Math Help - UFD or Not?

1. UFD or Not?

In Z_5[x] we have x^2 +3x +1 = (2x+3)(3x+2) but we also have x^2 +3x +1 = (4x+1)(4x+1) does this mean that Z_5[x] is not a UFD?

2. Originally Posted by Coda202

In Z_5[x] we have x^2 +3x +1 = (2x+3)(3x+2) but we also have x^2 +3x +1 = (4x+1)(4x+1) does this mean that Z_5[x] is not a UFD?
no! for any field F, the ring F[x] is always a UFD (even more, it's a PID). in your questions, modulo 5 we have:

$2x+3=2x - 2=2(x-1)$ and $4x + 1 = -x +1.$ also we have $3x+2=-2x + 2 = 2(-x+1).$

as you see the irreducible factors $2x+3$ and $3x+2$ are equal to $4x+1$ multiplied by a unit. this does not contradict the definition of a UFD.