# In ZFC there exists a perfect field of given positive characteristic and given infinite cardinality

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.

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.

• any chance to give an explicit construction? – user693936 Aug 19 '19 at 15:21
• @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. – Noah Schweber Aug 19 '19 at 15:23