# Prime extensions on $k=\mathbb{Q}(\alpha)$ with $\alpha$ is a root of $x^4-14$

I have been working on my homework like this:

Let $k=\mathbb{Q}(\alpha)$ with $\alpha$ a root of $f(x)=x^4-14$.

(1) Show that the prime $11$ has three extensions to prime $\mathfrak{p}_1,\mathfrak{p}_2,\mathfrak{p}_3$ of $k$ and $k_{\mathfrak{p}_1}=k_{\mathfrak{p}_2}=\mathbb{Q}_{11}$ while $[k_{\mathfrak{p}_3}:\mathbb{Q}_{11}]=2.$

(2) The prime $13$ has four extensions to prime of $k$.

(3) Show that prime $5$ has two extensions to primes $\mathfrak{p}_1,\mathfrak{p}_2$ of $k$ and $[k_{\mathfrak{p}_1}:\mathbb{Q}]=[k_{\mathfrak{p}_2}:\mathbb{Q}]=2.$

I am comfused why the local fields $k_{\mathfrak{p}_1},k_{\mathfrak{p}_2}$ and $k_{\mathfrak{p}_3}$ appear here and how to compute the extensions.

Every comments will be appreciated. Thanks in advance!

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I doubt that the extension might not be Galoisian. –  Qiang Zhang Jun 26 '12 at 3:07
You're not being asked to compute completions. The only aspect of the completions you're really being asked about is their degree over the subfield ${\mathbf Q}_p$, and that degree is the product of the ramification index and residue field degree. The ram. index and res. field degree of all primes in ${\mathcal O}_k$ lying over a prime $p$ can be read off from the way $p$ factors into prime ideals in ${\mathcal O}_k$. Any prime number $p$ besides 2 or 7 is unramified in $k$ (why?) and thus the shape of its prime ideal factorization in $k$ can be read off from how $x^4 - 14$ factors mod $p$. –  KCd Jun 26 '12 at 3:30
@KCd, thanks a lot for the hint! I have another question posted here, see if you are interested. –  Qiang Zhang Jun 26 '12 at 14:04
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