 Quote by WWGD
Hi, Algebraists:
Say V is finite-dimensional over F . Is there more than one way of defining the
action of F on V (of course, satisfying the vector space axioms.) By different
ways, I mean that the two actions are not equivariant.
Thanks.
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It depends upon the field F. If F is Q (the field of rationals) or more generally a prime field (i.e. either Q or Z_p for some prome p), then there is only one possible field action of F on V: The vector space axioms determine the value of nv for each natural number n and each v in V, and this can only be extended in one way to Z and then to Q if the axioms shall hold.
But in general, there can be more than one possibility. For example, If V is a vector space over C (the complex numbers), and if a' denotes the conjugate of a complex number a, then we define a new action x of C upon V by a x v = a'v (where a'v is computed by the original action). It can be shown that V with this new action and the same addition as before satisfies the vector space axioms.