Show that the set of all polynomials $f$ in $F[x]$ such that $f(A)=0$ is an ideal

Question

Let $A$ be an $n \times n$ matrix over a field $F$. Show that the set of all polynomials $f$ in $F[x]$ such that $f(A)=0$ is an ideal.

I don't understand how to apply this when it comes to matrices. The question comes from the book "Linear Algebra" by Hoffman and I am not finding the book particularly helpful regarding ideals.

Answer 1

I don't know much about linear algebra, but I do know abstract algebra, so this should be pretty easy.

There are two things we need to prove about $K$, the set of all $f$ in $F[X]$ such that $f(A)=0$:

1. $K$ is an additive subgroup of $F[X]$. This can be shown using the one-step subgroup test.
2. For all $g \in F[X]$ and $f \in K$, $g \cdot f \in K$.

Part 1: Let $f, g \in K$. Then, $f(A)=0$ and $g(A)=0$. Thus, $f(A)-g(A)=0$, so $(f-g)(A)=0$. Thus, $f-g \in K$, proving that $K$ is an additive subgroup of $F[X]$ which is Part 1.

Part 2: Let $g \in F[X]$ and $f \in K$. $f(A)=0$, so $g(A) \cdot f(A)=0$ since anything times $0$ is $0$, meaning $(g \cdot f)(A)=0$. Thus, $g \cdot f \in K$, proving Part 2.

Thus, we have proven both parts, which proves that $K$ is an ideal of $F[X]$.