Prove that $A$ is an ideal of $\Bbb R[x]$ and $A = \langle x\rangle$

Question

Let $\Bbb R[x]$ denote the set of all polynomials with real coefficients and let $A$ denote the subset of all polynomials with constant term $0$. Then $A$ is an ideal of $\Bbb R[x]$ and $A=\langle x\rangle$

So I actually have proved that $A$ is an ideal of $\Bbb R[x]$ and am not confused by that at all, but I am having difficulty proving that $A$ is the set generated by $x$. Not necessarily sure where to start here.

Answer 1

The ideal $\left<x\right>$ is the set of polynomials in $R[x]$ that are multiplies of $x$, i.e. $$\left<x\right> = \{xf(x) : f(x) \in R[x]\}.$$

Answer 2

$\bullet$ [$A\subset \langle x\rangle$]: If $p(x)\in A$ then by definition it is of the form $$p(x)=a_1x+a_2x^2+\cdots+a_nx^n$$ for some $a_1,\ldots,a_n\in \mathbb{R}$. Factoring $x$ we have $p(x)=x(a_1+a_2x+\cdots+a_nx^{n-1})$, so $p(x)\in \langle x\rangle$.

$\bullet$ [$\langle x\rangle\subset A$]: If $q(x)\in\langle x\rangle$ then $q(x)=xs(x)$ for some $s(x)\in\mathbb{R}[x]$ and it follows that $x\in A$.

Answer 3

Note that for:

$\langle x\rangle = \{xf(x)|f(x) \in \Bbb R[x]\}$, we have for, say:

$f(x) = a_0 + a_1x +\cdots + a_nx^n$, that:

$xf(x) = a_0x + a_1x + \cdots + a_nx^{n+1}$, which has constant term $0$, so that $\langle x\rangle \subseteq A$.

On the other hand, if $g(x) \in A$, then:

$g(x) = b_1x + b_2x^2 + \cdots + b_mx^m = x(b_1 + b_1x + \cdots + b_mx^{m-1}) \in \langle x\rangle$, so $\langle x\rangle = A$.