We are finally ready to talk about one of the most important types of rings, namely fields.

Let \(R\) be a ring, and let \(x \in R\). We say that \(x\) is a unit if there exists some \(y \in R\) such that \(xy = yx = 1\), that is, if \(x\) has a multiplicative inverse. For example, in any ring, \(1\) is always a unit, and \(0\) is never a unit.

Then, a field is defined as a commutative ring in which every nonzero element is a unit. We can equivalently say that a field is a commutative ring for which the nonzero elements form a group under multiplication. We usually use the notation \(x^{-1}\) for the multiplicative inverse of \(x\) in a field.

Here are some examples of fields which we have already seen:

  • The real numbers, \(\mathbb{R}\)
  • The rational numbers, \(\mathbb{Q}\)
  • The integers modulo \(2\), \(\mathbb{Z}_2\). Note that the multiplicative inverse for \(1\) in any ring necessarily exists (it is also \(1\)), and this ring has no other nonzero elements to consider, so it must be a field.

Here are some non-examples:

  • The ring of integers, \(\mathbb{Z}\). This fails to be a field because the only nonzero elements with multiplicative inverses are \(1\) and \(-1\); there is no integer which can be multiplied by \(2\) to yield \(1\), for example.

  • The ring of integers modulo \(4\), \(\mathbb{Z}_4\). This fails to be a field because the element \(\overline{2}\) does not have a multiplicative inverse. We can check this exhaustively:

    \[\begin{split}\overline{1} \cdot \overline{2} = \overline{2} \\ \overline{2} \cdot \overline{2} = \overline{0} \\ \overline{3} \cdot \overline{2} = \overline{2}\end{split}\]

    None of these are equal to \(\overline{1}\), so we can conclude that none of them is a multiplicative inverse of \(\overline{2}\).

  • The ring of \(2 \times 2\) matrices with entries in \(\mathbb{R}\). This fails to be a field because it is non-commutative, as we have seen, and also because there are nonzero elements which do not have multiplicative inverses.

We also have a name for rings in which all nonzero elements are units but multiplication is not necessarily commutative: these are called division rings, or sometimes skew fields. It just happens that most of the interesting examples of division rings are also fields, so we tend to spend more time thinking about fields. There is, however, one important example of a division ring which is not a field, which we will see later on.

A quick diversion into set theory

There are a couple of important results concerning fields which we will soon establish, but first we need another quick diversion into set theory. This builds upon the Permutations section in the chapter on groups, so if you need a refresher now might be a good time to revisit it.


Let \(A\) and \(B\) be sets. We say that \(A\) is a subset of \(B\) if \(x \in A \Rightarrow x \in B\), that is, every element of \(A\) is also an element of B. Symbolically, we write \(A \subseteq B\).

One consequence of this definition is that every set is a subset of itself. If we want to rule out this case, we would say that \(A\) is a proper subset of \(B\), and this is written \(A \subset B\).

Images of functions

We call the set of elements that can be produced as a result of applying a function \(f\) to an element of its domain the image of \(f\). Note that this set is necessarily a subset of the codomain; in fact, another way of defining a surjective function is one whose image is equal to its codomain.

Notationally, the image of a function \(f : X \rightarrow Y\) is written as \(f(X)\) — this is arguably a bit of an abuse of notation, as this looks like we’re applying a function to a set, which, if we’re being pedantic, doesn’t make sense — but it is defined as follows:

\[f(X) = \{\, f(x) \,|\, x \in X \,\}\]

So we have that \(f(X) \subseteq Y\) is true for any function \(f : X \rightarrow Y\), and also that \(f(X) = Y\) if and only if \(f\) is surjective.

Injectivity and surjectivity with finite sets

Here is an important result which we will need shortly:

  • Let \(X\) be a set with finitely many elements, and let \(f : X \rightarrow X\) be a function. Then \(f\) is injective if and only if it is surjective.

In this proof we will use \(n\) to refer to the size of the set \(X\), i.e. \(X\) has \(n\) distinct elements.

First, suppose \(f\) is injective. That is, if \(x \neq y\), then \(f(x) \neq f(y)\). It follows that \(f(X)\) has at least \(n\) elements, as each of the \(n\) elements of \(X\) which we can apply \(f\) to is mapped to a distinct element of the codomain of \(f\) (which, here, is also \(X\)). Since \(X\) is also the codomain of \(f\), we have that \(f(X) \subseteq X\), and in particular, \(f(X)\) can have no more than \(n\) elements (since \(X\) only has \(n\) elements). So \(f(X)\) has exactly \(n\) elements, and since each of them is an element of \(X\) we can conclude that \(f(X) = X\), i.e. \(f\) is surjective.

Conversely, suppose that \(f\) is surjective, i.e. each element of \(X\) can be obtained by applying \(f\) to some (possibly different) element of \(X\). In this case it must be injective; if it weren’t, there would be at least two elements of \(X\) which were mapped to the same thing by \(f\), and then of the remaining \(n - 2\) elements of \(X\), we have \(n - 1\) elements of \(X\) to reach, which is not possible.

Okay, that’s everything. Back to fields!

Every field is an integral domain

This is fairly straightforward to prove. Let \(F\) be a field, and let \(a, b \in F\), with \(a \neq 0\). Suppose \(ab = 0\). Since \(F\) is a field, \(a^{-1}\) exists. Multiplying both sides by \(a^{-1}\) yields \(a^{-1}ab = a^{-1}0\), which simplifies to \(b = 0\). That is, \(F\) has no zero-divisors. We have by assumption that \(F\) is commutative (since this is one of the requirements for a field) and therefore \(F\) is an integral domain.

This gives us a useful trick for determining whether some ring is a field or not: since all fields are integral domains, we can immediately deduce that a ring cannot be a field if it fails to be an integral domain, e.g. if it has any zero-divisors. Note that for two of the three non-examples of fields listed earlier, namely \(\mathbb{Z}_4\) and \(\mathrm{Mat}(2;\mathbb{R})\), it can be shown that they are not fields in this way.

Let’s do a quick recap on the hierarchy we have seen so far; we have:

  • rings \(\supset\) commutative rings \(\supset\) integral domains \(\supset\) fields.

That is, every commutative ring is a ring (but not every ring is commutative), every integral domain is a commutative ring (but not every commutative ring is an integral domain), and so on.

Every finite integral domain is a field

This is slightly more difficult to prove, so don’t worry if the proof doesn’t make complete sense to you at first.

Let \(R\) be a finite integral domain, and let \(a \in R\) with \(a \neq 0\). Now, define a function \(\lambda_a : R \rightarrow R\) by \(\lambda_a(x) = ax\), that is, the function \(\lambda_a\) represents multiplication by \(a\). Now let \(b, c \in R\), and notice that the cancellation law for integral domains tells us that \(ab = ac\) implies \(b = c\). That is, if \(\lambda_a(b) = \lambda_a(c)\), then \(b = c\). This is precisely what it means for the function \(\lambda_a\) to be injective.

Using our previously established result that an injective function on a finite set must also be surjective, we can deduce that \(\lambda_a\) is surjective, and consequently also bijective. Therefore, it must have an inverse function \(\lambda_a^{-1}\), and in particular if we let \(d = \lambda_a^{-1}(1)\), then we have that \(ad = 1\), i.e. \(d\) is a multiplicative inverse for \(a\).

We have now found a multiplicative inverse for every nonzero element of \(R\), and we have by assumption that \(R\) is commutative, so it follows that \(R\) is a field.

Look back now to exercise 6.4 in the previous chapter, which asks you to provide a rule for whether \(\mathbb{Z}_m\) is an integral domain given any \(m \geq 2\). This is quite a difficult exercise but the result is quite useful, so I recommend that you look at the solution now if you weren’t able to solve it yourself.

Using our new result that every finite integral domain is a field, we can now strengthen the result we found in exercise 6.4: since \(\mathbb{Z}_m\) is finite, if it is an integral domain, it must be a field. The field of integers modulo \(m\) for an appropriately chosen \(m\) (I’m deliberately being vague to avoid spoiling you for exercise 6.4 if you want to have another go at it) is generally my go-to example of a field, as these fields tend to be the simplest to deal with and can be faithfully represented on computers very easily — unlike, say, \(\mathbb{R}\).