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Stream: learning: questions

Topic: Integers and rational numbers as strict initial objects?

Madeleine Birchfield (Feb 22 2024 at 03:14):

Are the rational numbers a strict initial object in the category of ordered fields and strictly monotonic field homomorphisms? Similarly, are the integers a strict initial object in the category of ordered integral domains and strictly monotonic ring homomorphisms?

John Baez (Feb 22 2024 at 03:38):

How about this argument to show the first one:

Any ordered field $K$ has characteristic zero, and thus there is a unique ring homomorphism $f: \mathbb{Q} \to K$, since we must have $f(m/n) = f(m)/f(n)$ and $f(n)$ and $f(m)$ must be sums of $n$ (resp. $m$) copies of $f(1)$, and $f(1) = 1 \in K$.

John Baez (Feb 22 2024 at 03:41):

Thus the rational numbers are the initial object in the category of ordered fields and ring homomorphisms. But I claim $f$ above is order preserving (I'll try to think of a proof if you want), so the rationals are also initial in the category of ordered fields and strictly monotonic field homomorphisms.

(By the way, nobody talks about "field homomorphisms" because the only reasonable concept of a field homomorphism is just a ring homomorphism between fields.)

John Baez (Feb 22 2024 at 03:43):

Now, to show $\mathbb{Q}$ is a strict initial object we must show that given an ordered field $K$ and a strictly monotonic field homomorphism $g: K \to \mathbb{Q}$ this homomorphism must be an isomorphism.

John Baez (Feb 22 2024 at 03:46):

Every homomorphism of fields is one-to-one since its kernel must be trivial since a field has no ideals except $\{0\}$ and the whole field, and $1$ cannot be in the kernel.

John Baez (Feb 22 2024 at 03:47):

So $g$ is one-to-one, so we can think of $\mathbb{Q}$ as a subfield of $\mathbb{Q}$. So we need to show the only subfield of $\mathbb{Q}$ is $\mathbb{Q}$ itself.

John Baez (Feb 22 2024 at 03:48):

But such a subfield must contain $1$, and by addition also all integers $n$, and by division all rationals. So yes, every field homomorphism $g: K \to \mathbb{Q}$ is an isomorphism.

John Baez (Feb 22 2024 at 03:56):

My general impression from all this is that the ordering doesn't help us much beyond the fact that ordered fields have characteristic zero. The bulk of the argument is to show that $\mathbb{Q}$ is strictly initial in the category of fields of characteristic zero and ring homomorphisms.

Then we have an extra job, which I didn't actually do: to show that if $f: \mathbb{Q} \to K$ is a ring homomorphism and $K$ is an ordered field, then $f$ is order-preserving. Since $f$ is one-to-one it's a ring isomorphism between $\mathbb{Q}$ and $\mathrm{im}(f) \subseteq K$, which is a subfield of $K$ and thus an ordered field in its own right. So we just need to show that the axioms for an "ordered field" are strong enough to uniquely specify the usual ordering on $\mathbb{Q}$: that is, there's just one way to make $\mathbb{Q}$ into an ordered field. And this is very believable, since if there were another way we would have heard about it. :upside_down:

Madeleine Birchfield (Feb 22 2024 at 04:52):

My general impression from all this is that the ordering doesn't help us much beyond the fact that ordered fields have characteristic zero.

It does matter for ordered integral domains though, because Kevin Arlin showed in Ring homomorphisms between integral domains that there exist ordered integral domains with a weakly monotonic ring homomorphism to the integers. So the strictly monotonic requirement matters here, which requires the strict ordering to be defined.

James Deikun (Feb 22 2024 at 04:55):

I think though that it only or mostly matters because it forces all the morphisms to be one-to-one.

Madeleine Birchfield (Feb 22 2024 at 05:11):

So this is really a question about integral domains with characteristic zero and injective ring homomorphisms, of which ordered rings and strictly monotonic ring homomorphsims are only a special case.