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This question already has an answer here:

This question is in regards to separable field extensions. I am to show that this $\alpha$ is in the given field and verify by direct computation that the given generators for the extension of $\mathbb{Q}$ can indeed by expressed as formal polynomials in $\alpha$ with coefficient in $\mathbb{Q}$.

For $\mathbb{Q}(\sqrt2, \sqrt3)$, the given answer is as follows :

"We try $\alpha = \sqrt2 + \sqrt3$. Squaring and cubing, we find $\alpha^2 = 5 + 2\sqrt2\sqrt3$ and $\alpha^3 = 11\sqrt2 + 9\sqrt3$.

Because $\sqrt2 = \frac{\alpha^3 - 9\alpha}{2}$ and $\sqrt3 = \frac{11\alpha - 9\alpha^3}{2}$, we see that $\mathbb{Q}(\sqrt2 + \sqrt3)$ = $\mathbb{Q}(\sqrt2, \sqrt3)$"

I'm trying to understand the logical jump from knowing the $\alpha^2$ and $\alpha^3$ to the value of $\sqrt2 = \frac{\alpha^3 - 9\alpha}{2}$ and $\sqrt3 = \frac{11\alpha - 9\alpha^3}{2}$. Also, how is this relevant to the question and separable fields?

Thank you in advance.

(It isn't a duplicate, I believe. I'm trying to draw a connection between $\alpha^2$ and why $\alpha^3$ was even used in the first place when the the irreducible is of degree 2.)

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marked as duplicate by Dietrich Burde, Elaqqad, Ayman Hourieh, Aaron Maroja, Chappers Apr 9 '15 at 0:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Also, every separable finite degree field extension has a primitive element $\alpha$, see here. $\endgroup$ – Dietrich Burde Apr 8 '15 at 18:01
  • $\begingroup$ It isn't a duplicate, I believe. I'm trying to draw a connection between $\alpha^2$ and why $\alpha^3$ was even used in the first place when the the irreducible is of degree 2. $\endgroup$ – Squires McGee Apr 8 '15 at 18:11
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    $\begingroup$ We know that $\mathbb{Q}(\sqrt{2} + \sqrt{3}) \subset \mathbb{Q}(\sqrt{2},\sqrt{3})$. We need to show the other way containment too for which it is enough to show $\sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3}) $ and also for $\sqrt{3}$. By expressing each of these element as a polynomial expression in $\alpha$,we see that indeed both of these elements lie in the required field. $\endgroup$ – user2902293 Apr 8 '15 at 18:21
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    $\begingroup$ I agree, not a duplicate. You are trying to understand a particular proof you read, but the answers to math.stackexchange.com/q/93463/139123 use different proofs. However, you should still look at them, as they show a different insight. $\endgroup$ – David K Apr 8 '15 at 21:08
  • $\begingroup$ Thanks a lot, David. I did go through several of them. $\endgroup$ – Squires McGee Apr 8 '15 at 22:45
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The general principle is this: if $K \supseteq F$ are fields and $\alpha\in K$ then $F(\alpha) \subseteq K$, because $F(\alpha)$ is the smallest field containing $F$ and $\alpha$.

So, since $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ contains both $\sqrt{2}$ and $\sqrt{3}$, it contains their sum, so it contains $\mathbb{Q}(\sqrt{2}+\sqrt{3})$.

To show the other inclusion, it follows from the "general principle" (and the fact that $F(\alpha, \beta) = F(\alpha)(\beta)$) that it suffices to show that $\sqrt{2}$ and $\sqrt{3}$ are elements of $\mathbb{Q}(\sqrt{2}+\sqrt{3})$. Letting $\alpha = \sqrt{2}+\sqrt{3}$, we seek to express $\sqrt{2}$ and $\sqrt{3}$ as rational functions of $\alpha$. We compute its powers, starting with $$\alpha^2 = 5+2\sqrt{2}\sqrt{3}.$$ We can't isolate either surd from this so we go on to $$\alpha^3 = 11\sqrt{2} + 9\sqrt{3}.$$ Look! That has $11$ root twos in it. So $$\alpha^3 - 11\alpha = -2\sqrt{3}$$ from which we obtain the expression for $\sqrt{3}$. Similarly, $$\alpha^3 - 9\alpha = 2\sqrt{2}.$$

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