# Monic Polynomials

• Mar 12th 2009, 06:20 PM
Coda202
Monic Polynomials
Show that there is only one way (disregarding the order of the factors) to factor x^2 +x+3 as a product of monic irreducible polynomials in Z(sub5)[x].

So, I found that f(1) produces a zero for this polynomial, then I used to the division algorithm to get (x^2 +x+3)/(x-1) = (x+2). However, (x-1)(x+2) does not equal x^2 +x+3, so now I am stuck...
I also used the quadratic formula to get a pair of complex roots, but I'm unsure as how to prove that is the only way to factor x^2 +x+3 as a product of irreducible monic polynomials in Z(sub5)[z]
• Mar 12th 2009, 07:01 PM
GaloisTheory1
Quote:

Originally Posted by Coda202
Show that there is only one way (disregarding the order of the factors) to factor x^2 +x+3 as a product of monic irreducible polynomials in Z(sub5)[x].

So, I found that f(1) produces a zero for this polynomial, then I used to the division algorithm to get (x^2 +x+3)/(x-1) = (x+2). However, (x-1)(x+2) does not equal x^2 +x+3, so now I am stuck...
I also used the quadratic formula to get a pair of complex roots, but I'm unsure as how to prove that is the only way to factor x^2 +x+3 as a product of irreducible monic polynomials in Z(sub5)[z]

$
\mathbb{Z}_5$
is a field and hence a unique factorization domain [trivially]. also, $R[x]$ is a ufd if $R$ is a ufd, so this is easily seen this way. Also, $x^2 +x+3$ would only have linear factors, so just test $0, 1, 2, 3, 4$.
• Mar 12th 2009, 07:03 PM
bob000
(x-1)(x+2) DOES equal x^2 +x+3. Remember we a are in a modulo ring.
x-1 is the same as x+4
(x+4)(x+2) = x^2 + 6 x + 8 3