Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In a previous question, Algebra Fan said:

In general, if you have two fields $K\subset L$, $L$ can be viewed as a $K$− vector space with the obvious scalar multiplication: if $k\in K$ and $\ell \in L$, then $k⋅\ell =k\ell$ where $k\ell$ is just the product of $k$ and $\ell$.

I've been thinking about this for a while, and I'm not sure I understand it.

We say $V$ is a vectorspace over $R$ if it follows some rules, including that there be some multiplication operation $R\times V\to V$. Some questions:

  1. Is the analogous multiplication operation in a general field $K\times L \to L$ or $K\times L\to K$?
  2. Is the dimension of the vectorspace the order of $K$? The order of $L$? The index of $K$ in $L$?
  3. Are the basis units just the elements of $K$?

Alternatively, if someone could link me to lecture notes etc. giving some intuition about what's going on that would be appreciated.

share|improve this question
1  
Try the example with K being the field of rationals, and L being the field extension consisiting of all numbers of the form a + $\sqrt{2}$ b, where a and b are rationals. –  Joe Aug 15 '11 at 16:52
    
Hint: L has dimension 2 over K and a basis is {1, $\sqrt{2}$ } . –  Joe Aug 15 '11 at 17:01
1  
@Matt I like that we picked the same example. I thought about going for $\mathbf{Q}(i)$, but I thought, "No, too adventurous." –  Dylan Moreland Aug 15 '11 at 17:06
add comment

3 Answers 3

up vote 3 down vote accepted

Let's keep in mind a simple example of a field extension, like $$ \mathbf Q \subset \mathbf Q(\sqrt{2}) = \mathbf{Q}[\sqrt{2}] \approx \mathbf Q[X]/(X^2 - 2). $$

  1. With $K \subset L$, you're going to have $K \times L \to L$, and moreover this map is surjective because $1 \in K$.

  2. You want to talk about the dimension of $L$ as a $K$-vector space. As usual, that's the number of elements in a $K$-basis for $L$. In our example, we have a basis $\{1, \sqrt{2}\}$ over $\mathbf Q$, so $\dim_KL = 2$. Authors often denote this by $[L : K]$. The numbers you mentioned are all infinite.

    It's definitely possible for $[L : K]$ to be infinite; for example, take $L = K(X)$, the field of rational functions with coefficients in $K$.

  3. If the extension is non-trivial, i.e. $L \neq K$, then your basis for $L$ over $K$ will necessarily contain elements of $L \setminus K$.

I don't know of a reference that spends much time on this. My hope is that once the part of your brain that does linear algebra adjusts to this new setting, everything will seem very natural. I think that justifying the assertions I've made about $\mathbf Q(\sqrt{2})$ and then going on to do this for more interesting fields is time well spent. I listed some sources for exercises in Galois theory in an earlier answer, and those sources will all have problems about the degrees of field extensions.

share|improve this answer
    
I think I was psyching myself out: I was convinced that only finite fields were understandable, but I didn't know many examples of finite fields with non-trivial subfields. But now I see that of course an infinite field can have finite dimension... Thanks! –  Xodarap Aug 16 '11 at 14:37
add comment

Let $\mathbf{F}$ be a field. A vector space over $\mathbf{F}$ is an abelian group $(\mathbf{V},+)$ together with an operation $\mathbf{F}\times\mathbf{V}\to\mathbf{V}$, denoted by $\cdot$, such that:

  • $\alpha\cdot (\mathbf{v}+\mathbf{w}) = \alpha\cdot\mathbf{v} + \alpha\cdot\mathbf{w}$ for all $\alpha\in\mathbf{F}$, $\mathbf{v},\mathbf{w}\in\mathbf{V}$;
  • $(\alpha+\beta)\cdot\mathbf{v} = \alpha\cdot\mathbf{v}+\beta\cdot\mathbf{v}$ for all $\alpha,\beta\in\mathbf{F}$, $\mathbf{v}\in\mathbf{V}$;
  • $1\cdot\mathbf{v}=\mathbf{v}$ for all $\mathbf{v}\in\mathbf{V}$;
  • $\alpha\cdot(\beta\cdot\mathbf{v}) = (\alpha\beta)\cdot\mathbf{v}$ for all $\alpha,\beta\in\mathbf{F}$, $\mathbf{v}\in\mathbf{V}$.

Now, suppose that $K\subseteq L$ are fields. We "forget" about the multiplication of $L$ and consider $L$ merely as an abelian group under its usual addition. And we take the multiplication map of $L$, $L\times L\to L$, and restrict it to the subset $K\times L$ to get a function $K\times L\to L$.

Because $L$ is a field, multiplication distributes over the sum, multiplication by $1$ is the identity map, and multiplication is associative. That is, the restriction of the multiplication of $L$ to $K\times L$ satisfies the four condetions above, and so "works" as a scalar multiplication over $K$. That means that $(L,+)$ can be considered as a vector space over $K$ using this scalar product.

So to answer your questions:

  1. The operation needs to be $K\times L\to L$ because $L$ is the "vector space". More generally, note that $1\times \ell = \ell$ for all $\ell\in L$, so the range has to be $L$, not $K$.

  2. The dimension of $L$ as a vector space over $K$ is, by definition, the degree of the extension $[L\colon K]$. This is not equal to the index of $K$ in $L$. I'm cheating, though, because the degree of the extension is precisely defined to be the dimension of $L$ as a vector space over $K$.

    To see that it is not in general equal to the dimension, consider $K=\mathbb{R}$ and $L=\mathbb{C}$. Then $L$ is a $2$-dimensional vector space over $K$ (${1,i}$ is a basis), but the index of $K$ in $L$ is infinite, since $(a+bi)+\mathbb{R} = (x+yi)+\mathbb{R}$ if and only if $b=y$.

    For a simple extensions, $L=K(\alpha)$, then the dimension will be countably infinite if $\alpha$ is transcendental over $K$, and equal to the degree of the irreducible polynomial of $\alpha$ over $K$ if $\alpha$ is algebraic over $K$.

  3. No, a basis will generally contain elements not in $K$ (unless $L=K$); there are, of course, many, many, many bases. In the two simple cases I described above, if $L=K(\alpha)$ with $\alpha$ algebraic over $K$, if the monic irreducible polynomial of $\alpha$ over $K$ is of degree $n$, then one possible basis is $[1,\alpha,\alpha^2,\ldots,\alpha^{n-1}]$; if $\alpha$ is transcendental over $K$, then a basis is a bit more difficult to describe. You will have all powers of $\alpha$, and in addition all "fractions" of the form $\frac{\alpha^i}{g(\alpha)}$ where $g(x)$ is a monic irreducible over $K$ and $i\lt\deg(g)$ (this is basically the partial fractions decomposition).

share|improve this answer
add comment

I'll try out questions 1 and 3 on the most familiar example first: $\mathbb{Q}\subseteq \mathbb{R}$. The idea is that we want to view $\mathbb{R}$ as a vector space over $\mathbb{Q}$. (Unfortunately, while we can answer question 2 for this example, it's not terribly illuminating on what happens in general).

Q1. If you multiply a rational number and a real number, you know the answer will be real. Will it always be rational? No - anytime you multiply an irrational and a (nonzero) rational, you get an irrational.

Thus, the multiplication map cannot land in $\mathbb{Q}$, so the multiplication map $\mathbb{Q}\times\mathbb{R}\rightarrow \mathbb{R}$. In general, then, you must have $K\times L\rightarrow L$. In fact, we have $kl\in K$ iff $l\in K$ for $k\neq 0$ (just as we saw in the rational/irrational case).

Q3. If the basis elements were just the rationals, the whole thing would just be the rationals. This is because, by definition, every real number would be expressible as $\sum q_i b_i$ (a finite sum) where the $q_i$ are the scalars in the ground field (i.e., are rational numbers) and the $b_i$ are the basis elements (i.e., rational numbers). But a sum of products of rational numbers is always rational.

In the basis, we can choose a single rational number (typically people choose $1$.), but if we choose any more rationals for the basis, they will be dependent. To get a basis of $\mathbb{R}$, we could start by, say, adding $\sqrt{2}$ into our set of $\{1\}$ to get $\{1,\sqrt{2}\}$. This still wouldn't span all of $\mathbb{R}$ so we could add more, like $\pi$ or $e$ -and it still wouldn't span all of $\mathbb{R}$. In fact, it can be shown that the cardinality of the basis must be the same as the cardinality of $\mathbb{R}$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.