Is there a universal property in this proposition (which regards field extensions)?

Question

Since learning a bit of category theory, I am trying, as an exercise, to state results that I come across in categorical language. I am trying to do this with the following:

Let $\mathbf{K},\mathbf{K'}$ be two fields, let $f:\mathbf{K}\to\mathbf{K'}$ be a ring homomorphism, and let $\alpha\in \mathbf{K}_a$ (where $\mathbf{K}_a$ denotes the algebraic closure of $\mathbf{K}$). Also, let $\mu$ denote the minimal polynomial of $\alpha$ over $\mathbf{K}$. If $\beta$ is a root of $f(\mu)$ in $\mathbf{K}_a'$, then there exists a unique extension $\phi:\mathbf{K}(\alpha)\to \mathbf{K}_a'$ of $f$ such that $\phi(\alpha)=\beta$.

This looks like some kind of universal property, but I find it difficult to see exactly what happens here (if, of course, there is actually something happening in the categorical level).

Thanks for any help!

Answer 1

The universal property is that $K(\alpha)$ is the universal (initial) extension of $K$ together with a root of the minimal polynomial of $\alpha$. The algebraic closure has the property that every polynomial has a root, so it is in particular such an extension. (One might think it's the algebraic closure that's supposed to have a universal property here, but this isn't the case.)

Answer 2

Recall $K(\alpha)=K[x]/(f)$ and that polynomial algebras have a universal property (freely adjoin an element) and quotient rings have a universal property (kill elements in the ideal). It follows that for any $K$-algebra $R$ we have a bijection between $K$-algebra homomorphisms $K(\alpha) \to R$ and roots of $f$ in $R$. And yes, this is a universal property, because universal properties are basically just descriptions of representable functors, in this case $\hom_{K\mathsf{Alg}}(K(\alpha),-)$.