Something about rings It's been some time since I've done ring theory, so please bear with me. Suppose $I$ is an ideal generated by some $x\in R$ where $R$ is a ring -- so $I=(x)$, and the multiplicative identity $1\in (x)$ then does that mean that all elements in $I$ are units?
 A: No, that's not true - for example, $0$ is in any ideal of a ring, and $0$ is only a unit if the ring is trivial (i.e. has only one element).
However, what you can conclude is that any generator of $I$, i.e. an element $y\in R$ such that $I=(y)$, must be a unit. This is because, for any $x\in R$,
$$x\text{ is a unit}\iff (x)=R\iff 1\in (x).$$


*

*If $x$ is a unit, say with inverse $y$, then
$$(x)=\{xz\mid z\in R\}\supseteq\{x(yz)\mid z\in R\}=\{1\cdot z\mid z\in R\}=R$$
and because $(x)\subseteq R$ we must have $(x)=R$. 

*If $(x)=R$, we clearly must have $1\in R=(x)$.

*If $1\in (x)$, then $1\in(x)=\{xz\mid z\in R\}$, so $1=xz$ for some $z\in R$. This $z$ is an inverse for $x$, so $x$ is a unit.
A: No that is not true.
Consider $\mathbb{Z}$ as a ring. Then $(1)$ is the whole ring but it is not true that $4 \in (1)$ is a unit in $\mathbb{Z}$.
In fact your question is equivalent to asking if for all unital commutative rings  $R$ is it the case that every element is a unit. This is not true as pointed out above.
