field of scalars

Every vector space is built over a chosen field F, and its scalars are exactly the elements of F. A field is a set where you can add, subtract, multiply, and divide by anything nonzero, with all the usual rules — R and C are fields, but so are the rationals Q and the finite field F_p of integers mod a prime p. Pick F before you start; it is part of the data of the space, not an afterthought.

The field is not passive scenery — it controls what is true. The polynomial x^2 + 1 has no root over R, so a real operator can fail to have eigenvalues; over C it splits and every operator does. Over the finite field F_2 = {0, 1} there are only two scalars, so a 'line' through the origin contains just two points. Change the field and you change the answer to almost every downstream question: which matrices diagonalize, which polynomials factor, even how many vectors a finite-dimensional space contains.

Why insist on a field rather than just R? Because the proofs only ever use that you can divide by nonzero scalars. Gaussian elimination divides by pivots; finding a basis divides to normalize; the whole theory runs on division. Demand exactly that — a field — and you get the cleanest possible setting where all the familiar machinery still works, no matter how exotic the scalars.

A caution: drop the division requirement and you leave the world of vector spaces entirely. Scalars from a ring that lacks inverses (like the integers Z) give a module, not a vector space, and modules misbehave — bases can fail to exist. Keeping F a genuine field is precisely what guarantees the comfortable results of Vol I survive into Vol II.