Question

Prove that an element of an integral domain is a unit iff it generates the domain.

Answer 1

Answer to Question #82540 - Math / Abstract Algebra

Question. Prove that an element of an integral domain is a unit iff it generates the domain.

Answer. By " $r$ generates $I$ " here we mean that $r$ generates an ideal $I$, in other words, $I$ is the least ideal containing $r$.

Let $R$ be an integral domain, and $a \in R$.

- ( $\Rightarrow$ ) Assume that $a$ is a unit. Let $I$ be an ideal of $R$ containing $a$. Let $b \in R$. As $b = (ba^{-1})a$ and $a \in I$, by the properties of ideal, $b \in I$. Hence $I = R$. As $I$ was arbitrary, every ideal containing $a$ is equal to $R$. Hence $R$ is the least ideal containing $a$, in other words, $a$ generates $R$.

- ( $\Leftarrow$ ) Assume that $a$ generates $R$. The set $I = \{ra \mid r \in R\}$ is an ideal of $R$ as shown below.

- If $ra, sa \in I$ for some $r, s \in R$, then $ra + sa = (r + s)a \in I$.

- $0 = 0 \cdot a \in I$.

- If $ra \in I$ for some $r \in R$, then $-(ra) = (-r)a \in I$.

- If $ra \in I$ for some $r \in R$, and $s \in R$, then $s(ra) = (sr)a \in I$.

Also $a = 1 \cdot a \in I$. Hence $I$ is an ideal containing $a$. As $R$ is the least ideal containing $a$, $I$ includes $R$, so $I$ contains 1. Hence there is $r \in R$ such that $1 = ra$. In other words, $a$ is a unit.

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