# 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.

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