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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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: and also we have as you see the irreducible factors and are equal to multiplied by a unit. this does not contradict the definition of a UFD.

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