Let \(k\) be a field. The characteristic of \(k\) is the smallest integer \(p\) such that \(\sum_{i=1}^p 1=0\) in \(k\). Note that, since \(k\) is a field, if \(p\) is not a prime then \(p=ab\) and \(0=p\cdot 1=(a\cdot 1)(b\cdot 1)\), a contradiction. If no such \(p\) exists then the characteristic is defined to be \(0\).

We say that \(k\) is algebraically closed if every polynomial \(p(X)\) with coefficients in \(k\) has a zero in \(k\), i.e., there exists \(\alpha\in k\) such that \(p(\alpha)=0\). This is equivalent to the statement that every polynomial splits over \(k\), i.e., \(p(X)\) of degree \(n\) can be written as
\[ p(X)=\prod_{i=1}^n (X-\alpha_i)\]for some \(\alpha_i\in k\).

An example of an algebraically closed field is the complex numbers \(\mathbb{C}\), whereas \(X^2+1\) does not split over \(\mathbb{R}\) or \(\mathbb{Q}\). The polynomial \(X^q-X\) in \(\mathbb{F}_p[X]\) splits over the finite field \(\mathbb{F}_q\), but not over any smaller field, so no finite field \(\mathbb{F}_q\) is algebraically closed.

The most important statement from the theory of fields that we will need is the following.

Lemma 0.1. Any finite subgroup of \(k^\times\), the multiplicative group of a field, is cyclic.

Proof: Let \(G\) be any finite subgroup of \(k^\times\) of order \(n\), and note that \(G\) is an abelian group. Note also that the exponent of \(G\), i.e., the smallest integer \(m\) such that \(g^m=1\) for all \(m\in G\), is equal to \(n\) if and only if \(G\) is cyclic, so assume that \(m\lt n\). Therefore every element of \(G\) is a zero of the polynomial \(X^m-1\) in \(k[X]\). However, this polynomial can have at most \(m\) distinct zeros, a contradiction.

Hence \(m=n\) and so \(G\) is cyclic, as claimed. □

We see a few fields in this course. We normally work over \(\mathbb{R}\) and \(\mathbb{C}\), but do talk about finite fields. Recall that any finite field, denoted \(\mathbb{F}_q\), has order \(q\) for some power \(q\) of the characteristic of \(\mathbb{F}_q\). Up to isomorphism, there is exactly one for each prime power \(q\).