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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?

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Hint: $0 \in I$. –  Sebastian Feb 14 '12 at 10:23

2 Answers 2

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.

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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.

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