Ideal generated by an element 
Let $R\left [ x \right ]$ 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 $R\left [ x \right ]$ and$ A=\left \langle x \right \rangle.$

Attempt to verify the above:

Let $A =\left \{ p\left ( x \right ) \in R\left [ x \right ] \mid p\left ( x \right ) \text{has constant term 0} \right \}$

Using the Ideal test:
Suppose $X\left ( x \right ),Y\left ( x \right ) \in A.$
Then, by the ideal test, $X\left ( x \right )+\left ( -\left ( Y\left ( x \right ) \right ) \right ) \in A$
Now, suppose $R\left ( x \right ) \in R\left [ x \right ] and p\left ( x \right ) \in A.$
Then, $R\left ( x \right )\cdot p\left ( x \right ) \in A$ since the constant term $0$ in $p\left ( x \right )$ kills any constant term.
Thus, A is an ideal of $R\left [ x \right ]$.
To show that $A=\left \langle x \right \rangle, $
$x = \left \{ x^{1},x^{2},\cdot \cdot \cdot ,x^{n} \right \}$
Then, $\left \langle x \right \rangle=\left \{ \alpha_{1}x^{1}+\cdot \cdot \cdot +\alpha_{n}x^{n} \right \}$
and I suppose we would have shown that this is equivalent to A.
But what confuses me is the notation for $\left \langle x \right \rangle$.
Shouldn't $\left \langle x \right \rangle=\left \{ x^{1},x^{2},\cdot \cdot \cdot ,x^{n} \right \}$ according to cyclic notation?
 A: Whenever you're talking about an expression like $\langle x \rangle$ in an algebraic structure, it is always with respect to the operations in the structure. 
If $x$ is in a semigroup and you're only talking about a subsemigroup generated by $x$, then you only have multiplication, and $\langle x\rangle=\{x,x^2, x^3, x^3,\ldots \}$.
If $x$ is in a group and you're talking about the subgroup generated by $x$, then you need to incorporate multiplication and inverses, so not only do you have the powers of $x$, you also need their inverses (so that $\langle x\rangle$ is a group) and in that case $\langle x\rangle=\{\ldots,x^{-2},x^{-1},1,x,x^2, x^3, x^3,\ldots \}$.
If you're talking about an ideal of a ring $R$, then both addition AND multiplication need to be incorporated into generating the ideal. The set $\{\ldots, -3x, -2x, -x, 0, x, 2x, 3x,\ldots\}$ isn't necessarily closed under multiplication by elements of $R$, and the set $\{x, x^2, x^3, \dots\}$ is not necessarily closed under addition.
Here is a rundown of the definition of what a principal ideal is in rings:


*

*For a ring, not necessarily with unity:
$$
\langle x\rangle=\left\{zx+\sum_{i=1}^n r_ixs_i\mid r_i, s_i\in R, n\in \mathbb N, z\in \mathbb Z\right\}
$$

*If the ring has identity, then it is no longer necessary to have the  $zx$ part, since $1$ is available to be $r_i, s_i$ or both:
$$
\langle x\rangle=\left\{\sum_{i=1}^n r_ixs_i\mid r_i, s_i\in R, n\in \mathbb N\right\}
$$

*If $R$ is commutative with identity, this simplifies further since you can combine the $r_i$'s and $s_i$'s:
$$
\langle x\rangle=\left\{\sum_{i=1}^n xr_i\mid r_i\in R, n\in \mathbb N\right\}=\{xr\mid r\in R\}
$$

*If your ring is commutative without identity, you can still combine $r_i$'s and $s_i$'s, but you need to bring back the integer combinations of $x$:


$$
\langle x\rangle=\{xr+xz\mid r\in R, z\in \mathbb Z\}
$$
A: In a ring $R$, if $a$ is an element then $\langle a\rangle$ usually denotes the ideal generated by $a$, that is the smallest ideal of $R$ contaning $a$, which is $Ra$ (in particular it contains all $a^n$, but not only).
A: The notation $\langle f(x) \rangle$ refers to the ideal generated by the polynomial $f(x)$ and it is defined to be the set $\{ f(x) g(x): g(x) \in R[x]\}$ of all multiples of $f(x)$ in $R[x]$.  For example, $\langle x \rangle$ is the set of all polynomials of the form $x g(x)$, which is exactly the set of all polynomials in $R[x]$ whose constant term is 0.  
