# Cyclic modules over a polynomial ring

I think I proved the following proposition. Is this correct and well-known?

Proposition: Let $V$ be a finite dimensional vector space over a field $K$. Let $f: V \to V$ a $K$-linear map. Let $K[X]$ be the polynomial ring. $V$ can be regarded as a $K[X]$-module via $f$. If the characteristic polynomial of $f$ is the minimal polynomial, then $V$ is a cyclic $K[X]$-module, i.e. generated by a single element of $V$ over $K[X]$.

• Yes, this is a well-known result. Commented Apr 14, 2012 at 13:31
• Is there any reference? Commented Apr 14, 2012 at 14:11
• This should be covered in virtually any graduate-level algebra book. Check the section on modules, or the material leading up to the proof of Jordan canonical form. Commented Apr 14, 2012 at 16:13
• It will also appear in the material leading up to the Rational Canonical Form; e.g., in Friedberg, Insel, and Spence Commented Apr 14, 2012 at 21:39
• @JimBelk "This should be covered in virtually any graduate-level algebra book." I have Bourbaki and Lang on algebra. I'm afraid I cannot find it covered by them. Would you please tell me the names of some books that do so? Commented Dec 10, 2015 at 19:09

This is a consequence of the structure theorem for finitely generated modules over a principal ideal domain $R$, which says that every such module can be written a direct sum of cyclic modules $\bigoplus_{i=1}^lR/(d_i)$ where the generators $d_i$ (invariant factors) divide one another in order: $d_1\mid d_2\mid \cdots \mid d_l$. Take the module to be $V$ as in the question, then the final generator $d_l\in K[X]$ annihilates all the cyclic factors, so (assuming the $d_i$ are taken to be monic) it is the minimal polynomial of $f$, while the characteristic polynomial is the product of all polynomials $d_i$. Clearly the two are equal if and only if $l=1$ (the invariant factors $d_i$ are never $1$) which is the case if and only if $V$ is a cyclic module.