<Home>Qualifying Exams><Algebra>Spring 2016 - Problem 5

Spring 2016 - Problem 5

module theory

Let $A$ be the ring $\C\gen{u, v}/\gen{uv - vu - 1}$ , the quotient of the free associative algebra on two generators by the two-sided ideal.

Show that every nonzero $A$ -module $M$ has infinite dimension as a complex vector space.
Let $M$ be an $A$ -module with a nonzero element $y$ such that $uy = 0$ . Show that the elements $y, vy, v^2y, \ldots$ are $\C$ -linearly independent in $M$ .

Solution.

Let $y \in M$ be nonzero. If $uy = 0$ and $vy = 0$ , then

0 = u\p{vy} = \p{vu + 1}y = v\p{uy} + y = y.

Thus, at least one of $uy$ and $vy$ is nonzero. Without loss of generality, suppose $uy = 0$ so that $vy \neq 0$ . It remains to show that $y, vy, v^2y, \ldots$ are linearly independent.

Suppose $c_0y + c_1vy + \cdots + c_nv^ny = 0$ . First, note that for $j \geq 1$ ,

\begin{aligned} uv^j &= \p{vu + 1}v^{j-1} \\ &= vuv^{j-1} + v^{j-1} \\ &= v\p{vuv^{j-2} + v^{j-2}} + v^{j-1} \\ &= v^2uv^{j-2} + 2v^{j-1}. \end{aligned}

By induction, we see $uv^j = v^ju + jv^{j-1}$ .

We now proceed by induction. $\set{y}$ is linearly independent since $y \neq 0$ , so now assume $\set{y, vy, \ldots, v^ny}$ are all linearly independent. Suppose $c_0y + c_1vy + \cdots + c_nv^ny = 0$ , and left multiply by $u$ , which gives

\begin{aligned} 0 = u \cdot \sum_{j=0}^n c_jv^jy &= \sum_{j=0}^n c_j \p{uv^j} y \\ &= \sum_{j=1}^n c_j \p{v^ju + jv^{j-1}}y \\ &= \sum_{j=1}^n jc_j v^{j-1}y. \end{aligned}

By the inductive hypothesis, it follows that $c_1 = \cdots = c_n = 0$ , so we have $c_0y = 0$ . By the base case, $c_0 = 0$ as well, so we have linear independence, which means $M$ is infinite dimensional.