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Let $\mathbb{K}$ a field with a discrete non archimedean absolute value and $A$ its ring of integers. Then $A$ is Noetherian, every ideal is generated by a power of $\pi\in A$, a uniformizer, and its unique maximal ideal is $\mathfrak{M}=\pi A$.

For instance, $\mathbb{Q}_p$ is such a field. Its residue field is $A/\mathfrak{M}\cong\mathbb{Z}/p\mathbb{Z}=\mathbb{F}_p$, the finite field of $p$ elements.

Is there any field $\mathbb{K}$ whose residue field is not finite?

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Yes, you can form the maximal unramified extension of $\Bbb Q_p$. Its value group is the same as that of $\Bbb Q_p$, that is, $p$ is still a generator of the maximal ideal. If you want an “explicit” construction, just adjoin the $m$-th roots of unity for all $m$ prime to $p$. Slightly more efficiently, adjoin all $(p^r-1)$-th roots of unity. You see that the residue field is an algebraic closure of $\Bbb F_p$.

More generally, any finite extension of $\Bbb F_p$ can be lifted to an unramified extension of $\Bbb Q_p$, preserving the Galois group, which will be generated by the Frobenius automorphism of the field in characteristic zero.

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  • $\begingroup$ True! I had that example in mind but I didn't realize. Thanks! $\endgroup$ – Angelo Rendina Dec 30 '14 at 14:07
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Let $K = k(t)$, the field of rational functions over a field $k$. We can define a valuation $v_0$ on $K$ via $$v_0\left(t^n \frac{p(t)}{q(t)}\right) = n$$where $p,q \in k[t],\text{with} \ p(0), q(0) \ne 0$ - ie $v_0$ is the order of the zero/pole at $0$. This is a discrete valuation, with integer ring $$A = \left\{f(t) = \frac{p(t)}{q(t)}:q(0) \ne 0\right\}$$with maximal ideal $$\mathfrak M=\{f(t):f(0) = 0\}$$Then we have $k \cong A/\mathfrak M$ via $f\mapsto f(0)$. Since $k$ was an arbitrary field, it can be infinite.

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  • $\begingroup$ This response is perhaps better and more illuminating than mine. $\endgroup$ – Lubin Dec 30 '14 at 14:12
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    $\begingroup$ @Lubin : your answer presents a field which characteristic differs from the one of its residue class field, while Mathmo123's answer shows a field which characteristic equals the characteristic of its residue class field. Each example complement the other. Both are illuminating. $\endgroup$ – Chilote Nov 18 '16 at 20:08
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You can take ring of formal power series $K[[X]]$ over an infinite field $K$. It is a local ring, with maximal ideal generated by $X$, so that its residue field is (isomorphic to) $K$. The valuation of a power series is its order and it's easy to check it is non-archimedean.

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  • $\begingroup$ For anyone who doesn't see this, in this case there is an underlying field as well, which is the ring of Laurent series $K((X))$, which is by definition $\{ \sum_{i \geq n} a_i X^i: n \in \mathbb{Z}, a_i \in K \}$, with addition and multiplication defined analogously to $K[[X]]$ (en.wikipedia.org/wiki/Formal_power_series#Formal_Laurent_series). The valuation of a formal Laurent series is similarly its order, i.e. the smallest integer $i$ for which $a_i \not= 0$. Then $K[[X]]$ is its local ring, and Bernard's answer completes the requirements of the question. $\endgroup$ – Anamay Chaturvedi Apr 6 '17 at 16:41
  • $\begingroup$ @Anamay Chaturvedi:: Taken the other way, the field of Laurent series is an explicit description of the field of fractions of $K[[X]]$. $\endgroup$ – Bernard Apr 6 '17 at 17:01

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