Let $R$ be an integral domain and $F$ a finitely generated free module over $R$. For a linear transformation $\alpha\in\operatorname{End}_R(F)$, the characteristic polynomial is \begin{equation} p_\alpha(t)=\det(t-\alpha)\in R[t]. \end{equation}

Similar transformations have same characteristic polynomial. However, these polynomials fail to characterize similarity, in the sense that non-similar transformations can have the same characteristic polynomial.

So my question is: What can we say about two transformations $\alpha$ and $\beta$ that share the same polynomial?

Obviously, $\alpha$ and $\beta$ have the same spectrum and same algebraic multiplicity for each eigenvalue.

But unless $R$ is an algebraically closed field--in this case the polynomial is determined by the roots and their multiplicities--we should be able to say more about $\alpha$ and $\beta$. Also we should know more than traces and determinants since they are just two of the coefficients.

Can someone give a hint? Thanks!

  • $\begingroup$ +1 This is one of those questions I always wanted to get around to sorting out in my head. I'm pretty sure the answer is here somewhere in this elementary construction: given a transformation $T$ of a module $M$, you can look upon $M$ as a $K[T]$ module, where $T^n\cdot m:=T^n(m)$ and everything else is defined by linearity. The characteristic and minimal polynomials have to do with the kernel of $K[x]\rightarrow K[T]$, since $K[x]$ is a PID. I'm not sure why the two polynomials carry different information, and I'm also not sure of what happens if you replace $K$ with an integral domain. $\endgroup$
    – rschwieb
    Feb 19 '13 at 15:25
  • 1
    $\begingroup$ You also might want to try to solve the question for $R$ a field before you try for $R$ a domain. I know from your past questions that domains seem to be the star of the show for you, but seriously, fields are "the Cadillac of rings"! :) $\endgroup$
    – rschwieb
    Feb 19 '13 at 15:29
  • $\begingroup$ @rschwieb: If fields are the Cadillac, what are algebraically closed fields? $\endgroup$ Feb 19 '13 at 16:10
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    $\begingroup$ @MarcvanLeeuwen Even Cadillacs have special features! Leather interior maybe? $\endgroup$
    – rschwieb
    Feb 19 '13 at 17:26

The degree to which the characteristic polynomial fails to charcterise similarity can be illustrated, for the case where $R$ is a field, by the rational canonical form, which in this case does characterise similarity. The rational canonical form is determined by a uniquely defined sequence of monic polynomials (invariant factors), each one a multiple of any one of polynomials that follow (I prefer this to the more common requirement of dividing what follows); the sequence can have any length, but it can be thought of ending with an indefinite repetition of the constant polynomial $1$ (just like partitions of an integer are weakly decreasing sequences of integers that can be thought of as ending with endless occurrences of $0$). The first invariant factor (in the order I chose) is the minimal polynomial, and the product of all invariant factors is the characteristic polynomial.

Now if you fix the characteristic polynomial, its irreducible factors in $R[X]$ are determined; all of them must occur as a factor of the minimal polynomial. Indeed the multiplicity of a fixed irreducible polynomial $P$ in the invariant factors must be weakly decreasing, so these multiplicities form a partition of the multiplicity of $P$ in the characteristic polynomial. Choosing a partition for each occurring irreducible factor $P$ is exactly the freedom one has for choosing a rational canonical form, and therefore determines the number of similarity classes of transformations with a given characteristic polynomial.

The "regular" case is where the minimal polynomial is equal to the characteristic polynomial; this is forced when the characteristic polynomial is squarefree. In this case the centraliser of $\alpha$ (in the endomorpism algebra) is equal to the set of polynomials in $\alpha$, which is of dimension $n$ (the same as that of the vector space on which $\alpha$ acts). I think that in all other cases the dimension of the centraliser is strictly larger than $n$, whence the similarity class is smaller than in the regular case; this intuitively explains why there aren't any polynomial functions that can separate the "singular" cases from the regular case with the same characteristic polynomial.


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