# Find the discriminant of $\mathbb{Q}(\sqrt{3},\sqrt{5})$

So far I have that $$B = \{1,\sqrt{3},\sqrt{5},\sqrt{3}\sqrt{5}\}$$ is a $$\mathbb{Q}$$-basis for $$\mathbb{Q}(\sqrt{3},\sqrt{5})$$.

I think the discriminant of $$B$$ is $$2^83^25^2$$, which implies that, if $$B$$ is not an integral basis, then there is an algebraic integer of the form $$\frac{1}{p}(a + b\sqrt{3} + c\sqrt{5} + d\sqrt{3}\sqrt{5})$$ for integers $$0 \leq a,b,c,d \leq p-1$$ and $$p$$ one of $$2,3,5$$.

Beyond this, I'm not sure how to proceed.

Any help would be much appreciated.

• See this post for the discriminant of $\Bbb Q(\sqrt{m},\sqrt{n})$. – Dietrich Burde Jul 3 '19 at 15:11

The formula for the discriminant of $$\Bbb Q(\sqrt{m},\sqrt{n})$$ is given in Theorem $$3$$ of the paper Integers of biquadratic fields, page $$525$$. Theorem $$2$$ lists the integral bases for each case. Since $$\Bbb Q(\sqrt{3},\sqrt{5})=\Bbb Q(\sqrt{3},\sqrt{15})$$ we are in the case $$(3,3)\bmod 4$$.