# How are unramified extensions of number fields formed?

Every extension is formed by adjoining a root of a polynomial. E.g.:

• Totally ramified = root of Eisenstein polynomial.

• Unramified over a local field = root of cyclotomic polynomial.

• What about unramified over a number field?

• What about tamely ramified extensions?

I recall reading both of these in an algebraic number theory book, in the middle of chapter 2 or so, but I don't remember which one. :( I think Cassels or Frohlich was one of the authors.

• In the local setting, an extension $L/K$ is totally and tamely ramified of degree $n$ if and only if $\mu_n\subset K$ and $L = K(\sqrt[n]{\pi_K})$ where $\pi_K$ is a uniformiser, and $\mu_n$ are the $n^{\text{th}}$ roots of unity. – Mathmo123 Jan 8 '16 at 12:13

There aren't too many unramified extensions of a number field. In fact, if $K$ is a number field and $K^{unr}$ is the maximal unramified extension of $K$, then $G(K^{unr}/K)\simeq Cl(K)$, where $Cl(K)$ is the ideal class group of $K$. So, for instance, there are no unramified extensions of $\mathbb{Q}$. The inverse of this map behaves like $(\mathbb{Z}/n\mathbb{Z})^{\times}\rightarrow G(\mathbb{Q}(\zeta_n)/\mathbb{Q})$, in that we send the equivalence class of a prime $[\mathfrak{p}]$ into the Frobenius of $\mathfrak{p}$. This means that a prime will split completely in $K^{unr}$ if and only if it is principle. It is possible to compute these fields, and sometimes it's easy.
If we're okay with some ramification, say ramification at all primes dividing the ideal $\mathfrak{m}$, then we can create the Ray Class Group of $\mathfrak{m}$, denoted $Cl_{\mathfrak{m}}(K)$ which is like the class group but it avoids the primes that divide $\mathfrak{m}$. We can then use a similar idea to create all extensions that are only ramified at primes dividing $\mathfrak{m}$. Ie, there is a field $K^{unr}_{\mathfrak{m}}$ that is the maximal field unramified at every prime not dividing $\mathfrak{m}$, we call this the Ray Class Field of $\mathfrak{m}$. If a prime power divides $\mathfrak{m}$, then we can control ramification at that prime.
There's a little more we can do. Sometimes a field can split over the real/complex numbers. We can control that by adding an additional, artificial term to $\mathfrak{m}$, making it a Modulus. We can then get Ray Class Fields of this modulus as well.
• You are missing the adjective "abelian" throughout your answer. For example, the maximal unramified extension of an arbitrary number field can have degree much larger than $h_K$, although $\mathbf{Q}$ still has no unramified extensions. Also, every abelian extension is contained in a ray class field, but does not need be one. – Brandon Carter Jan 7 '16 at 23:53