Lecture. The subfield generated by a subset and simple extensions [l2]

  • Apr 1, 2026

Definition 1. generated fields [1301]

    Let $K / F$ be a field extension and $S$ be a subset of $K$. Let $$ F(S) = \text{the intersection of all subfields of } K \text{ which contains }F\text{ and }S. $$ One can check that $F(S)$ is a subfield of $K$ and is the smallest subfield of $K$ that contains both $F$ and $S$. We call $F(S)$ the field generated by $S$ over $F$.

    Notation. some notations [1301A]

      When $S = \{ \alpha_{1}, \dots, \alpha_{n} \}$, we write $F(S)$ as $F(\alpha_{1}, \dots, \alpha_{n})$.

      Given $S \subseteq K$, taking intersections is a general way to generate a smaller field extension of $F$, which also gives the smallest one that contains both $S$ and $F$.

      From this definition, it is hard to compute $F(S)$ because there might be infinite subfields containing $F$ and $S$. So we need a workaround to get $F(S)$.

      We know that taking field of fractions gives us a field, if we consider each element of $F(S)$ as a fraction of elements, what do we know about these elements?

      • They must belong to an integral domain.
      • They must be composed of elements from $F$ and $S$.

      Interestingly, $K$ is a field, so taking fractions of elements from $S$ simply brings another element of $K$, and it also belongs to a subfield of $K$ containing $S$. Actually, to construct a subfield of $K$ containing $S$, we first need to take multiples of any elements of $S$, then take inverses of those multiples, next, take multiples of those inverses and elements of $S$.

      Remember that the needed subfields also contain $F$, given a subfield containing $S$, how do we enlarge it so it contains $F$, too? Well, simply multiplying every element by elements of $F$ and taking sums of those multiples is a good way. Recall that $\operatorname{Frac}(F) = F$, hence we have the following proposition:

      Proposition 2. the field of fractions of $F[S]$ is $F(S)$ [1302]

        Given $K / F$ and $S \subseteq K$, let $$ F[S] = \{ \text{finite sums of }a \cdot \alpha_{1} \alpha_{2} \dots \alpha_{n} \mid a \in F, \alpha_{1}, \dots, \alpha_{n} \in S, n \in \mathbb{Z}_{\geq 0}\}. $$ Then $F[S]$ is a subring of $K$ and $$ F(S) = \left\{ \left. \frac{a}{b} \right\vert a, b \in F[S], b \neq 0 \right\}. $$ ($F(S)$ is the fraction field of $F[S]$, and it is exactly the same object as in generated fields.)

        Exegesis 2.1. why finiteness [1302A]

          TODO

          Taking multiples ensures $F[S]$ is closed under multiplication. Taking finite sums then ensures $F[S]$ is closed under addition. Moreover, we can take $a = 1$ and $n = 0$, then $F[S]$ contains $1$. It is easy to verify that $F[S]$ is actually an integral domain, thus we can take its field of fractions.

          Proposition 3. identity on generated fields [1303]

            Let $\varphi : F(S) \to F(S)$ be a field homomorphism such that $\varphi(\alpha) = \alpha$ for all $\alpha \in F \cup S$. Then $\varphi = id_{F(S)}$.

            Exegesis 3.1. why extends to $F(S)$ [1303A]

              By the construction of $F(S)$ from elements of $F \cup S$ in Proposition 1302, obviously $\varphi = id_{F(S)}$.

              Definition 4. finitely generated, simple extensions and primitive elements [1304]

                1. A field extension $K / F$ is called finitely generated if there exist $\alpha_{1}, \dots, \alpha_{n} \in K$ such that $K = F(\alpha_{1}, \dots, \alpha_{n})$.
                2. A field extension $K / F$ is called simple if $K = F(\alpha)$ for some $\alpha \in K$. In this case, $\alpha$ is called a primitive element for the field extension $K / F$.

                Example 5. $F(\theta), \mathbb{F}_p(t) / \mathbb{F}_p, F(x)$ [1305]

                  1. Let $p(x)$ be an irreducible element in $F[x]$. Then by Theorem 1206, we know that $K = F[x] / \left( p(x) \right)$ is a field extension of $F$, and $K = F + F\theta + \cdots + F\theta^{n-1}$ with $\theta = \overline{x} \in F[x] / \left( p(x) \right)$. Hence $K = F[\theta] = F(\theta)$, and $K / F$ is a simple extension, and $\theta$ is a primitive element for $K / F$.
                  2. $\mathbb{F}_{p}(t) / \mathbb{F}_{p}$ is simple and $t$ is a primitive element for $\mathbb{F}_{p}(t) / \mathbb{F}_{p}$. In this case, $\mathbb{F}_{p}(t) \neq \mathbb{F}_{p}[t]$.
                  3. Let $F$ be a field. Then $$F(x) = \left\{ \left. \frac{a(x)}{b(x)} \right\vert a(x), b(x) \in F[x], b(x) \neq 0 \right\}$$ is a field with $$\begin{align*}+ &: \frac{a(x)}{b(x)} + \frac{c(x)}{d(x)} = \frac{a(x) d(x) + b(x) c(x)}{b(x) d(x)} \\ \times &: \frac{a(x)}{b(x)} \cdot \frac{c(x)}{d(x)} = \frac{a(x) c(x)}{b(x) d(x)}\end{align*}.$$$F(x) / F$ is simple and $x$ is a primitive element for $F(x) / F$. $F[x] \neq F(x)$.