# Determine the irreducible polynomial for $\alpha=\sqrt{3}+\sqrt{5}$ over $\mathbb{Q}(\sqrt{10})$

I've already found that the irreducible polynomial of $$\alpha$$ over $$\mathbb{Q}$$ is $$x^4-16x^2+4$$. I've also found that $$\mathbb{Q}(\sqrt{3}+\sqrt{5})=\mathbb{Q}(\sqrt{3},\sqrt{5})$$ and that $$\mathbb{Q}(\sqrt{3},\sqrt{5},\sqrt{10})=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$$. Since $$[\mathbb{Q}(\sqrt{10}):\mathbb{Q}]=2$$ and $$[{\mathbb{Q}(\sqrt{3},\sqrt{5}):\mathbb{Q}}]=4$$, $$[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}(\sqrt{3},\sqrt{5})]$$ must be either 1 or 2.

I know it's 2 but I'm having a hard time proving that $$\mathbb{Q}(\sqrt{3},\sqrt{5})\neq\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$$. I'm trying to show that $$\sqrt{2}\not\in\mathbb{Q}(\sqrt{3},\sqrt{5})$$ but I'm not having much luck.

The solution in the link below uses a theorem of Galois theory we haven't covered yet so I don't feel comfortable using it. Here is what we have covered that I suspect is relevant but haven't figured out how to use yet:

Let $$K$$ and $$K^\prime$$ be extensions of the same field $$F$$. An isomorphism $$\varphi:K\to K^\prime$$ that restricts to the identity on $$F$$ is an isomorphism of field extensions.

Let $$F$$ be a field and $$\alpha$$ and $$\beta$$ be elements of field extensions $$K/F$$ and $$L/F$$. Suppose $$\alpha$$ an $$\beta$$ are algebraic over $$F$$. There is an isomorphism of fields $$\sigma:F(\alpha)\to F(\beta)$$ that is the identity on $$F$$ and that sends $$\alpha\leadsto\beta$$ if and only if the irreducible polynomials for $$\alpha$$ and $$\beta$$ over $$F$$ are equal.

Let $$\varphi:K\to K^\prime$$ be an isomorphism of field extensions of $$F$$, and let $$f$$ be a polynomial with coefficients in $$F$$. Let $$\alpha$$ be a root of $$f$$ in $$K$$, and let $$\alpha^\prime=\varphi(\alpha)$$ be its image in $$K^\prime$$. Then $$\alpha^\prime$$ is also a root of $$f$$.

If I start by assuming that $$\mathbb{Q}(\sqrt{3},\sqrt{5})=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$$ then I suspect the three statements above will lead to a contradiction somewhere. I just don't have a good firm grasp of how to put them into practice yet.

Any help is appreciated. Thanks,

• Well it suffices to show that $[\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}):\mathbb{Q}] = 8$, which is answered in math.stackexchange.com/questions/36129/… – ADF Mar 25 '12 at 2:43
• In what this gives you the minimal polynomial of $\alpha$ ? I'm also looking for this minimal polynomial, but I don't understand how you can find it. – MSE Apr 20 '17 at 14:59

$$\sqrt{2}=a+b\sqrt{3}+c\sqrt{5}+d \sqrt{15} \,.$$

Squaring both sides and using the fact that $1, \sqrt{3}, \sqrt{5}, \sqrt{15}$ are linearly independent over Q you get

$$2=a^2+3b^2+5c^2+15d^2 \,.$$ $$ab+5cd =0 \,.$$ $$ac+3bd=0 \,.$$ $$ad+bc=0 \,.$$

From the last two equations we get

$$3bd^2=-acd=bc^2 \,.$$

Since $3d^2=c^2$ has no rational roots we get that $b=0$.

It follows that

$$2=a^2+5c^2+15d^2 \,.$$ $$5cd =0 \,.$$ $$ac=0 \,.$$ $$ad=0 \,.$$

From the last two equations it follows that two of $a,c,d$ must be $0$, and then you get from the first equation that $x^2 \in \{ 2, \frac{2}{5}, \frac{2}{15} \}$ where $x$ is the nonzero one... Contradiction with $x \in Q$.