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Let $V$ denote the vector space of polynomials in one variable with coefficients in $\mathbb{R}$, and let $T(f(x))=x(f(x))$.

Prove that if $W\subset V$ is a subspace $\neq 0$ such that $T(W)\subset W$, then $V/W$ is finite dimensional.

My attempt: We know that $\forall p\in W$, $xp\in W$ because $T(W)\subset W$, so given that $1$ is in our subspace $W$, so must $x$ be, and hence every linear combination of $x$ and coefficients in $\mathbb{R}$. Thus, $V/W$ is simply the constant functions (is this true?)

It this is the case, how do I show the constant functions are finite dimensional?

I understand this may have been solved before, but I would like insight on whether my line of reasoning works.

Any help appreciated!

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    The problem is not quite correct as stated. $W$ could equal the zero subspace, and then $T(W)\subset W$, but $V/W \cong V$ is the full space of polynomials, which is not finite-dimensional. I think it becomes correct if you assume $W$ contains at least one non-zero polynomial. – 2017-02-10

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First, your proof doesn't quite work. Unless you assume $1 \in W$, this does not follow. If that is an assumption, then $1 \in W$ along with $T(W) \subset W$ implies that $W=V$ (all polynomials) since $1 \in W$ would imply that $T(1)=x1=x \in W$ and so $T(x)=x^2 \in W$ etc.

Second, the statement is not true for $W=\{0\}$. Notice that $T(\{0\}) \subset \{0\}$ but $V/\{0\} \cong V$ (which is not finite dimensional).

Now, suppose $W \not= \{0\}$ and that $T(W) \subset W$. Since $W \not= \{0\}$, you have some $f(x) \in W$ with $f(x) \not=0$. Since $T(f(x)) = xf(x) \in W$ and by induction $T^k(f(x))=x^kf(x) \in W$, you have that any linear combination of such polynomials lie in $W$. Therefore, $g(x)f(x) \in W$ for all $g(x) \in V$.

Now take an arbitrary $h(x)+W \in V/W$. Then you can divide $h(x)$ by $f(x)$ (since it's non-zero). Thus $h(x)=f(x)q(x)+r(x)$ for some polynomials $q(x),r(x)$ where either $r(x)=0$ or the degree of $r(x)$ is less than the degree of $f(x)$. But then $h(x)+W=f(x)q(x)+r(x)+W =r(x)+W$ (since $f(x)q(x) \in W$). This means that every element of $V/W$ can be represented by a polynomial $r(x)$ of degree less than $f(x)$ (or is zero). This means that if $f(x)$ has degree $n$, then $1+W, x+W, \dots, x^{n-1}+W$ span $V/W$ so it's finite dimensional.

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    You don´t even need the fact that $\mathbb{R}[X]$ is a principal ideal domain – 2017-02-10
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    Yes and no. The first bit of my argument is pretty close to showing W is an ideal. The next part of my argument when I call on the division algorithm to reduce the degree of the representative (to get from f to r), I'm close to showing W is a principal ideal. :) – 2017-02-10
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    Yes thats true but the argument still holds if $f$ is not the generator as You don´t say that $1+W,...,x^{n-1}+W$ are linearly independant. – 2017-02-10
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There is no reason for 1 to be in W. The exercise as stated is in fact not quite true. $W=0$ is a counterexample (in cat the only counterexample). W is a module over $\mathbb{R}[x]$ and as such generated by a single polynomial $$ f(x)=\sum_{i=0}^n a_ix^{i} $$ The vector-space $V/W$ is then $n$ dimensional with basis the image of $1,x\cdots, x^{n-1}$.