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?

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- Apr 6th 2009, 05:41 PM #1

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- Apr 6th 2009, 06:08 PM #2

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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:

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

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