Here $F$ is a field, $F[x]$ is the ring of polynomials in $x$ with coefficients in $F$, $V$ is a finite dimensional vector space over $F$ of dimension $n$, $T$ is a fixed linear transformation of $V$ by which we make $V$ into an $F[x]$-module.
The following is an excerpt from Section $12.3$, The Jordan Canonical Form of Abstract Algebra, $\mathit{3^{rd}}$ edition by Dummit and Foote:
For the $F[x]$-module $V$ the invariant factors were monic polynomials $a_1(x), a_2(x),\dots, a_m(x)$ of degree at least one (with $a_1(x)|a_2(x)|\dots|a_m(x)$), so the associated elementary divisors are the powers of the irreducible polynomial factors of these polynomials. These polynomials are only defined up to multiplication by a unit and, as in the case of the invariant factors, we can specify them uniquely by requiring that they be monic.
To obtain the simplest possible elementary divisors we shall assume that the polynomials $a_1(x), a_2(x),\dots, a_m(x)$ factor completely into linear factors, i.e., that the elementary divisors of $V$ are powers $(x-\lambda)^k$ of linear polynomials....
Under this assumption on $F$, it follows immediately from Theorem $6$ that $V$ is the direct sum of finitely many cyclic $F[x]$-modules of the form $F[x]/(x-\lambda)^k$ where $\mathbf{\lambda\in F}$ is one of the eigenvalues of $\mathbf{T}$, corresponding to the elementary divisors of $\mathbf{V}$.
My Question: I don't quite follow the boldface texts. Indeed, each $a_i(x)$ can be written as product of irreducible elements in $F[x]$ since $F[x]$ is a Unique Factorization Domain, and the irreducibles are certainly of the form $x-\lambda$; but why can we assume that $\lambda$ is an eigenvalue? Any help would be greatly appreciated.
If there is a submodule of $V$ isomorphic to $F[x]/(x-\lambda)^k$ (for $k>0$), then $\lambda$ must be an eigenvalue. Indeed, consider the element $v=(x-\lambda)^{k-1}\in F[x]/(x-\lambda)^k$. This is a nonzero vector and $(x-\lambda)v=0$, which says exactly that $v$ is an eigenvector with eigenvalue $\lambda$.