Fix a prime number $p$. In ZFC does there exist a perfect field of characteristic $p$ of any infinite cardinality? I know some constructions of fields of characteristic $0$ of arbitrary cardinality but not of positive characteristic.
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Here's an overkill response:
The Lowenheim-Skolem theorem tells us that there are models of the theory $T_{perf, p}$ of characteristic $p$ perfect fields of every infinite cardinality. Now a field of characteristic $p$ is perfect iff every element of the field is a $p$th power, and this is clearly a first-order condition, so every model of $T_{perf, p}$ is in fact a perfect field of characteristic $p$.
As a less silly answer, just note that the perfect closure of an infinite field $k$ has the same cardinality as $k$.
Given that that's obviously the right answer, why did I bother with the LS-approach? Well, LS is a very useful hammer to have - in "equational" situations it's more-or-less pointless since there's usually a straightforward "closure" construction (or similar) which does the job, but with more complicated properties it lets one address "coarse" set-theoretic questions without having to dive into the messy details, at least right away.
Also it's funny.
answered Aug 19 '19 at 15:20
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$\begingroup$ any chance to give an explicit construction? $\endgroup$ – user693936 Aug 19 '19 at 15:21
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$\begingroup$ @hello We can just take an arbitrary infinite field and "close under $p^n$th roots." This doesn't blow up the cardinality at all. $\endgroup$ – Noah Schweber Aug 19 '19 at 15:23
