# Spaces that satisfy closure with respect to addition and scalar multiplication but aren't vector spaces?

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.

• 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. Commented Mar 8, 2015 at 19:42
• @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). Commented Mar 8, 2015 at 19:45
• 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 $a\cdot(x_1,\ldots,x_n)=(a+x_1,\ldots,a+x_n)$.
– user856
Commented Mar 8, 2015 at 19:52
• @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! Commented Mar 8, 2015 at 19:55

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 $b$.
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.
• 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! Commented Mar 8, 2015 at 19:50
• @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. Commented Mar 8, 2015 at 19:58
• @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 $(a,0,0,\ldots,0)$. Commented Mar 8, 2015 at 20:02