A Field of Polynomials of a Linear Operator 
Consider a finite-dimensional vector space $V$ over the field $\mathbb{F}.$ Consider a linear operator $T : V \to V$ such that no nonzero subspace of $V$ is mapped into itself by $T.$ Let $\mathbb{F}[T\,]$ denote the ring of polynomials in $T.$

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*Prove that the null space of any nonzero $S \in \mathbb{F}[T\,]$ is trivial.


*Prove that $\mathbb{F}[T\,]$ is a field.


*Prove that $[\mathbb{F}[T\,] : \mathbb{F}\,] = \dim_\mathbb{F}(V).$

We were asked the above question on a recent qualifying exam in algebra, and at the time, I thought I had a solution, but thinking about it more afterward, I realized that my attempts were incorrect. From the first part of the question, it follows that $T$ is invertible with unique inverse $T^{-1}.$ But why does this guarantee that any polynomial $f(T) = a_n T^n + \cdots + a_1 T + a_0 I$ has an inverse in $F[T]?$ For the third part of the question, I would like to establish that the minimal polynomial for $T$ is identical to the characteristic polynomial for $T,$ but I am not entirely sure if this is possible. I would greatly appreciate any suggestions or comments.
 A: An answer has already appeared. But I found this a good exercise for myself; it's possible that this answer is more elementary, being expressed in terms even I can understand.
If $p(T)x=0$ then $p(T)(Tx)=Tp(T)x=0$. So the null space of $p(T)$ is $T$-invariant, so it is either $V$, meaning $p
(T)=0$, or $\{0\}$.
So if $p(T)\ne0$ then $p(T)$ is invertible. (Detail: In general it's easy to see that if $A$ is invertible then $A^{-1}$ is a polynomial in $A$; for example both the minimal polynomial and the characteristic polynomial must have non-zero constant term.) So it's a field - every non-zero element is invertible.
Now note this: If $p(T)=0$ and $\deg(p)=d<\dim(V)$ then $T$ has non-trivial invariant subspaces. Because for any $x$ the span of $x,Tx,\dots T^{d-1}x$ is invariant.
So the minimal polynomial has degree $\dim(V)$ (and equals the characteristic polynomial), hence the dimension is $\dim(V)$.
We didn't use the fact mentioned in the other answer, that there are no invariant subspaces if and only if the characteristic polynomial is irreducible. That fact interested me more than the original question:
If $W$ is an invariant subspace then looking at a basis for $V$ that includes a basis for $W$ shows that the the characteristic polynomial is not irreducible. Conversely, suppose the characteristic polynomial is $pq$. If $p(T)=0$ or $q(T)=0$ then there is an invariant subspace, as above. If not then $W=q(T)V$ is an innvariant subspace (invariant because $Tq(T)x=q(T)(Tx)$; $W\ne V$ because $p(T)W=0$ and $p(T)\ne0$; $W\ne\{0\}$ because $q(T)\ne0$.)
A: .Let $S \in \mathbb F[T]$ be non-zero as a linear transformation. Note that $ST = TS$, since $S$ is just a polynomial in $T$. Therefore, if $Sv = 0$, then $S(Tv) = T(Sv) = 0$, so $T$ maps $\ker S$ to itself i.e. if $v \in \ker S$ then $Tv \in \ker S$. However, this can happen precisely when $\ker S = \{0\}$(as $S$ is non-zero). Hence, $S$ is invertible.
Now, the characteristic polynomial of $T$ is zero in $\mathbb F[T]$. Therefore, $p_T(T) = \sum_{i=0}^n a_iT^i = 0$.

Claim to verify yourself : the characteristic polynomial of a linear operator is irreducible if and only if it has no invariant subspaces.

Now, the minimal polynomial must be the characteristic polynomial itself, and therefore $\mathbb F[T] \cong \frac{\mathbb F[x]}{p_T(x)}$. Hence, since $p_T$ is of degree the dimension of $V$, part two and three follow. 
