Let X be a reduced algebraic space with 1 point. Then is it true that X is representable by the spectrum of a field?

The answer is no in general. The space [Spec(\bar{Q})/Gal(\bar{Q}/Q)] of Example Tag 02Z6 in the stacks project is a counter example.

Currently the best positive result (in the stacks project) is Lemma Tag 047Z which says that this holds when X is a decent algebraic space.

An algebraic space X is called *decent* if every point of X corresponds to a quasi-compact monomorphism Spec(k) —> X with k a field. This is not currently the definition in the stacks project, but it can be shown to be equivalent. It turns out that this is a convenient class of algebraic spaces to work with. It contains all schemes, all quasi-separated algebraic spaces, and likely all locally separated algebraic spaces (David Rydh, private communication). On the other hand a decent algebraic space is a bit like a schemeĀ and has “enough points” in some sense.

Because there are non-representable 1-point spaces it turns out that the notions of “radicial” and “universally injective” are not the same for morphisms of algebraic spaces. Namely, the morphism [Spec(\bar{Q})/Gal(\bar{Q}/Q)] —> Spec(Q) is universally injective but not radicial (with any reasonable definition of radicial I can think of). Again for decent morphisms this does not happen, see Lemma Tag 0484.