I want to prove that $\mathbb{Q}(\sqrt p_{1}+\sqrt p_{2})=\mathbb{Q}(\sqrt p_{1},\sqrt p_{2})$ for $p_{1},p_{2}$ primes. I know this was proved before for the generalized case of $p_{1},....,p_{n}$ primes here:

How to prove that $\mathbb{Q}[\sqrt{p_1}, \sqrt{p_2}, \ldots,\sqrt{p_n} ] = \mathbb{Q}[\sqrt{p_1}+ \sqrt{p_2}+\cdots + \sqrt{p_n}]$, for $p_i$ prime?

And every proof of this kind of equality involves Galois theory but I still don't know Galois theory and I'm supposed to prove this with only basic Field theory(field extensions, irreducible polynomials, algebraic extensions, etc).

By definition $\mathbb{Q}(\sqrt p_{1}+\sqrt p_{2})$ is the smallest field containing $\mathbb{Q}$ and $\sqrt p_{1}+\sqrt p_{2}$, also $\mathbb{Q}(\sqrt p_{1},\sqrt p_{2})$ is the smallest field containing $\sqrt p_{1}$ and $\sqrt p_{2}$. Or

$$\mathbb{Q}(\sqrt{p_{1}},\sqrt{p_{2}})=\{a+b\sqrt{p_{1}}+c\sqrt{p_{2}}+d\sqrt{p_{1}p_{2}} \mid a,b,c,d\in\mathbb{Q}\}$$

$$\mathbb{Q}(\sqrt{p_{1}}+\sqrt{p_{2}}) = \lbrace a+b(\sqrt{p_{1}}+\sqrt{p_{2}}) \mid a,b \in \mathbb{Q} \rbrace $$

Also I know that $[\mathbb{Q}(\sqrt p_{1},\sqrt p_{2}):\mathbb{Q}]=4$ from this:

Proving that $\left(\mathbb Q[\sqrt p_1,\dots,\sqrt p_n]:\mathbb Q\right)=2^n$ for distinct primes $p_i$.

Still don't know how to proceed proving this two extensions are the same.



Step 1 : $$\sqrt{p_1}-\sqrt{p_2}=\frac{p_1-p_2}{\sqrt{p_1}+\sqrt p_2}\in \mathbb Q(\sqrt{p_1}+\sqrt{p_2})$$

Step 2 :

$$\sqrt{p_1}=\frac{(\sqrt{p_1}+\sqrt p_2)+(\sqrt{p_1}-\sqrt{p_2})}{2}\in \mathbb Q(\sqrt{p_1}+\sqrt{p_2}).$$

I let you manage to show that $\sqrt{p_2}\in \mathbb Q(\sqrt{p_1}+\sqrt{p_2})$ as well and conclude the equality.

  • 1
    $\begingroup$ @Cos We can view this a bit more conceptually - see my answer. $\endgroup$ – Bill Dubuque Apr 9 at 18:17

Hint $ $ If field F has $\,2\,$ F-linear independent combinations of $\rm\, \sqrt{a},\ \sqrt{b}\, $ then we can solve for $\rm\, \sqrt{a},\ \sqrt{b}\, $ in F. For example, the Primitive Element Theorem works that way, obtaining two such independent combinations by Pigeonholing the infinite set $\rm\ F(\sqrt{a} + r\ \sqrt{b}),\ r \in F,\ |F| = \infty,\,$ into the finitely many fields between F and $\rm\ F(\sqrt{a}, \sqrt{b}),\,$ e.g. see this proof.

In this case it is simplest to notice $\rm\ F = \mathbb Q(\sqrt{a} + \sqrt{b})\ $ contains the independent $\rm\ \sqrt{a} - \sqrt{b}\ $ since

$$\rm \sqrt{a}\ -\ \sqrt{b}\ =\ \dfrac{\ a\,-\,b}{\sqrt{a}+\sqrt{b}}\ \in\ F = \mathbb Q(\sqrt{a}+\sqrt{b}) $$

To be explicit, note that $\rm\, u = \sqrt{a}+\sqrt{b},\ v = \sqrt{a}-\sqrt{b}\in F\, $ so solving the linear system for the roots yields $\rm\, \sqrt{a}\ =\ (u+v)/2,\ \ \sqrt{b}\ =\ (u-v)/2,\, $ both of which are clearly $\rm\,\in F,\,$ since $\rm\,u,v\in F\,$ and $\rm\,2\ne 0\,$ in $\rm\:F,\:$ so $\rm\,1/2\:\in F.\,$ This works over any field where $\rm\,2\ne 0,\,$ i.e. where the determinant (here $\,2)\,$ of the linear system is invertible, i.e. where the linear combinations $\rm\:u,v\:$ of the square-roots are linearly independent over the base field.

  • $\begingroup$ The property in the OP is a general one, see math.stackexchange.com/a/2457638/300700 $\endgroup$ – nguyen quang do Apr 10 at 21:27
  • $\begingroup$ @nguyenquangdo Yes, there are various ways to prove these and related results using field theory, but most are beyond the knowledge of users who ask these types of questions. $\endgroup$ – Bill Dubuque Apr 10 at 21:32

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