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However, we know&nbsp;p = b<sup>2</sup> + 4, so p = b<sup>2</sup> (mod 4). The only squares mod 4 are 0 and 1 and 4 does not divide p, so p = 1 (mod 4).  
 
However, we know&nbsp;p = b<sup>2</sup> + 4, so p = b<sup>2</sup> (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&nbsp;<math>N(a+b\sqrt{p})=a^2-b^2p=a^2-b^2</math>&nbsp;(mod 4). But since the only squares mod 4 are 0 and 1, we see there are no elements with N(x) = 2. Thus&nbsp;<math>(b+\sqrt{p})(b-\sqrt{p})=2*2</math>&nbsp;shows two different factorizations of 4 into irreducibles. So Z<sub>p</sub> is not a UFD.<br> <br> [[Assignment|Back to Assignment]]  
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Thus we can see&nbsp;<math>N(a+b\sqrt{p})=a^2-b^2p=a^2-b^2 \pmod{4}</math>. But since the only squares mod 4 are 0 and 1, we see there are no elements with N(x) = 2. Thus&nbsp;<math>(b+\sqrt{p})(b-\sqrt{p})=2*2</math>&nbsp;shows two different factorizations of 4 into irreducibles. So Z<sub>p</sub> is not a UFD.<br> <br> [[Assignment|Back to Assignment]]  
  
 
[[Category:Assignment]]
 
[[Category:Assignment]]

Revision as of 07:07, 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 \pmod{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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