<Home>Qualifying Exams><Algebra>Spring 2016 - Problem 3

Spring 2016 - Problem 3

ring theory

Let $R$ be a ring which is left artinian (that is, artinian with respect to left ideals). Suppose that $R$ is a domain, meaning that $1 \neq 0$ in $R$ and $ab = 0$ implies $a = 0$ or $b = 0$ in $R$ . Show that $R$ is a division ring.

Solution.

Let $a \in R$ be non-zero, and consider the (left) ideal $\gen{a}$ is not all of $R$ . Since $R$ is artinian, the chain

\gen{a} \supseteq \gen{a^2} \supseteq \gen{a^3} \supseteq \cdots

stabilizes, so there exist $r \in R$ and $n > 1$ such that $ra^n = a$ . But

ra^n - a = 0 \iff \p{ra^{n-1} - 1}a = 0

and $a \neq 0$ , so because $R$ is a domain, we see $ra^{n-1} = 1$ . Since $n > 1$ , this means $ra^{n-2} \cdot a = 1$ , so $a$ is left-invertible. Let $b = ra^{n-2}$ so that

aba = aab = a \implies aba - aab = 0 \implies a\p{ba - ab} = 0.

Again, because $R$ is a domain, we have $ab = ba$ , so $b$ is also a right inverse. Hence, $R$ is division.