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The following question is based on material from the first section on module theory in Hungerford's algebra and is a possible test question for a qualifying exam in linear algebra. I have been able to prove the first part of the problem which is an easy exercise in checking bijectivity of the map $h$ defined below. I am stuck on the second part of the question.

Given two $A$-modules $E,F$ and a linear map $u: E \rightarrow F$.

$h:E \times F \rightarrow E \times F$ is an automorphism of $E \times F$ given by $h(x,y) = (x,y-u(x))$. Suppose further that there is a linear mapping $v:F \rightarrow E$ and an $ a \in E$ such that $v(u(a)) = a$.

Show there exists an automorphism $w:E \times F \rightarrow E \times F$ such that $w(a,0) = (0,u(a))$

Is the map given by $w(x,y):=(x-v(u(x)),x+u(y))$ or $w(x,y):=(x-v(u(x)),u(x)-y)$ the correct automorphism? I am having trouble figuring out where to put the linear map $v$ even after considering the image of (a,0) and (0,a) under various combinations of $h$ and $h^{-1}$ with $v$.

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    Your first option for $w$ does not make sense, because you have an $x$ in the second component. On the other hand, your second option for $w$ vanishes on $(a,u(a))$, so it is clearly not an isomorphism (presumably, $a$ is not zero...)2011-09-15

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