Show that $Q[\sqrt3]$ is a field

Question

I'm trying to show that $Q[\sqrt3]$ is a field. I am particularly struggling with showing that every element is a unit.

So for some element, we want to show that for some element $a+b\sqrt3$, there exists $c+d\sqrt3$ in $Q[\sqrt3]$ such that $(a+b\sqrt3)(c+d\sqrt3) = 1$.

Is that the right direction? I am not really sure of how to come up with that inverse though. Any help would be appreciated.

Answer 1

Hint: start by showing that if $a+b\sqrt{3}$ is a non-zero element of $\mathbb{Q}(\sqrt{3})$ then $a^2-3b^2\neq 0$, and then show that $$ \frac{a-b\sqrt{3}}{a^2-3b^2} $$ is the inverse of $a+b\sqrt{3}$.

By the way, this is essentially the same procedure as the one used to find the multiplicative inverse of a complex number. $a^2-3b^2$ is the norm of the element $a+b\sqrt{3}$, and $a-b\sqrt{3}$ is its conjugate.

Answer 2

Hint: $$ (a + b\sqrt{3})^{-1} = \frac{1}{a + b\sqrt{3}} = \frac{1}{a + b\sqrt{3}} \cdot \frac{a -b \sqrt{3}}{a -b \sqrt{3}} $$