Rings, Fields, and Vector Spaces — Studies in Algebra and Group Theory

6 Rings, Fields, and Vector Spaces

15 September 2025 · Denis Auroux

Rings and fields

Definition 6.1 (Ring). A (commutative) ring is a set with two operations such that

is an abelian group with identity ;
is a (commutative) semigroup with identity , namely
- for all ,
- for all ,
- ( for all , if commutative);
distributive law: for all .

Definition 6.2 (Field). A field is a commutative ring such that for all there exists such that , i.e. is an abelian group rather than a semigroup.

Remark.

The ring axioms imply for all . (, plus cancellation.)
The trivial ring is the only case where . By convention this is not a field.
Most rings of interest to us are commutative. (Matrices are the main exception.)
In a field, or . This is not necessarily true in a ring.
Hence, in a field, we have the usual properties of cancellation (simplification) for both addition and multiplication.

Definition 6.3 (Ring / field homomorphism). A ring/field homomorphism is a map that respects both operations:

(we’ve seen this implies , );
;
(this doesn’t follow from , even for fields: consider !).

Proposition 6.4. If is a field homomorphism, then is injective.

Proof. If then there exists such that , so , which implies . So (as an additive group homomorphism), hence is injective. ∎

Example 6.5.

, are rings.
are fields. So is for prime!! This is denoted when viewed as a field.

Why is a field: if in , then its order is (it divides and is ), so . Hence there exists such that . This gives the inverse!

Polynomials and rational functions

Definition 6.6 (Polynomial ring). Given a field , the ring of polynomials in one formal variable is

Remark. is a formal variable, i.e. not an element of anything, though we can evaluate a polynomial at an element of or of any field containing .

Definition 6.7 (Field of rational functions). isn’t a field, but it can be turned into a field by considering fractions (just like the ring the field ): the field of rational functions is

(This generalizes to polynomials and rational functions in any number of variables.)

Power series and Laurent series

Definition 6.8 (Formal power series). The ring of formal power series in is (Add and multiply just like polynomials, term by term. Check that each coefficient in is a finite expression.)

Lemma 6.9. has a multiplicative inverse in iff .

Proof. We want such that . This gives If , clearly there is no solution; if , we can solve inductively: , , … (each step is ✓). ∎

Since every nonzero element of is of the form where is the first nonzero coefficient, to get a field we just need to allow :

Definition 6.10 (Laurent series). The field of Laurent series is

Adjoining roots

Given a field and a polynomial (of degree ), we can evaluate , , and look for roots such that . If there are none in , we can form a field in which has a root.

Example 6.11. : has no roots, but we can form which is a field: .

Example 6.12. , , where is usually called .

On the other hand, over an algebraically closed field such as , every nonconstant polynomial already has a root, and there are no further algebraic extensions.

Characteristic

Given a field , we always have a ring homomorphism (This determines , since generates and , .)

Is this injective? For most fields we’ll consider (e.g. ), it is. If so, say has characteristic zero. Otherwise:

Proposition 6.13. for some prime .

Proof. is a subgroup of , hence of the form . If is not prime, write for . Then , but this implies or (if , multiply by to get ). Since by assumption is the smallest positive integer such that , this is a contradiction. ∎

Definition 6.14 (Characteristic ). Say has characteristic if . (This means !)

So far our only example of such a field is , but there are more.

Theorem 6.15. For all and prime , there exists a unique field with elements (up to isomorphism), and these are all the finite fields.

(There are also infinite fields of characteristic , for example !)

Vector spaces

Definition 6.16 (Vector space). Fix a field . A vector space over is a set with two operations:

addition ,
scalar multiplication ,

such that

is an abelian group (denote by the identity element);
for all ;
for all , ;
for all , ;
for all , .

(2) and (3) are identity and associativity for ; (4) and (5) are the distributive property.

(Note: for all , using the distributive property.)

Definition 6.17 (Subspace). A subspace of a vector space is a nonempty subset that is preserved by addition and scalar multiplication: , (in fact both are ). The second condition, with nonempty, implies .

(So is also a vector space!)

Example 6.18.

with componentwise addition / scalar multiplication.
(sequences in ) {sequences which are eventually zero}.
(isomorphic to the previous example!).
Given any set , ( the case ).
{maps } {continuous maps} {differentiable maps }.

Notes not reached in this lecture, covered at the start of Lecture 7

Span, linear independence, basis

Let be a vector space over .

Definition 6.19 (Span). Given , the span of is the smallest subspace of which contains . Concretely,

Say span if .

Definition 6.20 (Linear independence). We say are linearly independent if

Equivalently, given , we have a linear map and

are linearly independent is injective;
span is surjective.

Definition 6.21 (Basis). are a basis of if they are linearly independent and span .

Then any element of can be expressed uniquely as for some .

Example 6.22. and are a basis of . So are and for most fields . (What if ?)

Note. We will see soon: if has a basis with elements, then every basis of has elements. We say the dimension of is .

One can also consider infinite-dimensional vector spaces:

Definition 6.23 (Span, independence, basis for arbitrary subsets). For any subset:

= the smallest subspace of containing (all finite linear combinations of elements of ).
The elements of are linearly independent if there are no finite linear relations:
is a basis of if its elements are linearly independent and span .

Example 6.24.

is a basis of .
Does have a basis? What is it?

Linear maps

Definition 6.25 (Linear map). Let be vector spaces over . A homomorphism of vector spaces, or linear map, , is any map that is compatible with the operations:

Proposition 6.26. The set of linear maps is itself a vector space over , denoted .

Proof. Given and , define One can check that

and defined in this way are linear maps (so we do have operations on );
these operations on satisfy the axioms of a vector space.

(Rather boring, but worth checking if you’re not sure!) ∎

We’ll soon see: if and , then . (In bases for and , linear maps become matrices!)

How does the choice of the field matter?

Given a subfield (e.g. or ), a vector space over can also be viewed as a vector space over , by “restriction of scalars” (namely, only look at scalar multiplication restricted to the domain ). In particular, itself is a vector space over !

Example 6.27. is a vector space over itself (of dimension 1; is a basis). It is also a vector space over (of dimension 2, with basis ).

If are -vector spaces, hence also -vector spaces, any -linear map is also -linear, but the converse isn’t true: .

For example, complex conjugation is -linear: and for all ; but it is not -linear ().

So: the choice of field matters.