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. It holds if and only if ∣(a+db)(a−db)∣=0. It is true iff (a+db)(a−db)=0. Since Z[d] has no divisors of zero, we have either a+db=0 or a−db=0. And finally this holds if and only if db=a=0.
iii) If N(x)=1 and x=a+db, we clearly have ±N(x)=(a+db)(a−db)=a2−db2=±1. Then x is unit.
Now suppose that x is unit. Then we have N(x)N(x−1)=N(xx−1)=N(1)=1. But then the only possible value for N(x) is 1, since N(x)∈N.
iv) Let N(x) be prime and x be not irreducible. Then x can be represented as x=ab, where a and b are not equal to one. Then N(x)=N(ab)=N(a)N(b). By iii) we have that N(a)N(b)=1. And N(x)=1. Contradiction.
b.
We will prove that C and R are not isomorphic.
Suppose the contrary: there exists an isomorphism f:C→R. Then we must have f(1)=1. Also f(−1)2=f((−1)2)=1. Hence we have f(i)2=f(i2)=f(−1)=−1, contradiction.
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