Suppose I have a local field $K$ with discrete valuation $v$, and we denote $k$ to be its residue field. Furthermore, suppose $K$ and $k$ to be perfect. I am having trouble understanding the following two things.

1) How do we know that the algebraic closure $\bar{K}$ is also a local field?

2) And how does the Galois group for $\bar{K}/K$ act on $\bar{k}$, the residue field of $\bar{K}$.

Any comments or explanation is appreciated. Thank you very much.

PS Please ignore the first one as it is not true as explained in the comment.

  • $\begingroup$ The algebraic closure is not a local field, since its residue field is not finite. Is your question actually about what the valuation on $\overline K$ is? $\endgroup$ – Mathmo123 Dec 11 '17 at 19:11
  • $\begingroup$ Oh, I see. Let me fix the problem. But I guess it is a field with valuation? Then how does one defines the residue field here? $\endgroup$ – Johnny T. Dec 11 '17 at 20:43

The algebraic closure is not a local field since its residue field is not finite.

Here’s what is going on. We can define a valuation $\overline v$ on $\overline K$ as follows: for each $x\in \overline K$, $x$ must lie in some finite extension $L/K$. If $v_L$ is the unique extension of $v$ to $L$, we define $\overline v(x)$ to be $v_L(x)$. This gives a valuation $v:\overline K\setminus\{0\}\to \mathbb Q$, which is not discrete, which is another way of seeing that $\overline K$ is not a local field.

Let $\overline{\mathcal O}$ denote the ring of integers of $\overline K$ - i.e. the ring of elements of non-negative valuation. Then $\overline O$ is still a local ring, with unique maximal $\overline{\mathfrak p}$ consisting of the elements with positive valuation.

The Galois group $\mathrm{Gal}(\overline K/K)$ acts on $\overline {\mathcal O}$ and fixes $\overline{\mathfrak p}$. Hence, it acts on the residue field $\overline k = \overline{\mathcal O}/\overline{\mathfrak p}$.


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