This is my confusion when I read dummit and foote. If $V$ is a finite dimensional vector space over field $F$, and we consider it as $F[x]$-module via a linear transformation $T$ (i.e. $p(x)\cdot v=p(T)v$), why $V$ is a torsion $F[x]$-module? How to construct nonzero $p(x)$ such that $p(x)\cdot v = 0$ for each $v \in V$?enter image description here

  • $\begingroup$ This is not true in general. $\endgroup$ Commented Nov 9, 2020 at 5:36
  • 4
    $\begingroup$ Maybe what is meant is that if V is finite dimensional then it has nontrivial torsion as an $F[x]$-module. $\endgroup$ Commented Nov 9, 2020 at 5:52
  • $\begingroup$ Yeah, $V$ is finite dimensional. I forgot to mention it. $\endgroup$ Commented Nov 9, 2020 at 6:35
  • 1
    $\begingroup$ $V$ is finite dimensional, therefore so is $\operatorname{End}(V)$. In $\operatorname{End}(V)$ there must be a linear dependence between $1,x,x^2,x^3,\ldots$, which immediately gives a polynomial $p(x) = 0$. $\endgroup$
    – Joppy
    Commented Nov 9, 2020 at 11:16

1 Answer 1


Without an explicit understanding of $V$ itself we have no hope of an explicit construction of $p$. However, this can still be done as follows. Consider the ring $End_F(V)$ of linear endomorphisms of $V$. As $V$ is an $F[x]$ module via $T$, we consider the map $\phi: F[x] \longrightarrow End_F(V)$ via $x \mapsto T$. This map is a ring homomorphism and is furthermore $F$ linear. As $V$ is finite dimensional over $F$, $End_F(V)$ is as well (just square the dimension). However, $1, x, x^2, x^3, \dots$ forms an infinite linearly independent subset of $F[x]$. Thus, it is infinite dimensional so $\phi$ must have a nonzero kernel. Call this kernel $I$. It is an ideal of $F[x]$, so using the Euclidean division algorithm, we can find a generator $I = (p)$, i.e. $F[x]$ is a PID. Without loss of generality, we can assume $p$ to be monic by dividing out its leading coefficient. This will be the minimal polynomial. If you're uncomfortable with this ring theory, we don't need $(p) = I$ to finish the proof, only that $I$ contains a nonzero polynomial.

Now, let's think about what all of this meant. We have a nonzero monic polynomial $p$ such that $ker(\phi) = (p)$. By definition, $\phi(p) = p(T) = 0$. Furthermore, by definition, $F[x]$ acts on $V$ via $p \cdot v = p(T) v$. But as we just said, $p(T) = 0$ so $p \cdot v = 0$ for all $v$. Thus, $V$ is a torsion $F[x]$ module.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .