Definition 1. [Bourbaki] A splitting field of $f\in \Bbbk[x]$ is an extension $\Bbbk\subset \mathbb K$ which splits $f$ (into possibly repeated linear factors) and satisfies $\mathbb K=\Bbbk(Z(f,\mathbb K))$ where $Z(f,\mathbb K)$ is the set of roots of $f$ in $\mathbb K$.
Definition 2. A splitting field of $f\in \Bbbk[x]$ is a weakly initial object in the category of field extensions of $\Bbbk$ which split $f$.
We have $1\implies 2$ by mapping zeros to zeros. Conversely, suppose $\Bbbk\subset \mathbb K$ is a splitting field for $f\in \Bbbk[x]$. We wish to prove it is isomorphic over $\Bbbk$ to $\Bbbk(Z(f,\mathbb K))$. By assumption we have a field extension morphism $\mathbb K\subset \Bbbk(Z(f,\mathbb K))$. As a $\Bbbk$-algebra morphism out of a field, this is a $\Bbbk$-linear injection. Moreover, the target is a finite dimensional $\Bbbk$-linear space since $f$ has finitely many zeros. So the $\Bbbk$-algebra morphism is a $\Bbbk$-linear injection of $\Bbbk$-linear spaces of equal $\Bbbk$-dimension whence bijective and therefore a $\Bbbk$-algebra isomorphism.
The proof of $2\implies 1$ seems to fail for infinite families of polynomials, where we cannot know the extension generated by all their zeros is finite.
Question. What is an example where $2\implies 1$ fails? Is there any reason to take the seemingly weaker definition 2 in the case of infinitely many polynomials?
Best Answer
Let $\{f_i\}_{i\in I}$ be a family of polynomials in $\Bbb k[X]$. Let $\mathcal C$ be the category of field extensions of $\Bbb k$ in which all $f_i$ split into linear factors. Let $\Bbb K$ be weakly initial in $\mathcal C$. Let $Z=\{\,a\in\Bbb K\mid \exists i\in I\colon f(a)=0\,\}$. Let $\Bbb L=\Bbb k[Z]$. Then each $f_i$ splits in $\Bbb L$ because $\Bbb L$ contains the zeroes we know to be in $\Bbb K$. So $\Bbb L$ is an object in $\mathcal C$. Hence there exists a homomorphism $ \Bbb K\to\Bbb L$ (which is the identity on $\Bbb k$!). Together with the inclusion $\Bbb L\to \Bbb K$, this induces an endomorphism of $\Bbb L$, which again is determined by a permutation of $Z$ (which in fact permutes zeroes only per $f_i$). The inverse permutation does define an endomorphism of $\Bbb L$ as well, which is of course the inverse of the above (so it turns out to be an automorphism). We can use this to obtain a homomorphism $\Bbb K\to\Bbb L$ that is the identity on $\Bbb L$. As homomorphisms of fields are injective, we conclude that $\Bbb K=\Bbb L$.