2
$\begingroup$

This question is probably a silly one, but I'm not the greatest at coming up with counterexamples. Perhaps someone can lend some insight.

The usual (I think?) definition of a vector space $V$ is a space which is closed with respect to (vector) addition and scalar multiplication (with scalars taken from some field $\mathbb{K}$) and upon which these operations satisfy a number of algebraic axioms (which I won't list here but which are listed elsewhere). So my question:

Can someone give an example of a pair $(V,\mathbb{K})$ consisting of a space $V$ which is closed with respect to addition and multiplication by elements of a field $\mathbb{K}$ but which isn't a vector space due to it not satisfying any/all of the vector space axioms? I tried Googling such a thing as well as searching for various related phrases here in SXE but I had no luck.

Any information would be hugely appreciated!

Edit: In the third paragraph, I do still want $\mathbb{K}$ to be a field. I've added that to the paragraph itself.

$\endgroup$
4
  • 1
    $\begingroup$ I wonder. Do you still want $\mathbb{K}$ to be a field in your third paragraph? If not, you could search about R-Modules or group actions. $\endgroup$ Commented Mar 8, 2015 at 19:42
  • $\begingroup$ @DavidMolano - Thank you for pointing out my carelessness! I've edited the question to indicate my intention (which was for $\mathbb{K}$ to be a field). $\endgroup$
    – cstover
    Commented Mar 8, 2015 at 19:45
  • 1
    $\begingroup$ Just define any two functions $V\times V\to V$ and $\mathbb K\times V\to V$ and call them addition and scalar multiplication. Odds are, they won't satisfy the axioms. For example, take $V=\mathbb R^n$, $\mathbb K=\mathbb R$, $(x_1,\ldots,x_n)+(y_1,\ldots,y_n)=(x_1-y_1,\ldots,x_n-y_n)$, and $a\cdot(x_1,\ldots,x_n)=(a+x_1,\ldots,a+x_n)$. $\endgroup$
    – user856
    Commented Mar 8, 2015 at 19:52
  • $\begingroup$ @Rahul - I knew that my question must have been a silly one, but your answer has illustrated how unequivocally true that sentiment is. Thanks for taking the time to reply! $\endgroup$
    – cstover
    Commented Mar 8, 2015 at 19:55

2 Answers 2

2
$\begingroup$

For example $V$ could be the set of all ordered pairs $(a,b)$ where $a$ and $b$ are real, under the usual addition, and we could have $r(a,b)=(0,0)$ for all reals $r$, $a$, and $b$.

$\endgroup$
1
  • $\begingroup$ This is indeed a very worthwhile and very general answer. I of course didn't mention triviality of the example, etc., because I wanted to see a wide variety of counterexamples; even so, I'm going to upvote but save "accepting" it to see whether there are less-trivial examples provided. Many thanks! $\endgroup$
    – cstover
    Commented Mar 8, 2015 at 19:47
1
$\begingroup$

Let $K$ be $\mathbb{Z}/3\mathbb{Z}$ and consider the action on $\mathbb{Z}/6\mathbb{Z}$ where $1\cdot a=4a$, $2\cdot a=2a$ for $a\in\mathbb{Z}/6\mathbb{Z}$. This turns $\mathbb{Z}/6\mathbb{Z}$ into a $\mathbb{Z}/3\mathbb{Z}$ module, but this module is not unital hence is not a vector space.

$\endgroup$
5
  • $\begingroup$ This is precisely the sort of answer I'd hoped for! I notice that this is a very algebraic counterexample: I wonder, then, if this can be generalized for general fields $\mathbb{K}=\mathbb{Z}/p\mathbb{Z}$, for various rings $\mathbb{Z}/n\mathbb{Z}$ ($p$ = prime), and for miscellaneous operations between them? I'm choosing this as the best answer regardless, but if you happen to know of generaliations of your counterexample, I'd love to hear them! $\endgroup$
    – cstover
    Commented Mar 8, 2015 at 19:50
  • 1
    $\begingroup$ @twigg1313 I've always thought this example was cool, but I've never looked for generalizations. $\endgroup$ Commented Mar 8, 2015 at 19:55
  • 1
    $\begingroup$ @twigg1313 There are other nontrivial examples. We could take any $\mathbb{R}\times \mathbb{R}$ module and have $\mathbb{R}$ act on it by having $a\in\mathbb{R}$ act as $(a,0)\in\mathbb{R}\times\mathbb{R}$. Again, it isn't unital. $\endgroup$ Commented Mar 8, 2015 at 19:58
  • 1
    $\begingroup$ @twigg131 Actually now that I think about it that is the same as my answer, where we treat $\mathbb{Z}/6\mathbb{Z}$ as $\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$. Take a module over a direct product ring where one of the factors is a field, and have an element $a$ of the field act as $(a,0,0,\ldots,0)$. $\endgroup$ Commented Mar 8, 2015 at 20:02
  • $\begingroup$ Excellent observation! I'm going to be getting on a plane soon and I'm going to spend that time tinkering with generalizations (of the flavor I mentioned) as well because I'm sure they must exist. :D $\endgroup$
    – cstover
    Commented Mar 8, 2015 at 20:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .