# Is a vector space only defined for one field?

Suppose a vector space V is defined on a field F. Does this at all imply that V is also defined on all fields, or does it only dictate that V is defined on F (and could also work with other fields if proven)?

I realize it's sort of silly to assume anything in math, but my confusion comes from examples of vector spaces that I've seen, such as n-tuples of a field with coordinate-wise addition and scalar multiplication holding for any arbitrary field.

Thanks a lot.

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a finite vector space (finite dimensional over a finite field) cannot be a vector space over, say, $\mathbb{R}$. – yoyo Mar 30 '11 at 15:16
Ah, that makes sense. Thanks. – Nick Van Hoogenstyn Mar 31 '11 at 8:03

## 2 Answers

A priori, if we have an abelian group $V$ (the abelian group structure provides the addition of a vector space), and we give it the structure of a vector space over a field $F$, then we only know how to make $V$ a vector space over $F$, and over any subfield of $F$. This is because when we give $V$ the structure of a vector space over $F$, the information we have specified is how to multiply elements of $V$ by elements of $F$. If $L\subset F$ is a subfield of $F$, then we already know how to define multiplication of elements of $V$ by elements of $L$: elements of $L$ are also elements of $F$, and we just use our definition for them!

For example, the collection of ordered pairs of complex numbers, $V=\mathbb{C}^2$, is an abelian group under the usual addition $$(\alpha_1,\alpha_2)+(\beta_1,\beta_2)=(\alpha_1+\beta_1,\alpha_2+\beta_2) \text{ for all }(\alpha_1,\alpha_2),(\beta_1,\beta_2)\in V .$$ It can be given the structure of a vector space over $\mathbb{C}$ by defining $$\lambda(\alpha_1,\alpha_2)=(\lambda\alpha_1,\lambda\alpha_2)\text{ for all }\lambda\in\mathbb{C},\,\,(\alpha_1,\alpha_2)\in V.$$ But, now that we've done that, it is also a vector space over $\mathbb{R}$, which is a subfield of $\mathbb{C}$ - we know how to multiply elements of $V$ by real numbers because we already have specified how to multiply by complex numbers.

However, the abelian group $V$ cannot be given the structure of a vector space over $\mathbb{Z}/p\mathbb{Z}$ where $p$ is a prime number, which is a field that is not a subfield of $\mathbb{C}$. This is because we would have to have $$p\cdot (\alpha_1,\alpha_2)=(p\alpha_1,p\alpha_2)=0$$ for any $(\alpha_1,\alpha_2)\in V$, which is false.

Finally, I would point out that even if $L$ is not a subfield of $F$, that doesn't prevent $V$ from also being able to be given the structure of a vector space over $L$. In our example of $V=\mathbb{C}^2$, suppose we had originally specified that $V$ was to be considered as a vector space over $\mathbb{R}$. That is, suppose we had said, "Here is our abelian group $V=\mathbb{C}^2$, and we make it into a vector space over $\mathbb{R}$ by defining $$c(\alpha_1,\alpha_2)=(c\alpha_1,c\alpha_2)\text{ for all }c\in\mathbb{R},\,\,(\alpha_1,\alpha_2)\in V."$$ This wouldn't change the fact that it can also be given the structure of a vector space over $\mathbb{C}$, in a way that agrees with the original structure over $\mathbb{R}$, even though $\mathbb{C}$ is a larger field than $\mathbb{R}$.

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I didn't totally follow the example you gave of a field for which the vector space did NOT hold, but I get what you're trying to say and it answered my question. Thanks! – Nick Van Hoogenstyn Mar 30 '11 at 8:37

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I don't understand what it means to say that "$V$ is not a vector space over $\mathbb{Z}/p\mathbb{Z}$": I would say that $V$ cannot naturally be given the structure of a vector space over $\mathbb{Z}/p\mathbb{Z}$. – Qiaochu Yuan Mar 30 '11 at 14:48
@Qiaochu - well, I would agree that the set $V$ can non-naturally be made into a $\mathbb{Z}/p\mathbb{Z}$-vector space, but what I was saying was that the abelian group $V$ is torsion-free - doesn't that mean it's impossible to give it a $\mathbb{Z}/p\mathbb{Z}$-vector space structure? – Zev Chonoles♦ Mar 30 '11 at 17:37
sure, but I didn't see you say "this abelian group" or "this set." In either case I am somewhat uncomfortable with the use of the word "is" in such situations. It hides a lot of subtleties. – Qiaochu Yuan Mar 30 '11 at 17:43
@Qiaochu: Fair point; I was glossing over this point given the level of the OP, but I'll edit. – Zev Chonoles♦ Mar 30 '11 at 18:01

Many of the constructions of vector spaces do not depends on the specific field. Category theory can help to explain why this is since the constructions are really category theory constructions which don't use any property of the field. However, some constructions do use properties of the field, and there isn't any obvious analogue over other fields. For example, smooth functions $\mathbb{R} \to \mathbb{R}$ form a real vector space, and there isn't a natural analogue over the field with $2$ elements.

If you have fields $K \subset L$ (such as $\mathbb{R} \subset \mathbb{C}$) then for any $K$-vector space $V$, there is a natural way to construct an $L$-vector space, $V \otimes_K L$. This is generated by elements of the form $v \otimes \ell$ where $v\in V$ and $\ell \in L$, with rules including $v_1 \otimes \ell + v_2 \otimes \ell = (v_1+v_2)\otimes \ell$, $v \otimes \ell_1 + v\otimes \ell_2 = v\otimes (\ell_1+\ell_2)$, and $k v \otimes \ell = v \otimes k \ell$. Tensor products over $K$ of $K$-vector spaces are always $K$-vector spaces, but these also have the structure of an $L$-vector space because you can multiply the second coordinate by a scalar from $L$. You can identify $V$ with the set of elements of the form $v \otimes 1$.

Also, if you have fields $K \subset L$, then any $L$-vector space is also a $K$-vector space.

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I didn't give this answer because the OP specified that he wanted to know what fields $V$ was defined over, not ways of making new vector spaces from $V$ that would be defined over other fields. But nonetheless it is certainly worth mentioning. – Zev Chonoles♦ Mar 30 '11 at 6:46
And I also know nothing of category theory! :) But I'll take your word that this is valid. – Nick Van Hoogenstyn Mar 30 '11 at 7:51
– Douglas Zare Mar 30 '11 at 22:18
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