Submodules and $T$-invariant subspaces.

Question

My professor wrote this on the board:

Claim: there is a bijection between submodules and $T$-invariant subspaces {$T(w) \subset W$}.

Proof:

If $W \subset V$ is $T$-invariant, then $f(X).w = f(T)(w) \in W.$ so $W$ is a submodule. Conversely, let $W \subset V$ be a submodule, then for $w \in W,$ $X.w = T(w) \in W$ and hence $W$ is $T$-invariant.

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Before that she wrote this:

Let $k$ be a field, $V$ a $k-$ vector space, $n = \dim_kV < \infty.$$R = k[x], T \in L(V)=$ linear operators on $V.$

$T$ has minimal polynomial and characteristic polynomial in $R.$

$V$ has the structure of an $R$-module, a structure induced by $T.$

$$\varphi : R \rightarrow \operatorname{End_k(V)} = L(V)$$ $$1 \mapsto I$$ $$X \mapsto T$$

So $\varphi(f(X)) = f(T)$

Scalar multiplication in $V$ is $f(X).v = f(T)(v).$ then she asked what are the submodules of $V$ and she began to prove the above claim.

But I do not understand the following:

1- Why she defined $\varphi$ like that?

2- why the scalar multiplication in $V$ looks like that?

3- I do not understand at all how the forward direction proves that it is a submodule, why we are applying $f(X)$ to $w$ to prove this?

4- why in the backward direction we are applying $X$ to $w$?

could anyone help me understand all these points?

Answer 1

1. $k[X]$ is a free $k$-algebra on one generator and it has the universal property that $k$-algebra homomorphisms from $k[X]$ into any $k$-algebra is defined by the choice of the image of the indeterminate $X$. Also note that $\operatorname{End}_k(V)$ is a $k$-algebra, in which multiplication is simply the composition of endomorphisms.
2. Scalar multiplication by a polynomial results from considering a homomorphism from $k[X]$ into $\operatorname{End}_k(V)$, i.e. from the choice of an endomorphism $T$ associated to $X$. If you're interested in the properties of a given endomorphism, you'll choose for the image of $X$ this endomorphism.
3. We actually apply $f(T)$ to a vector in $V$ because by definition, scalar multiplication by the polynomial $f(X)$ consists in applying the endomorphism $f(T)$ to this vector.
4. I don't understand what you call the backwards direction is.