Ken wrote:

> So the field k itself is some kind of trivial vector space? I feel like there's a notational pun I'm missing.

Yes, every finite-dimensional vector space over a field \\(k\\) is isomorphic to \\(k^n\\), the set of \\(n\\)-tuples of elements of \\(k\\), with the vector space operations defined as follows:

* Addition: \\( (x_1, \dots, x_n) + (y_1 , \dots y_n) = (x_1 + y_1 , \dots x_n + y_n) \\).

* Scalar multiplication: \\( a(x_1, \dots, x_n) = (ax_1, \dots, ay_n) \\).

When \\(n = 1\\) we get \\(k^1\\), which is just \\(k\\). A 1-tuple of elements of \\(k\\) is just an element of \\(k\\).

You may have seen this stuff when \\(k = \mathbb{R}\\), the field of real numbers. They say the real line is a 1-dimensional real vector space.

(I wouldn't say the case \\(n = 1\\) is trivial: the _really_ trivial case is \\(n = 0\\).)

> So the field k itself is some kind of trivial vector space? I feel like there's a notational pun I'm missing.

Yes, every finite-dimensional vector space over a field \\(k\\) is isomorphic to \\(k^n\\), the set of \\(n\\)-tuples of elements of \\(k\\), with the vector space operations defined as follows:

* Addition: \\( (x_1, \dots, x_n) + (y_1 , \dots y_n) = (x_1 + y_1 , \dots x_n + y_n) \\).

* Scalar multiplication: \\( a(x_1, \dots, x_n) = (ax_1, \dots, ay_n) \\).

When \\(n = 1\\) we get \\(k^1\\), which is just \\(k\\). A 1-tuple of elements of \\(k\\) is just an element of \\(k\\).

You may have seen this stuff when \\(k = \mathbb{R}\\), the field of real numbers. They say the real line is a 1-dimensional real vector space.

(I wouldn't say the case \\(n = 1\\) is trivial: the _really_ trivial case is \\(n = 0\\).)