I'm taking a basic course in linear algebra and struggling to understand the definition of the annihilator of a set given in my lecture notes.
In these notes, the definition of a module over a ring is given in the first place, then the following operation over a vector space $V$ is defined:
$$K[x] \times V \rightarrow V$$
$$(p,v) \rightarrow pv$$
where $K[x]$ is the polynomial ring over the field $K$, $p\in K[x]$, $v\in V$ and
$$pv=(\sum^m_{i=0}a_ix^i)v=(\sum^m_{i=0}a_ih^i)v=p(h)(v)\in V,$$
where $h$ is an endomorphism of the vector space $V$. Later, it is shown that $V$ is a module over $K[x]$ with this associated operation. Finally, the annihilator of a non-null set $S\subseteq V$ is defined as
$$\operatorname{Ann}_R(S):=\{ p\in R \mid pv=0 \quad\forall v\in S\}$$
where $R=K[x]$ in this case.
What isn't clear to me is, wouldn't that definition also depend on the choice of the endomorphism $h$? For instance, if $V=\mathbb{R}^3$ and we choose $h=2v \ \ \forall v\in V$, then the polynomial $x-2$ annihilates the whole space (because $(x-2)v=h(v)-2v=2v-2v=0 \ \ \forall v\in\mathbb{R}^3)$ and then $x-2 \in \operatorname{Ann}_R(\mathbb{R}^3)$, but if $h \neq 2v$, then it follows that $(x-2)v\neq0$ and $x-2 \notin \operatorname{Ann}_R(\mathbb{R}^3)$.
I've been unable to find an answer in this Wikipedia article or elsewhere, so I'd appreciate any help.
Best Answer
Yes, this definition highly depends on the choice of $h$.
To be more precise: The whole module-structure highly depends on the choice of $h$.
Maybe the following will clarify a little: // Edit: I just realized I wrote $k$ instead of $K$... So $k := K$.
You are actually interested in studying an endomorphism $h : V \to V$ most of the time. In order to study $h$, you equip $V$ with the structure of an $k[X]$-module essentially by defining that $X \cdot v := h(v)$ and then extending the way you wrote down in your question for an arbitrary polynomial $p \in k[X]$. Let's call $V$ equipped with this kind of action $V_h$.
This way you've kind of "embedded" the "behaviour" of $h$ into the module structure of $V_h$.
Now, if you have a good understanding of the module $V_h$, it essentially gives you interesting information related to $h$.
Luckily, $k[X]$ is a principal ideal domain and if now furthermore $V$ is a finite-dimensional $k$-vector space, it is finitely generated as a $k[X]$-module (as $k$ injects into $k[X]$).
Now one has fairly strong structure theorems on finitely generated modules over a principal ideal domain.
Therefore, one understands $V_h$ pretty well from the module-theoretic point of view and one can use that knowledge to gain understanding of $h$.
Also note that if you equip $V$ with any kind of $k[X]$-module structure, you can show that $V = V_h$ for $h : V \to V, v \mapsto X\cdot v$.
Therefore: Studying a $k[X]$-module structure on a $k$-vector space $V$ amounts to studying a fixed endomorphism $V \to V$. You neither really change any information whatsoever when changing from $V$ with a fixed endomorphism $h$ to $V_h$ or the other way around. The perspective on $V_h$ is particularly useful though, because it allows you to formulate lots of interesting phenomena related to $h$ using the abstract framework of modules (maybe f.g. over a p.i.d.)., which is interesting and very important for its own sake.
Also regarding the definition of the annihilator:
If $R$ is any ring, $M$ an $R$-module and $S \subseteq M$ one defines $\mathrm{Ann}_R(S) := \{ r \in R \mid r \cdot s = 0 \ \forall s \in S\}$.