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Is it true that given any vector space $V$ over any field $F$ we have $(\alpha-\beta)(x-y)=\alpha x-\alpha y-\beta x+\beta y$ for all $x,y \in V$ and $\alpha,\beta \in F$?

I figured that, for example, if, say, $a$ and $b$ are contained in a field $F$, which—by definition—is a commutative associative ring, which, similarly, is—by definition—an additive abelian group, then the equation \begin{eqnarray} x+a & = & b \end{eqnarray} has a unique solution. This is because if we consider $s$ a solution to the above equation—that is, $s+a=b$, then \begin{eqnarray} (s+a)+(-a) & = & b+(-a), \end{eqnarray} so \begin{eqnarray} (s+a)+(-a) & = & s+(a+(-a)) & = & s+0 & = & s, \end{eqnarray} thus \begin{eqnarray} s & = & b+(-a). \end{eqnarray}

This being the case—that is, this uniqueness, then

\begin{eqnarray} (\alpha-\beta)= \alpha+(-\beta) \end{eqnarray}

should be a unique scalar in $F$, so by the multiplicative distributivity of vectors in a vector space shouldn't the proposed operation hold for all vector spaces of and arbitrary non-single-element field? I don't even know if my thoughts are in the right place really. Any help is appreciated.

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What about $GF(2)$? –  Trancot Apr 8 '13 at 6:25
Vinberg says that "[a] ring that consists of only zero is not regarded as a field," but what about non-zero single-element rings? Is that even possible? –  Trancot Apr 8 '13 at 6:28
No. A ring, by definition, has a zero element; if the ring has only one element, it's got to be zero. –  Zev Chonoles Apr 8 '13 at 6:29
@Trancot You can edit your question, no need to leave the above remarks as comments. –  Git Gud Apr 8 '13 at 6:29
@ZevChonoles How should I approach proving this? –  Trancot Apr 8 '13 at 6:36
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1 Answer

In a word, yes. Wikipedia's entry on modules says this explicitly among the module axioms in the formal definition. A vector space is a module over a field.

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