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Let $\alpha $ be a root of the polynomial ${x^5} + 6{x^3} + 8x + 10$. How many $\mathbb{Q}$-embeddings of $\mathbb{Q}\left[ {\alpha ,\sqrt 7 } \right]$ (the least field extension of $\mathbb{Q}$ which contains elements $\alpha $ and $\sqrt 7 $) into $\mathbb{C}$ does there exist?

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    $\begingroup$ Certainly Eisenstein's criterion shows irreducibility of that quintic in $\mathbb Z[x]$. In the extension obtained by adjoining $\sqrt{7}$ the prime $2$ splits, since $3^2=2$ mod $7$. Thus (localizing if necessary to get a PID) Eisenstein still implies irreducibility of the quintic... $\endgroup$ – paul garrett Nov 18 '12 at 18:20
  • $\begingroup$ I must admit that i don't understand the answer, though I'd like to, since it seems to me that this particular approach would work in a wider set of problems than my solution. I think the one I managed to get is correct, though. I'd appreciate your comment if there is a flaw that I didn't notice $\endgroup$ – Alen Nov 18 '12 at 20:38
  • $\begingroup$ Oop, yes, you're right, I was too eager to do something a little more complicated... Your argument using the point that 2 does not divide 5 is right on the mark. The view I noted would still work without that non-divisibility property, if you understand how the prime used in the Eisenstein criterion splits in the extension. Thus, a little bit of "algebraic number theory" would be required, but/and is very helpful in understanding such examples. $\endgroup$ – paul garrett Nov 18 '12 at 23:09
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Since $p\left( x \right) = {x^5} + 6{x^3} + 8x + 10$ is irreducible over $\mathbb{Q}$ by Eisenstein criterion, $p$ is minimal polynomial for $\alpha $ over $\mathbb{Q}$, from which it follows that $\left[ {\mathbb{Q}\left[ \alpha \right]:\mathbb{Q}} \right] = 5$. Furthermore, $\sqrt 7 \notin \mathbb{Q}\left[ \alpha \right]$, otherwise we would have $\underbrace {\left[ {\mathbb{Q}\left[ \alpha \right]:\mathbb{Q}} \right]}_{ = 5} = \left[ {\mathbb{Q}\left[ \alpha \right]:\mathbb{Q}\left[ {\sqrt 7 } \right]} \right]\underbrace {\left[ {\mathbb{Q}\left[ {\sqrt 7 } \right]:\mathbb{Q}} \right]}_{ = 2}$ which is impossible. So, $\left[ {\mathbb{Q}\left[ {\alpha ,\sqrt 7 } \right]:\mathbb{Q}\left[ \alpha \right]} \right] \geqslant 2$. Also, ${x^2} - 7 \in \left( {\mathbb{Q}\left[ \alpha \right]} \right)\left[ x \right]$ (polynomials with coefficients in ${\mathbb{Q}\left[ \alpha \right]}$) which gives us $\left[ {\mathbb{Q}\left[ {\alpha ,\sqrt 7 } \right]:\mathbb{Q}\left[ \alpha \right]} \right] \leqslant 2 \Rightarrow \left[ {\mathbb{Q}\left[ {\alpha ,\sqrt 7 } \right]:\mathbb{Q}\left[ \alpha \right]} \right] = 2$. We conclude that $\left[ {\mathbb{Q}\left[ {\alpha ,\sqrt 7 } \right]:\mathbb{Q}} \right] = \left[ {\mathbb{Q}\left[ {\alpha ,\sqrt 7 } \right]:\mathbb{Q}\left[ \alpha \right]} \right]\left[ {\mathbb{Q}\left[ \alpha \right]:\mathbb{Q}} \right] = 2 \cdot 5 = 10$.

So, there are 10 different $\mathbb{Q}$-embeddings of ${\mathbb{Q}\left[ {\alpha ,\sqrt 7 } \right]}$ into $\mathbb{C}$.

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