This is a follow-up to my previous question on why the direct product $\mathbb{R}\times\mathbb{R}$ is not well-defined, where a proof was given to show that there is no way to construct it without reaching a contradiction. However, to me it seems like the main method in the proof could be used to prove something more general:
"Given a field $k$, let us assume that the direct product $k\times k$ is well-defined in the category of fields, with canonical projections $\pi_1$ and $\pi_2$ (which are automatically injective since morphisms in the category of fields are always injective). By the universal property defining $k\times k$, there exists a morphism $\Delta:k\rightarrow k\times k$ that is the inverse of both of them, implying that $\pi_1=\pi_2$, and hence we may set $k\times k=k$, with $\pi_1=\pi_2$ being the identity map on $k$, without loss of generality. Moreover, using the universal property once more, we see that this immediately implies that for any field $k'$, there is at most one morphism $k'\rightarrow k$."
Right now I don't see any error in the above line of reasoning, and if I'm correct we can make the following conclusions:
Let $k$ be a field. If $k\times k$ is well-defined as a field, then
- $k\times k=k$, and
- For every field $k'$, there is at most one morphism $k'\rightarrow k$.
However, if we make the assumption that for every field $k'$, there is at most one morphism $k'\rightarrow k$, then if I'm not mistaken we can pretty easily just define the direct product $k\times k$ to be $k$ without any issue. As a result, we would get the following.
Let $k$ be a field. Then the following are equivalent:
- $k\times k$ is well-defined in the category of fields.
- For every field $k'$ there is at most one morphism $k'\rightarrow k$.
From my previous question I was given that if $k=\mathbb{Q}$ then $k\times k$ is indeed $k$, but I'm not sure exactly how one would prove this. Thus, I have the following questions:
- Is my reasoning and conclusions about the direct product of fields correct?
- If so, how can we most easily prove that there is at most one morphism $k\rightarrow \mathbb{Q}$ for any field $k$?
- Are there any other well-known examples of fields that have a well-defined direct product (as a field)?
Best Answer
Given any fields $k,K$ and two distinct field inclusions $\phi_1,\phi_2:K\mapsto k,$ we can preclude that $k\times k$ exists.
This is because, if $k\times k$ exists, $\phi_1\times\phi_2:K\to k\times k.$ By your argument, since $\pi_1=\pi_2,$ we must get $$\phi_1=\pi_1\circ (\phi_1\times \phi_2)=\pi_2\circ(\phi_1\times \phi_2)=\phi_2,$$ contradicting that the $\phi_i$ are distinct.
Now, every field contains either $\mathbb F_p$ for some prime or $\mathbb Q.$
If $k\cong \mathbb F_p$ or $k\cong\mathbb Q,$ the only field maps $K\to k$ are isomorphisms, and the only automorphism is the identity, so $k\times k$ can be shown to exist.
If $k$ contains $\mathbb F_p$ but is not isomorphic, then there is an non-trivial endomorphism $k\to k$ defined as $\phi:\alpha\mapsto \alpha^p.$ It is non-trivial because $x^p=x$ can have at most $p$ roots, and all the elements of $\mathbb F_p$ are roots.
But this means we have two distinct morphisms $k\to k,$ with one the identity, the other $\phi.$
So this leaves fields which contain $\mathbb Q.$
We know that given any field containing $\mathbb Q$ and an element $\alpha$ transcendental over $\mathbb Q$ gives two morphisms:
$$\mathbb Q(x)\to k$$
one sending $x\mapsto \alpha,$ and one $x\mapsto \alpha^{-1}.$
So $k$ cannot contain any transcendental over $\mathbb Q.$
Also, if $\alpha$ is algebraic over $\mathbb Q,$ with minimal rational polynomial $p(x),$ then $\alpha$ can be the only root of $p(x)$ in $k.$ If there is another root $\beta\in k,$ then there are two inclusions $\mathbb Q[x]/\langle p(x)\rangle\to k,$ sending $x\mapsto \alpha,\beta.$
So we’ve reduced to cases like $k=\mathbb Q(\sqrt[3]2),$ where no algebraic conjugates of any element is contained in the field.
I think in these cases, we actually get $k\times k$ existing, because there is at most one morphism $K\to k$ for each field $K,$ and thus the axioms of the product are confirmed.
I’m not sure how to characterize these fields more precisely. My Galois theory is rusty - maybe there is a term for such a field.