irreducible polynomial is said to be reducible. – TheGeekGreek Sep 12 '16 at 14:13
  • How does $2 \cdot (x^2 + 1) = x^2 + 1$ in $\Bbb Z_3$? – Oliver G Sep 12 '16 at 14:16
  • I think that was dumb. $2$ is for sure a unit of $\mathbb{Z}_3$, but the product is wrong. It is not $x^2 + 1$. Sorry. – TheGeekGreek Sep 12 '16 at 14:18
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    Modulo $5$: $x^2+1=x^2-4=(x-2)(x+2)$. – egreg Sep 12 '16 at 14:27
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Here's a proof of the simple result described in the comments: $ x^2 + 1 $ is a quadratic, so it is irreducible over a field $ K $ if and only if it has no roots there. Thus, $ x^2 + 1 $ is irreducible in $ \mathbf F_p $ for an odd prime $ p $ if and only if $ -1 $ is not a quadratic residue in $ \mathbf F_p $. Since the group $ \mathbf F_p^{\times} $ is cyclic, this happens if and only if $ |\mathbf F_p^{\times}| = p-1 $ is not a multiple of $ 4 $, that is, iff $ p \equiv 3 \pmod{4} $.

Ege Erdil
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$ x^2\equiv \text{ either } 1,0 \pmod 3 $

So adding $1$, you can't get zero remainder.

But that is not so in case of 5.

$ 2^2 + 1 =0$ (mod $5$)

jnyan
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