What does an ideal generated by a subset look like?

Solution 1:

Let $R$ be a ring, and $\{r_\alpha\}_{\alpha\in I}\subset R$. The ideal generated by $\{r_\alpha\}$ consists of all the finite sums$$\sum_{i=1}^na_ir_{\alpha_i},$$where the coefficients $a_i$ are just elements of $R$.

Take for example $\mathbb{Z}[x]$, the ring of polynomials in one variable with integer coefficients, and consider the ideal generated by $\{2,x\}$. It consists of all the sums $2a+bx,\quad a,b\in\mathbb{Z}[x]$. Note that this ideal cannot be generated by a single element.

Solution 2:

Hint $ $ By definition, the ideal generated by a set is the smallest ideal containing the set. Now applying the closure propertes of an ideal we deduce

$$ \begin{array}{} & I \,\supseteq \,\{ a,\,\ b,\,\ldots\, \}\\[.2em] \iff & I \,\supseteq\, a R,\, b R,\, \ldots\\[.2em] \smash[t]{\overset{\rm\color{#c00}U\!}\iff} & I \,\supseteq\, a R + b R + \ldots \end{array}\qquad\quad$$

But the latter set is already an ideal, so necessarily the smallest such ideal.

Remark $\ $ The final equivalence uses $\,\rm\color{#c00}U = $ universal property of the ideal sum, which specializes to the gcd universal property in principal ideal domains, since there "contains = divides", i.e.

$$\begin{align} I\supseteq A,B,\ldots\!\! &\!\! \overset{\rm\color{#c00}U\!\!}\iff I\,\supseteq\ A+B+\cdots\\[.3em] \leadsto\ \ i\mid a,b,\ldots\!\! &\iff i\mid \gcd(a,b,\ldots) \end{align}$$

when specialized to principal ideals$\, A = (a),\ B = (b)\,$ in a PID.