I'm trying to show that the minimal polynomial of a linear transformation $T:V \to V$ over some field $k$ has the same irreducible factors as the characteristic polynomial of $T$. So if $m = {f_1}^{m_1} ... {f_n}^{m_n}$ then $\chi = {f_1}^{d_1} ... {f_n}^{d_n}$ with $f_i$ irreducible and $m_i \le d_i$.
Now I've managed to prove this using the primary decomposition theorem and then restricting $T$ to $ker({f_i}^{m_i})$ and then using the fact that the minimal polynomial must divide the characteristic polynomial (Cayley-Hamilton) and then the irreducibility of $f_i$ gives us the result.
However I would like to be able to prove this directly using facts about polynomials/fields without relying on the primary decomposition theorem for vector spaces. Is this fact about polynomials true in general?
We know that $m$ divides $\chi$ and so certainly $\chi = {f_1}^{m_1} ... {f_n}^{m_n} \times g$ but then how do we show that $g$ must have only $f_i$ as it's factors? I'm guessing I need to use the fact that they share the same roots. And I'm also guessing that it depends on $k$, i.e. if $k$ is algebraically closed then it is easy because the polynomials split completely into linear factors.
Help is much appreciated, Thanks
One can reason using only some field theory, rather than linear algebra, if one prefers. I will assume the Cayley-Hamilton theorem, and the fact that all eigenvalues are roots of the minimal polynomial$~\mu$, which holds because an eigenvector for$~\lambda$ is killed by $P[T]$ (if and) only if $P$ is a multiple of $X-\lambda$.
First observe that the result is true when $\chi$ splits over $k$ (and hence so does$~\mu$, which divides$~\chi$ by the Cayley-Hamilton theorem), since the factors $X-\lambda$ of either polynomial are precisely those with $\lambda$ an eigenvalue. This in particular takes care of the case where $k$ is algebraically closed; the remainder deals with the case where $\chi$ does not split into linear factors in $k[X]$.
If $K/k$ is a field extension, one can extend scalars to obtain $\def\ext{\otimes_kK}T\ext: V\ext\to V\ext$. Since one gets identical matrices$~A$ for $T$ and for $T\ext$, using some $k$-basis of $V$ and the corresponding $K$-basis of $V\ext$, they have the same characteristic polynomials. They have the same minimal polynomials as well, since the minimal polynomial can be found from the unique solution of the linear system $x_0I+X_1A+\cdots+x_{d-1}A^{d-1}=A^d$ (one equation for each matrix coefficient) for the smallest $d$ for which this system has a solution at all; solving a linear system with coefficients in$~k$ over a larger field$~K$ will not change the existence of a solution, or the solution in case it is unique.
So over a sufficiently large field$~k$ (a splitting field of$~\chi$ will do), the polynomials $\mu$ and $\chi$ split, and give rise to the same set of linear factors $X-\lambda$ (though some may occur with larger multiplicity for$~\chi$ than for$~\mu$). In order to conclude that they also have the same set of irreducible factors in $k[X]$, I propose two arguments. One is that the presence of an irreducible factor$~f$ in $k[X]$ can be read off from the presence of any one of the factors into which $f$ splits in $K[X]$, because no two (monic) irreducible polynomials over$~k$ share a root in$~K$: if $\alpha\in K$ is a root of a monic irreducible $f\in k[X]$, then the minimal polynomial of$~\alpha$ over$~k$ is $f$ (this is actually the minimal polynomial of multiplication by$~\alpha$ viewed as $k$-linear map $K\to K$). The other argument is that, having the same set of (linear) factors, $\chi$ divides in $K[X]$ some power of $\mu$. But this division is valid in $k[X]$ as well (where both $\mu$ and $\chi$ live), which shows that the irreducible factors of$~\chi$ occur among those of$~\mu$ (and C-H gives the converse).