I've been studying polynomial rings for a few weeks now and I've hit a bit of a stump. Originally, I was interpreting $\mathbb{Q}(\sqrt{a} + \sqrt{b}) = \{x + y(\sqrt{a} + \sqrt{b}) |x,y \in \mathbb{Q}\}$, but I've come to the realization that this may be incorrect. I was using this idea in attempts to demonstrate that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$, but I'm now seeing that my interpretation must be false since I seem to be encountering a contradiction using this method. Is there a good way to interpret this field with set builder notation? If there is not, or if there is a better way I ought to interpret the field without set notation, what would that be?
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1Related. – Git Gud Jun 05 '18 at 00:20
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1Do you have notes/a book/a website where you can look up the definition of that notation for (an algebraic) field extension? I feel like the precise definition might clear this up. – Mark S. Jun 05 '18 at 00:22
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If you look, you will see that $(\sqrt 2+\sqrt 3)^2$ cannot be expressed in the form you wrote down.
As part of showing $\mathbb Q(\sqrt 2 +\sqrt 3) = \mathbb Q(\sqrt 2, \sqrt 3),$ you will see that the degree of the extension is $4.$ The elements can be represented as $$ a+b\sqrt 2+c\sqrt 3 + d\sqrt 6$$ for $a,b,c,d\in \mathbb Q.$
spaceisdarkgreen
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1FTR, it is believable but a bit ugly to show that $[\Bbb Q(\sqrt p,\sqrt{p_1},\ldots,\sqrt{p_n}):\Bbb Q(\sqrt{p_1},\ldots,\sqrt{p_n})]=2$ whenever $p\notin {p_1,\ldots, p_n}$, for primes $p,p_i$. – pancini Jun 05 '18 at 00:13
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@ElliotG $\Bbb Q^\times$ is $\Bbb {\pm 1} \oplus \bigoplus p^{\Bbb Z}$ by unique factorization which directly implies that a set of distinct primes will be linearly independent in $\Bbb Q^\times/(\Bbb Q^\times)^2$, so the result follows from Kummer theory, which also gives you the Galois group. (I know that uses a lot more than you need, but at least you don't have to do ugly computations) – Lukas Heger Jun 05 '18 at 02:04
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$F(u)$ normally means the (nonsingular) rational expressions of $u$ with coefficients in $F$ (compare $F[u]$, the polynomials in $u$ with coefficients in $F$.
If $u$ is the root of an irreducible polynomial of degree $d$ with coefficients in $F$, one can in fact show that $F(u) \cong F[u]$, essentially by exploiting the division algorithm. Indeed, one easily shows that there is a unique expression $a_0+a_1u+\dotsb+a_{d-1}u^{d-1}$, which reduces the problem to one of finding expressions in a finite-dimensional vector space over $F$, which is much easier than the general case.
But essentially we can boil it down to $$ (\sqrt{2}+\sqrt{3})^3 = 11\sqrt{2}+9\sqrt{3}, $$ so you can find rationals $a,b$ for which $a(\sqrt{2}+\sqrt{3})^3+b(\sqrt{2}+\sqrt{3})$ is equal to either $\sqrt{2}$ or $\sqrt{3}$.
Chappers
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