Let, $\mathbb{R}[x]$ be a polynomial ring and let $J = (x)$. True/false: $J$ consists of all the polynomials of $\mathbb{R}[x]$ whose constant terms are $0$.
I know $J=(x)$ is a maximal ideal of $\mathbb{R}[x]$, and the statement is true. But, I don't exactly understand why. Could someone please help understand this? Thanks.
On
An Ideal $I$ in a commutative (for simplicity) Ring $R$ is a subset which is an additive subgroup of $R$ and for which $ir\in I$ whenever $i\in I, r\in R$. The ideal generated by an element $x\in R$ is the smallest ideal of $R$ containing $x$ (the intersection of all such ideals is itself an ideal).
To sketch the ideas here, suppose $p(x)$ has constant term $0$, then $p(x)=xq(x)$ for some polynomial $q(x)$, and we therefore have $p(x)\in I$ by setting $i=x, r=q(x)$. An ideal containing $x$ must therefore contain all the polynomials with zero constant term.
Then the polynomials with zero constant term are an additive subgroup, satisfy the multiplicative rule for an ideal and contain $x$. So the Ideal of polynomials with zero constant term contains the ideal generated by $x$.
Therefore the ideal generated by $x$ must be equal to the ideal of polynomials with zero constant term.
There are some details to fill in.
$\mathbb{R}[x]/(x)$ is isomorphic to $\mathbb{R}$. so every polynomial with constant $0$ term must be contained in the ideal. And, as $\mathbb{R}$ is isomorphic to the quotient, nothing else can be contained in the ideal.