Given f(x) in k[x], is it possible to find two field extensions K/k and K'/k such that f has two different factorizations as linear polynomials?

Question

By Kronecker's theorem, given a field $k$ and $f(x)$ a polynomial with coefficients in $k$, there exists a field $K$ containing $k$ as a subfield and with $f(x)$ a product of linear polynomials in $K[x]$; \begin{equation*} f(x)=a(x-\alpha_{1})\cdots (x-\alpha_{n}) \end{equation*} with $\alpha_{i} \in K$ for all $i$ and $a \in k$ (Rotman, Advanced Modern Algebra, p. 191).

Since $K$ is a field, $\alpha \in K$ is a root of $f$ (i.e. $f(\alpha)=0$) iff the polynomial $x- \alpha$ divides the polynomial $f(x)$ (in $K[x]$), then the above can be restated as "$K$ contains all the roots of $f$".

My question is if it can happen to exist a second field extension $K'/k$ such that $f$ has a factorization in $K'[x]$ of the form \begin{equation*} f(x)=b(x-\beta_{1})\cdots (x-\beta_{n}) \end{equation*} with $\beta_{j} \in K'$ for all $j$ and $b \in k$, with at least one $\beta_{j} \neq \alpha_{i} $ for some $i,j$.

if I'm not mistaken it would mean then that the polynomial $f$ has two different collections of $n$ roots.

[Obviously if $K \subseteq K'$ such thing cannot happen, since it would imply $f$ to be a polynomial with coefficients in $K´$ of degree $n$ with more than $n$ roots].

I started recently to study field extensions then I don't know if this is a silly question or not, sorry in advance if it is :-)

Answer 1

You can have $K$ and $K'$ constructed as distinct sets, and thus in the strictest sense no element of $K$ is equal to any element of $K'$.

However, the smallest subfield of $K$ that contains all the roots of $f$ (i.e. $k(\alpha_1,\ldots,\alpha_n)$) is necessarily isomorphic to the smallest subfield of $K'$ that contains all the roots of $f$ (i.e. $k(\beta_1,\ldots,\beta_n)$). And there is an isomorphism that sends each $\alpha_i$ to some $\beta_j$.

So up to isomorphism, there is only one splitting field of $f$.

Answer 2

Be careful when talking about whether two things are equal in different extensions. Of course, they might just be the same thing with different names so they are not strictly equal.

However, you can take the subfield generated by the roots. This is known as the splitting field and is unique up to isomorphism. See Theorem 3.13

Answer 3

In the join of $K''$ of $K$ with $K'$ we'd have two distinct factorisations of the same polynomial, a contradiction since $K''$[x] is a UFD. You can have two distinct factorisations in irreducible non-linear polynomials though, for intermediate fields, coming from distinct "rearrangements" of the linear factors.