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Show that if p is a prime such that there is an integer b with p = b<sup>2</sup><sup></sup> + 4, then Z<sub>p</sub> is not a unique factorization domain.<br> <br> Proof: First notice that in Z<sub>p</sub>, the norm,<sup></sup> <math>N(a-b\sqrt{p})=a^2-b^2p</math>, is multiplicative and if N(x) = 1 for some x in Z<sub>p</sub>, x is a unit. | Show that if p is a prime such that there is an integer b with p = b<sup>2</sup><sup></sup> + 4, then Z<sub>p</sub> is not a unique factorization domain.<br> <br> Proof: First notice that in Z<sub>p</sub>, the norm,<sup></sup> <math>N(a-b\sqrt{p})=a^2-b^2p</math>, is multiplicative and if N(x) = 1 for some x in Z<sub>p</sub>, x is a unit. | ||
| − | Now <math>N(b+\sqrt{p}) = N(b-\sqrt{p}) = N(2) = 4</math>, so if any of these are reducible, say xy = 2, then N(x)N(y) = 4 and since x | + | Now <math>N(b+\sqrt{p}) = N(b-\sqrt{p}) = N(2) = 4</math>, so if any of these are reducible, say xy = 2, then N(x)N(y) = 4 and since x and y not units, N(x)=N(y)=2. |
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Revision as of 03:48, 26 June 2013
Problem 73 Solution #3
Show that if p is a prime such that there is an integer b with p = b2 + 4, then Zp is not a unique factorization domain.
Proof: First notice that in Zp, the norm, $ N(a-b\sqrt{p})=a^2-b^2p $, is multiplicative and if N(x) = 1 for some x in Zp, x is a unit.
Now $ N(b+\sqrt{p}) = N(b-\sqrt{p}) = N(2) = 4 $, so if any of these are reducible, say xy = 2, then N(x)N(y) = 4 and since x and y not units, N(x)=N(y)=2.
However, we know p = b2 + 4, so p = b2 (mod 4). The only squares mod 4 are 0 and 1 and 4 does not divide p, so p = 1 (mod 4).
Thus we can see $ N(a+b\sqrt{p})=a^2-b^2p=a^2-b^2 $ (mod 4). But since the only squares mod 4 are 0 and 1, we see there are no elements with N(x) = 2. Thus $ (b+\sqrt{p})(b-\sqrt{p})=2*2 $ shows two different factorizations of 4 into irreducibles. So Zp is not a UFD.
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