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| − | == <span style="line-height: 1.5em;">Problem 73 Solution #3</span> == | + | == <span style="line-height: 1.5em;">Problem 73 Solution #3</span> == |
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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, <math>N(x)=N(y)=\pm 2</math>. |
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However, we know 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 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 <math>N(a+b\sqrt{p})=a^2-b^2p=a^2-b^2</math> | + | Thus we can see <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 <math>N(x) = \pm 2</math>. Thus <math>(b+\sqrt{p})(b-\sqrt{p})=2*2</math> 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]] | ||
Latest revision as of 07:26, 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)=\pm 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) = \pm 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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