0
$\begingroup$

The dimension of a vector space is the number of its basis. And this field is defined over a field.

I am figuring out how a field influence the dimension of vector space. For example $\mathbb{F}[x]$ is finite-dimensional because the degree could approach infinity.

However, $\mathbb{F}_n[x]$ is finite-dimensional since the set of $N+1$ vectors $\{x^n\}^N_{n=0}$ is a basis.

So, a vector space could be finite dimension over a field like the former one however, be infinite dimensional in the later one. Is it true?

$\endgroup$
1
$\begingroup$

Yes, this is possible, at least if we identify the vector spaces (see below). For example, the dimension of $\mathbb{C}$ over itself is $1$ (this is true for any vector space), the dimension of $\mathbb{C}$ over $\mathbb{R}$ is $2$, and the dimension over $\mathbb{Q}$ is infinite.

Similarly, the splitting field $\mathbb{K}'$ of an irreducible polynomial in $\mathbb{K}[x]$ of degree $n$ has dimension $n$ over $\mathbb{K}$ but dimension $1$ over itself.

Strictly speaking, the underlying field is part of the data defining a given vector space, so, e.g., $\mathbb{C}$ viewed as a complex vector space is different from $\mathbb{C}$ viewed as a real vector space, despite that the underlying addition operation is the same. (Note that the scalar multiplication maps are, however, different.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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