Question #87227
a) Let d∈N , where d ≠1 and d is not divisible by the square of a prime.
Prove that N:Z[√d]→N∪{0}:N(a+b√d)=|a²-db²| satisfies the following properties for x,y∈Z[√d] :
i) N(x) = 0⇔x=0
ii) N(xy) = N(x)N(y)
iii) N(x) =1⇔ x is a unit
iv) N(x) is prime ⇒x is irreducible in Z[√d ] .
b) Prove or disprove that C≃ R as fields.
1
Expert's answer
2019-04-01T11:52:40-0400

a.

i) Let N(a+db)=0.N(a + \sqrt d b) = 0. It holds if and only if (a+db)(adb)=0.|(a + \sqrt d b) (a - \sqrt d b)| = 0. It is true iff (a+db)(adb)=0.(a + \sqrt d b) (a - \sqrt d b) = 0. Since Z[d]\mathbb{Z}[\sqrt d] has no divisors of zero, we have either a+db=0a + \sqrt d b = 0 or adb=0.a - \sqrt d b = 0. And finally this holds if and only if db=a=0.\sqrt d b = a = 0.

ii) N(xy)=xyxy=xyxˉyˉ=xxˉyyˉ=xxˉyyˉ=N(x)N(y).N(xy) = |xy \overline{xy}| = |x y \bar{x} \bar{y}| = |x \bar{x} y \bar{y}| = |x \bar{x}||y \bar{y}| = N(x)N(y).

iii) If N(x)=1N(x) = 1 and x=a+db,x = a + \sqrt d b, we clearly have ±N(x)=(a+db)(adb)=a2db2=±1.\pm N(x) = (a + \sqrt d b) (a - \sqrt d b) = a^2 - db^2 = \pm 1. Then xx is unit.

Now suppose that xx is unit. Then we have N(x)N(x1)=N(xx1)=N(1)=1.N(x)N(x^{-1}) = N(x x^{-1}) = N(1) = 1. But then the only possible value for N(x)N(x) is 1,1, since N(x)N.N(x) \in \mathbb{N}.

iv) Let N(x)N(x) be prime and xx be not irreducible. Then xx can be represented as x=ab,x = ab, where aa and bb are not equal to one. Then N(x)=N(ab)=N(a)N(b).N(x) = N(ab) = N(a) N(b). By iii) we have that N(a)N(b)1.N(a)N(b) \not= 1. And N(x)1.N(x) \not= 1. Contradiction.

b.

We will prove that C\mathbb{C} and R\mathbb{R} are not isomorphic.

Suppose the contrary: there exists an isomorphism f:CR.f: \mathbb{C} \to \mathbb{R}. Then we must have f(1)=1.f(1) = 1. Also f(1)2=f((1)2)=1.f(-1)^2 = f((-1)^2) = 1. Hence we have f(i)2=f(i2)=f(1)=1,f(i)^2 = f(i^2) = f(-1) = -1, contradiction.


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