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Let $A$ be a (not necessarily commutative) ring, $E$ a left $A$-module and $E^*$ the dual module of $E$. For an automorphism $u$ of $E$, let $\check{u}$ be the transpose of $u^{-1}$.

In Algebra, Chapter 2, Bourbaki makes the following statement:

The mapping $u\mapsto\check{u}$ is an isomorphism of the linear group $\mathbf{GL}(E)$ onto a subgroup of the linear group $\mathbf{GL}(E^*)$.

Maybe it's obvious, but I don't understand why this mapping is injective.

It is easy to see that $\check{u}=\text{id}_{E^*}$ is equivalent to $(u-\text{id}_E)(E)\subset M$, where $M$ is the orthogonal submodule to $E^*$. Is it known that $M=\{0\}$?

Can someone help me along?

Edit: (at the request of rschwieb)

Let $E,F$ be left $A$-modules, $u$ a linear mapping of $E$ to $F$. The transpose of $u$ is the mapping of $F^*$ into $E^*$ defined by $y^*\mapsto y^*\circ u$. It is a homomorphism of right $A$-modules.

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Please include the definition of "transpose". –  rschwieb Aug 2 '12 at 12:26

1 Answer 1

up vote 1 down vote accepted

There's something wrong: take $A=\mathbb{Z}$ and $E=\mathbb{Z}/3$, so that $GL(E)$ is cyclic of order $2$ but $E^*=Hom(\mathbb{Z}/3,\mathbb{Z})$ and $GL(E^*)$ are trivial.

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Thank you! Must be an error in the book, though I can't imagine what was intended. –  Stefan Walter Aug 2 '12 at 13:56
From the context, I'd say Bourbaki wanted to note it was a morphism, and he got a bit carried away. –  PseudoNeo Aug 2 '12 at 14:22

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