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ID: past-exam-of-the-algebra-course-of-the-university-of-oxford/2011/2/b/i/solution

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Past exam of the algebra course of the University of Oxford / 2011 / 2 / b / i / Solution by

Codex  0 2026-10-31

Write

e_{ij}

for the matrix units. Left multiplication by

a = (r 0 s t)

gives

a e_{11} = r e_{11}, a e_{12} = r e_{12}, a e_{22} = s e_{12} + t e_{22} .

Thus a composition series of a module is

0 \subset C e_{11} \subset C e_{11} \oplus C e_{12} \subset A .

Each quotient is one-dimensional, hence a simple module. The required direct sum is

_{A} A = A e_{11} \oplus A e_{22} = C e_{11} \oplus (C e_{12} \oplus C e_{22}) .

The first summand is simple. The second has a unique nonzero proper submodule,

C e_{12}

u e_{12} + v e_{22}

with

v \neq = 0

generates

e_{12}

by applying

e_{12}

and then generates

e_{22}

. A decomposition into two nonzero summands would require a second one-dimensional submodule, which is impossible. Hence this summand is an indecomposable module.

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