For 3, as you say there is a nilpotent element in the tensor product.
This cannot happen in a product of fields. If we have $R=F_1\times\cdots\times F_n$,
a product of fields, and $(a_1,\ldots,a_n)^2$ is zero in $R$ then $a_i^2=0$ in each $F_i$ so that $a_i=0$ (as $F_i$ is a field). In this example, then the tensor product is
not a product of fields.
For field extensions $F_1/F$, $F_2/F$ the tensor product $F_1\otimes_F F_2$ can only fail to be a product of fields when both $F_1/F$ and $F_2/F$ are inseparable extensions, and that is precisely the case here.
But in case 4, you have separable extensions. Indeed $F_1/F$ is a Kummer extension
here as $F=\Bbb F_4(T)$ has three cube roots of unity: $1$, $\omega$ and $\omega^2$.
Then
$$F_1\otimes _F F_1\cong F_1\times F_1\times F_1$$
via
$$a\times b\mapsto (ab,a\sigma(b),a\sigma^2(b))$$
where $\sigma:F_1\to F_1$ is the automorphism taking $\sqrt[3]T$ to $\omega\sqrt[3]T$.
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.
Best Answer
$(1,0)$ isn't a multiplicative identity because it doesn't work. For instance, $(1,0) \cdot (0,1) = (0,0)$, not $(0,1)$.
Direct products of fields are never fields (if you require $0 \neq 1$ for a field, which most sensible people do.) The reasons are plentiful: zero divisors will exist and not everyone is invertible.
Your confusion might lie here: $\mathbb{C}$ may be thought of as $\mathbb{R} \times \mathbb{R}$ either as a vector space over $\mathbb{R}$ or as a topological space. In these senses it IS a cross product, but it is NOT so as a ring.