Irreducible polynomials of $\sqrt{3}+\sqrt{5}$ for $\mathbb{Q}(\sqrt{5})$ and $\mathbb{Q}(\sqrt{15})$

Question

Could someone please verify the answers for me? Or let me know how to compute these correctly?

Question: Determine the irreducible polynomials of $\sqrt{3}+\sqrt{5}$ for $\Bbb Q(\sqrt{5})$ and $\Bbb Q(\sqrt{15})$.

Take $\alpha = \sqrt{3}+\sqrt{5}$, I got that $\alpha$ satisfies $x^4 - 16x^2 + 4 = 0$.

So for $\Bbb Q(\sqrt 5)$, we get that $x^4 - 16x^2 + 4$ should be the irreducible polynomial because $\sqrt {15} = \sqrt{5}\cdot\sqrt{3}$ is not in $\Bbb Q(\sqrt{5})$ (espepcially since $\sqrt3$ not in $\Bbb Q(\sqrt5)$.

For $\Bbb Q(\sqrt{15})$: We can reduce the polynomial to $(x-(\sqrt5+\sqrt3))^2\cdot (x+(\sqrt5+\sqrt3))^2$ and this will be the irreducible polynomial because these are the linear factors and cannot be further reduced.

Thank you in advance for your comments.

Answer 1

Let $L=\Bbb{Q}(\alpha)=\Bbb{Q}(\sqrt3,\sqrt5)$ (ask, if you don't know how to show that these two fields are equal). Then the fields $K_1=\Bbb{Q}(\sqrt5)$ and $K_2=\Bbb{Q}(\sqrt{15})$ are quadratic subfields of $L$. This implies that $$ [L:K_1]=\frac{[L:\Bbb{Q}]}{[K_1:\Bbb{Q}]}=2, $$ and similarly $$ [L:K_2]. $$ As $L=K_1(\alpha)=K_2(\alpha)$ you can tell without any calculations that the minimal polynomials of $\alpha$ over either $K_1$ or $K_2$ are quadratic.

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This is also easy to see directly:

• Show that $\alpha^2\in K_2$ to find a quadratic polynomial with coefficients in $K_2$ that has $\alpha$ as a zero.
• Use the fact that $(\alpha-\sqrt5)^2=3$ to find a quadratic polynomial with coefficients in $K_1$ that has $\alpha$ as a zero.

The answers are: $x^2-8-2\sqrt{15}\in K_2[x]$ and $x^2-2\sqrt5x+2\in K_1[x]$.