Discrete non archimedean valued field with infinite residue field

Solution 1:

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.

Solution 2:

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.

Solution 3:

You can take the 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.