Conditions for Which $\mathbb{F}[T]$ Is a Field

Question

Given a linear operator $T$ on a finite-dimensional vector space $V$ over a field $\mathbb{F},$ consider the ring $\mathbb{F}[T]$ of polynomials in the linear operator $T.$ Under what conditions is $\mathbb{F}[T]$ a field? One sufficient condition with which I am already familiar is that $T$ maps no nonzero subspace of $V$ into itself.

Proof. Given an element $p(T) \in \mathbb{F}[T],$ let $v$ be a vector in $\ker p(T).$ Considering that $T$ commutes with $p(T),$ we have that $p(T)(T(v)) = T(p(T)(v)) = T(0) = 0,$ from which it follows that $T(\ker p(T)) \subseteq \ker p(T).$ Our assumption that $T$ maps no nonzero subspace of $V$ into itself implies that $\ker p(T) = \{0\},$ i.e., $p(T)$ is injective. Of course, $p(T)$ is a linear operator on $V,$ hence $p(T)$ is invertible. We conclude that every element of $\mathbb{F}[T]$ is invertible. Furthermore, $\mathbb{F}[T]$ is an integral domain since if $p(T)$ is nonzero and $p(T) q(T) = 0,$ then we must have that $q(T) = I q(T) = [p(T)]^{-1} p(T) q(T) = 0.$

Lately, I have been thinking about whether it is sufficient that $T$ is invertible. Using the evaluation map $\varphi_T : \mathbb{F}[x] \to \mathbb{F}[T]$ defined by $\varphi_T(p(x)) = p(T)$ in tandem with the fact that $\mathbb{F}[x]$ is a PID, it is clear that $\mathbb{F}[T] \cong \mathbb{F}[x]/(\mu_T(x)),$ where $\mu_T(x)$ is the minimal polynomial of $T.$ We have therefore that $\mathbb{F}[T]$ is a finite-dimensional vector space over $\mathbb{F}$ with a basis $\{I, T, T^2, \dots, T^{d-1} \},$ where $d = \deg \mu_T.$ We could conclude that $\mathbb{F}[T]$ is a field if we could show that $\mathbb{F}[T]$ is an integral domain since every integral domain that is a finite-dimensional vector space over a field is itself a field. But I am not entirely sure how to prove that $\mathbb{F}[T]$ is an integral domain in this case.

Generally, are there any other conditions than the one I initially stated for which $\mathbb{F}[T]$ is a field?

Answer 1

Given a linear operator $T$ on a finite-dimensional vector space $V$ over a field $k,$ consider the ring $k[T]$ of polynomials in $T.$ Observe that there is a surjective ring homomorphism $\mathbf{ev}_T : k[x] \to k[T]$ defined by $\mathbf{ev}_T(p(x)) = p(T).$ Considering that $k[x]$ is a principal ideal domain (PID), we have that $k[T] \cong k[x]/(\mu_T(x)),$ where $\mu_T(x)$ is the minimal polynomial of $T.$

Question: Under what conditions is $k[T]$ a field?

Like I have mentioned in the original post and the subsequent comments, there are two sufficient conditions with which I am already familiar.

• If $T$ maps no nonzero subspace of $V$ into itself, then $k[T]$ is a field.

Proof. Polynomials in $T$ commute with $T,$ hence $T$ maps $\ker(p(T))$ into $\ker(p(T))$ for all $p(T)$ in $k[T].$ Consequently, every nonzero element of $k[T]$ is invertible, hence $k[T]$ is a field. QED.

• If $\mu_T(x)$ is irreducible, then $k[T]$ is a field.

Proof. If $\mu_T(x)$ is invertible, then it is prime, hence the ideal $(\mu_T(x))$ of $k[x]$ is maximal. QED.

One observation that I made above is that it is necessary but not sufficient that $T$ is invertible. For instance, the linear operator $T : k^2 \to k^2$ defined by $T(x, y) = (x, -y)$ is invertible; however, the linear operator $I + T$ satisfies $(I + T)(x, y) = (2x, 0)$ and hence is not invertible.

If anyone would like to suggest any additional sufficient conditions, please feel free!