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I am going through my number theory notes and have got on to the bit about the ring $\mathbb{Z}_p[x]$, where $p$ is prime, and unique factorisation domains. The example I am looking at is to do with irreducible and reducible polynomials. It says

e.g in $\mathbb{Z}_3[x]$, $x^2 + x + 1 = (x + 2)(x+2) = (x-1)(x-1)$ because $x^2 + x + 1 = x^2 - 2x + 1$. So $x^2 + x + 1$ is reducible in $\mathbb{Z}_3[x]$. But $x^2 + x + 1$ is irreducible in $\mathbb{Z}_2[x]$ or $\mathbb{Z}_2[x]$ (for example).

I don't get how my lecturer has done this. How can she write $x^2 + x + 1 = (x + 2)(x + 2)$ when $(x + 2) (x+2) = x^2 + 4x + 4$ and how can she say that $x^2 + x + 1 = x^2 - 2x + 1$? Also, why does this only work in $\mathbb{Z}_3[x]$ and not say $\mathbb{Z}_2[x]$ or $\mathbb{Z}_5[x]$?

EDIT: In case it helps, my definition of $\mathbb{Z}_p[x]$ is given by:

The proof of the Primitive Element Theorem uses the fact that if $p \in \mathbb{Z}_+$ is prime, then the ring $\mathbb{Z}_p[x] = \{a_0 + a_1x + \cdots + a_nx^n: n \in \mathbb{Z}, a_i \in \mathbb{Z}_p, 0 \leq i \leq n\}$ is a unique factorisation domain (UFD). This means that $\mathbb{Z}_p[x]$:

  • is a commutative ring with identity
  • has no zero divisors
  • has unique factorisations into irreducibles - which are also primes sinc this is a UFD. The "units" in $\mathbb{Z}_p[x]$ are the constant polynomials $a_0 \in \mathbb{Z}_p^*$.
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    $\begingroup$ In the ring $\mathbb{Z}_3$ we have $4=1$, and that's what your teacher used. Also $-2=1$. $\endgroup$ Commented Apr 10, 2013 at 10:16
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    $\begingroup$ So in $\mathbb{Z}_7[x]$ you would get $$x^2+x+1=x^2-6x+8=(x-2)(x-4).$$ All because in that case $-6=1=8$. $\endgroup$ Commented Apr 10, 2013 at 10:19
  • $\begingroup$ What exactly does $\mathbb{Z}_3$ mean then? I have written down that it is a commutative ring with an identity element, has no zero divisors, has unique factorisation into irreducibles. That doesn't really tell me anything about what the numbers mean $\endgroup$ Commented Apr 10, 2013 at 10:19
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    $\begingroup$ So far all the commenters and answerers seem to be assuming that (as usually in first course on abstract algebra) $\mathbb{Z}_p$ denotes the ring of residue classes of integers modulo $p$. The alternative would the so called $p$-adic integers, but don't worry about those, yet. The facts that you list are probably meant to refer to the polynomial ring $\mathbb{Z}_p[x]$. They do apply to the field $\mathbb{Z}_p$ except that unique factorization in a field is sorta vacuous (true for all fields). $\endgroup$ Commented Apr 10, 2013 at 10:21
  • $\begingroup$ @JyrkiLahtonen: I wonder if the property "unique factorisation into irreducibles" hints to $\mathbb Z_p$ being the $p$-adics. (Of course a field is also an UFD, but quite a trivial one.) $\endgroup$ Commented Apr 10, 2013 at 10:24

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When we write $(x+2)(x+2)$ in $\Bbb Z_3[x]$, we formally write $(x+[2])(x+[2])$ where $[2]$ is the equivalence class $2 \pmod 3$. Now by definition of multiplication by polynomials:

$$(x+[2])(x+[2]) = x^2 + ([2]+[2])x + [2][2] = x^2 + [4]x + [4] = x^2 + [1]x + [1]$$

where the last step follows as the equivalence classes $4 \pmod 3$ and $1 \pmod 3$ are equal.

If OTOH we were working in $\Bbb Z_5$ or some other $\Bbb Z_p$, then the meaning of $[2]$ would change to $2 \pmod 5$ or $2 \pmod p$, which obviously behave differently under multiplication.

Once we have a solid understanding of the ring $\Bbb Z_p$ that we are working in, we may choose to drop the square brackets as an abuse of notation (because mathematicians are lazy).

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  • $\begingroup$ I just saw another example in my lecture notes where it said $(x + 1)(x + 1) = x^2 + 1$ in $\mathbb{Z}_2[x]$. Is this because $(x + [1])(x + [1]) = x^2 + [2]x + [1]$. So we want some number that is $2 \mod 2 \equiv 0 \mod 2 = 0$ and so we leave out the $x$? $\endgroup$ Commented Apr 10, 2013 at 11:21
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    $\begingroup$ Precisely! $[0]$ is the zero of both the rings $\Bbb Z_p$ and $\Bbb Z_p[x]$, so $[0]x = [0]$ and we can leave that out, obviously. $\endgroup$ Commented Apr 10, 2013 at 11:23
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In $\mathbb{Z}_{3}$, we have that $-2 = 1$. This is because $ -2 = 3(-1) + 1 $.

Therefore, since $\mathbb{Z}_{3}[X]$ consists of polynomials with coefficients from $ \mathbb{Z}_{3} $, we have that $x^2 + (1)x + 1 = x^2 + (-2)x + 1 $

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Hint: A monic polynomial of degree $2$ over a domain $R$ is reducible iff it has a root in $R$.

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  • $\begingroup$ @Kaish: Yes, the polynomial comes from $R[x]$ where $R$ is a domain. $\endgroup$ Commented Apr 10, 2013 at 10:18
  • $\begingroup$ Sorry, I accidently deleted the comment. The comment I wrote was "Is $R$ the ring?$ $\endgroup$ Commented Apr 10, 2013 at 10:23

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