Let m be a square free negative integer that is not equal to -1 or -3. Let R be the ring of algebraic integers in  \mathbb {Q} [ \sqrt {m} ] . Show that the only units in R are 1, -1.

Proof.

Let [tex] x = a+b \sqrt {m} be a unit in R. Then the norm of x in R is:

 N_{R} (x) = N_{R} (a+b \sqrt {m} ) = a^2 - mb^2 = \{ -1,+1 \}

Now, I know that b must equal to 0 and a must equal to +1,-1, since anything else would not be able to eliminate m into 1, but how do I prove it?

Thanks.