Answer to Question #87227 in Abstract Algebra for albert

Question

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.

Answer 1

a.

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

ii) $N(xy) = |xy \overline{xy}| = |x y \bar{x} \bar{y}| = |x \bar{x} y \bar{y}| = |x \bar{x}||y \bar{y}| = N(x)N(y).$

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

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

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 $N(a)N(b) \not= 1.$ And $N(x) \not= 1.$ Contradiction.

b.

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

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