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,

  • 1
    $\begingroup$ 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/… $\endgroup$ – ADF Mar 25 '12 at 2:43
  • $\begingroup$ 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. $\endgroup$ – MSE Apr 20 '17 at 14:59

Suppose by contradiction that

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

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 $x \in Q$.

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