Unramified Extensions of Local Fields

Recall that every finite extension L of a local field K admits a unique extension of the valuation on K, with respect to which L is complete. That makes L a local field.

Now, the discrete valuation ring of K, R, has a unique prime ideal P. Then L is called unramified if P doesn’t ramify in the integral closure of R in L; otherwise, it’s called ramified. It’s fairly straightforward to show that n = ef, where e is the ramification index of P (i.e. the highest power of the prime of L that divides P), and f is something called the inertial degree that measures the enlargement of the quotient ring of R by P. Since P is maximal, R/P is a field; if S is the integral closure of R in L and Q is the prime of S, then S/Q is a field extension of R/P of degree f.

As usual, unramified extensions are nice in a very precise sense. For a start, suppose that K is one of the base local fields, i.e. F((x)) where F = F(p), or Q(p). Then R is F[[x]] or Z(p) and P is (x) or (p), and the quotient ring is F(p).

I’m going to handwave another result from Galois theory, namely that for each integer f, there’s exactly one field extension of F of degree f. Further, that extension, call it F(p^f), has multiplicative group C(p^f – 1), so it’s generated by a primitive (p^f – 1)st root of unity. So all finite extensions of F are generated by roots of unity of order coprime to p.

The thing about unramified extensions of local fields is that they satisfy the exact same condition. To see why, first suppose that two extensions of K = F((x)) or Q(p), L and L’, are unramified and have quotient rings S/Q and S’/Q’. Suppose further that S/Q = S’/Q’, and g is a suitable isomorphism. If w generates S/Q over F(p), then w‘ = g(w) generates S’/Q’.

Let h be any monic polynomial that reduces mod P to the minimal polynomial of w over F(p). Obviously, in L, h has a root mod P, so by Hensel’s lemma, h has a root, say r. Clearly, r generates L over K. Similarly, L’ is generated by r‘, whose minimal polynomial reduces to this of w‘. So we can define a ring homomorphism G from L to L’ by fixing all elements of K and sending r to r‘. It’s not especially difficult to see that G is an isomorphism, so L = L’.

This theorem requires L to be unramified, because otherwise P is not prime in L, which makes Hensel’s lemma inapplicable.

Conversely, if F(p^f) is an extension of F(p), then let w generate F(p^f) and let h‘ be the minimal polynomial of w. If h reduces mod P to h‘, then extracting a root of h will yield an extension of K of degree f.

In slightly less technical language, the unramified extensions of Q(p) and F((x)) are all generated by roots of unity of order coprime to p.

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