0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

Hint

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.

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

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.