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Is it possible to construct a quasi-vectorial space without an identity element?

In Apostol Multivariable Calculus, $1.5$ exercise $30 b$, he asks the reader to prove that Axiom $10$ is independent of the other axioms for vector spaces. Axiom $10$ says that $$1 * x=x$$

To do this, it is equivalent to prove that $(-1)*x\not = -x$, since $(-1)*x=-x$ implies that $1*x=x$ (we can prove it using the other axioms. My approach is to find $(V,+,*)$ that satisfies the first $9$ axioms and contradicts the axiom $10$.

My attempts so far:

First, I tried setting $V=\mathbb{R}$, redefining $a*x$ to be variations on multiplication. For example, I tried $a*x=|ax|$ but ran into problems with $|a(x+y)|\not=|ax|+|ay|$ in general.

Then, I tried setting $V=\mathbb{R^2}$ and $a*x=ax^T$ if $a$ is positive and $a*x=ax$ if $a$ is not positive. However, I ran into problems with $(a+b)*x\not=a*x+b*x$ for $a=-1$,$b=1$,$x=(1,0)$.

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marked as duplicate by Gerry Myerson, Jennifer Dylan, Asaf Karagila, Matt N., fpqc Sep 4 '12 at 13:52

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1 Answer

up vote 3 down vote accepted

Take any additive group $V$ and let $F$ be a field. Define a scalar product by $k \cdot v = 0$ for all $k \in F$ and $v \in V$, I believe this satisfies all the other axioms except $1 \cdot v = v$.

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Soo simple...I can't believe I missed it. –  Andrew Salmon Jun 23 '12 at 0:15
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