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?
Number of Q-embeddings into C
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0Oop, 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. – 2012-11-18
1 Answers
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 $\mathbb{C}$.