# Why vector space is torsion $F[x]$-module?

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 $$v \in V$$?

• This is not true in general. Commented Nov 9, 2020 at 5:36
• Maybe what is meant is that if V is finite dimensional then it has nontrivial torsion as an $F[x]$-module. Commented Nov 9, 2020 at 5:52
• Yeah, $V$ is finite dimensional. I forgot to mention it. Commented Nov 9, 2020 at 6:35
• $V$ 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 $p(x) = 0$. Commented Nov 9, 2020 at 11:16

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.