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Why the category of Field has no initial object? (see page $47$ of Horst Herrlich, George E. Strecker "Category Theory")

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Think about what the characteristic of an initial object must be. –  Qiaochu Yuan Nov 12 '11 at 21:52
Since there are no maps between fields of different characteristic, it is not very useful to lump them all together into a single category. –  GEdgar Nov 12 '11 at 22:49
If there were a field that mapped into all finite fields, what kind of problems might you have (especially since all those maps must be injective)? –  Jon Beardsley Nov 12 '11 at 23:21

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Note that in the category of all fields, given any field $\mathbf{F}$ there is always a different field $\mathbf{K}$ such that there are no field homomorphisms from $\mathbf{F}$ to $\mathbf{K}$ nor from $\mathbf{K}$ to $\mathbf{F}$. Therefore, there can be neither an initial nor a terminal object (because the former requires maps into every object, and the latter requires maps from every object).

And the reason this happens is that there can be no field homomorphisms between fields of different characteristics.

Now, if you change to the category of $\mathbf{Field}_{p}$, where $p$ is either a prime or $0$, and this is the category of all fields of characteristic $p$, then this category does have an initial object, namely the prime field of characteristic $p$: $\mathbf{F}_p$ for $p\gt 0$, and $\mathbb{Q}$ for $p=0$.

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Is there a nice functor from $\mathbf{Field}_p\to\mathbf{Field}_{p+1}$ that takes $\mathbb{F}_p$ to $\mathbb{F}_{p+1}$, and you could somehow (ignoring cardinality and universe issues) form the category of fields of characteristic zero as the colimit of these categories (I'm not even sure we have colimits in $\mathbf{Cat}$). –  Jon Beardsley Nov 12 '11 at 23:58
@JBeardz First, I assume you mean somehow go from $\mathbf{Field}_{p(k)}$ to $\mathbf{Field}_{p(k+1)}$, where $p(k)$ is the $k$th prime? (Otherwise, that only makes sense for $p=2$...). I don't think so, but I could be wrong. Note, though, that you can construct fields of characteristic zero with fields of finite characteristic: if you take a nontrivial ultraproduct of all the prime fields of positive characteristic, then you get a field of characteristic zero... (this was one of the questions in my Oral Qualifying...) –  Arturo Magidin Nov 13 '11 at 0:56
Thanks Arturo, yeah that's what I meant. Interesting. –  Jon Beardsley Nov 13 '11 at 2:15

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