1. ## Need HW help about norm and prime numbers

Let p be either a prime number, or -1. Prove that the following norm defined on Z[sqrt(p)] (a + bsqrt(p), a,b are in Z )

N(a+bsqrt(p))= /a^2 -pb^2/. ( the bars mean absolute value)

If N(a+bsqrt(p)) is a prime number, then a+bsqrt(p) is irreducible.

2. Originally Posted by JCIR
Let p be either a prime number, or -1. Prove that the following norm defined on Z[sqrt(p)] (a + bsqrt(p), a,b are in Z )

N(a+bsqrt(p))= /a^2 -pb^2/. ( the bars mean absolute value)

If N(a+bsqrt(p)) is a prime number, then a+bsqrt(p) is irreducible.
This norm has the following property: $\displaystyle N(\alpha \beta) = N(\alpha)N(\beta)$, you can prove this directly by letting $\displaystyle \alpha = a+b\sqrt{p}$ and $\displaystyle \beta = c+d\sqrt{p}$ and multipling everything out.

Let $\displaystyle N(\alpha)$ is a prime number. Suppose that $\displaystyle \alpha = \beta \gamma$ where neither $\displaystyle \beta,\gamma$ are units. Then it would mean $\displaystyle N(\alpha) =N(\beta)N(\gamma)$ with neither $\displaystyle N(\beta),N(\gamma)$ being $\displaystyle 1$. Thus, we have factored $\displaystyle N(\alpha)$ non-trivially, and so it cannot be a prime number. A contradiction. Thus, $\displaystyle \alpha$ is irreducible.