MATH111C Abstract Algebra [111c]

Mar 30, 2026 - Present

Lecture 1. Field extensions [l1]

Mar 30, 2026

Definition 1.1. field extension [1201]

Let $F$ be a field. A field containing $F$ is called a field extensions of $F$ or an extension of $F$.

Notation. some notations [1201A]

We use

$$ K / F \qquad \text{or} \qquad \begin{aligned} &K \\ &| \\ &F \end{aligned} $$

to mean that $K$ is a field extension of $F$.

When considering field extensions of $F$, the field $F$ is sometimes called the base field.

Example 1.2. examples from characteristics [1202]

Every field of characteristic $0$ is a field extension of $\mathbb{Q}$.
Every field of characteristic $p$ is a field extension of $\mathbb{F}_{p}$.

Definition 1.3. degree of field extensions [1203]

1.3.1. degree [1203A]

If $K$ is a field extension of $F$, then $K$ can be viewed as an $F$-vector space.

Exegesis. vector space [1203AA]

TODO

The degree of $K / F$ is $$ [K : F] := \dim_{F}K .$$

1.3.2. finiteness [1203B]

A field extension $K / F$ is called finite if $[K : F] < \infty$.

Example 1.4. complex and real numbers, rationals and polynomials [1204]

$\mathbb{C}$ is a field extension of $\mathbb{R}$ and $[\mathbb{C} : \mathbb{R}] = 2$.
$\mathbb{R}$ is a field extension of $\mathbb{Q}$ and $[\mathbb{R} : \mathbb{Q}] = \infty$.
Let $F$ be a field and $p(x)$ be an irreducible polynomial of degree $\geq 1$ in $F[x]$. Then $K = F[x] / (p(x))$ is a field extension of $F$ and $[K : F] = \deg p(x)$.
$F = \mathbb{Q}, p(x) = x^{2} - 2, K = \mathbb{Q}[x] / (x^{2} - 2)$. Then $K \cong \mathbb{Q}[\sqrt{ 2 }] = \{ a + b\sqrt{ 2 } \mid a, b \in \mathbb{Q} \}$.
$[\mathbb{F}_{p}(t), \mathbb{F}_{p}] = \infty$. $1, t, t^{2}, \dots, t^{n}, \dots$ are linearly independent over $\mathbb{F}_{p}$, so $\dim_{\mathbb{F}_{p}}\mathbb{F}_{p}(t) = \infty$.

Remark 1.5. [1205]

Theorem 1.6. [1206]

Let $F$ be a field and $p(x)$ be an irreducible polynomial in $F[x]$. Then

$K = F[x] / (p(x))$ is a field extension of $F$,
$\theta = \overline{x} \in K$ is a root of $p(x)$ in $K$,
$[K : F] = \deg p(x)$. Let $n = \deg p(x)$. Then $$1, \theta, \theta^{2}, \dots, \theta^{n-1}$$ is a basis of $K$ as an $F$-vector space. (Hence, $K = \{ a_{0} + a_{1} \theta + \cdots + a_{n-1}\theta^{n-1} \mid a_{0}, \dots, a_{n-1} \in F\}$.)

Proof. [1206A]

TODO

Theorem 1.7. [1207]

Let $F \subseteq K \subseteq L$ be fields. Then $[L:F]$ is finite if and only if $[L:K]$ and $[K:F]$ are both finite, and $$ [L:F] = [L : K][K : F]. $$

Proof. sketch [1207A]

$\implies$: Given basis of $L$ over $K$ and $K$ over $F$, denote by $\alpha_{i}, 1 \leq i \leq n$ and $\beta_{j}, 1 \leq j \leq m$ respectively, then we can prove $\alpha_{i}\beta_{j} \in L, \forall i, j$ are linearly independent over $F$. It follows that $[L : F] \geq mn$. So if $[L : F] < \infty$, it must follow that $m < \infty$ and $n < \infty$.

$\impliedby$: Assume now $m, n < \infty$. We can prove that $\alpha_{i}\beta_{j} \forall i, j$ actually spans $L$ over $F$ by showing that every element of $L$ can be written in the form $\sum F \alpha_{i} \beta_{j}$. Then $\alpha_{i} \beta_{j}$ is a basis of $L$ over $F$, hence $[L : F] = mn < \infty$.

When $[L : F] = \infty$, at most one of $[L:K]$ and $[K:F]$ can be finite. For example, $[\mathbb{C} : \mathbb{Q}] = \infty$ and $[\mathbb{R} : \mathbb{Q}] = \infty$ while $[\mathbb{C} : \mathbb{R}] = 2 < \infty$.

Lecture 2. The subfield generated by a subset and simple extensions [l2]

Apr 1, 2026

Definition 2.1. 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 2.2. 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 2.2.1. 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 2.3. 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 2.3.1. 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 2.4. finitely generated, simple extensions and primitive elements [1304]

A field extension $K / F$ is called finitely generated if there exist $\alpha_{1}, \dots, \alpha_{n} \in K$ such that $K = F(\alpha_{1}, \dots, \alpha_{n})$.
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 2.5. $F(\theta), \mathbb{F}_p(t) / \mathbb{F}_p, F(x)$ [1305]

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 $K / F$.
$\mathbb{F}_{p}(t) / \mathbb{F}_{p}$ is simple and $t$ is a primitive element for $\mathbb{F}_{p}(t) / \mathbb{F}_{p}$. In this case, $\mathbb{F}_{p}(t) \neq \mathbb{F}_{p}[t]$.
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 $F(x) / F$. $F[x] \neq F(x)$.

Lecture 3. Algebraic extensions [l3]

Apr 6, 2026

Definition 3.1. algebraic over $F$ [1401]

Let $K / F$ be a field extension.

An element $\alpha \in K$ is said to be algebraic over $F$ if $\alpha$ is a root of some nonzero polynomial in $F[x]$.
An element $\alpha \in K$ is called transcendental over $F$ if it is not algebraic.
The extension $K / F$ is called algebraic if every element of $K$ is algebraic over $F$.

Example 3.2. $\mathbb{C}, \mathbb{Q}$ and $F(t)$ [1402]

$\sqrt{ 2 }, \sqrt[3]{ 2 }, i \in \mathbb{C}$ are algebraic over $\mathbb{Q}$.
$e, \pi \in \mathbb{C}$ are transcendental over $\mathbb{Q}$. (nontrivial to prove)
Let $F$ be a field. $t \in F(t)$ is transcendental over $F$.

Proposition 3.3. properties of $\alpha$ being algebraic over $F$ [1403]

Let $K / F$ be a field extension and $\alpha \in K$. The following are equivalent:

$\alpha \in K$ is algebraic over $F$,
the ring homomorphism $\varphi : F[x] \to K, \varphi \left( a(x) \right) = a(\alpha)$ is not injective,
$[F(\alpha) : F] < \infty$,
$[F[\alpha] : F] < \infty$,
$F(\alpha) = F[\alpha]$.

Proof. sketch [1403A]

(1) $\iff$ (2) is trivial.

(5) $\implies$ (1): Since $F(\alpha) = F[\alpha]$, there exists $c_{n}\alpha^{n} + \cdots + c_{1}\alpha + c_{0} \in F[\alpha]$ with at least one nonzero coefficient such that it equals $\alpha ^{-1} \in F(\alpha)$. Hence $$ \begin{align*} &\alpha ^{-1} = c_{n}\alpha^{n} + \cdots + c_{1}\alpha + c_{0} \\ \iff &1 = c_{n}\alpha^{n+1} + \cdots + c_{1}\alpha^{2} + c_{0}\alpha \\ \iff &0 = c_{n}\alpha^{n+1} + \cdots + c_{1}\alpha^{2} + c_{0}\alpha - 1 \in F[\alpha] \end{align*} $$ Therefore, $\alpha$ is algebraic over $F$.

(2) $\implies$ (3)(4)(5): Since $\varphi$ is not injective, $\ker \varphi$ is non-trivial. By the First Isomorphism Theorem, $F[x] / \ker\varphi \cong \mathrm{Im} \varphi$. Here, $\mathrm{Im}\varphi$ is a subring of $K$ containing $1$. Now $K$ is a field $\implies$ $\mathrm{Im} \varphi$ is an ID $\implies$ $F[x] / \ker \varphi$ is an ID $\implies$ $\ker\varphi$ is a PID, i.e., $\ker \varphi = \left( p(x) \right)$ for some $p(x) \in F[x]$ $\implies$ $p(x)$ is prime in $F[x]$ $\implies$ $p(x)$ is irreducible in $F[x]$. By Theorem 1206, $F[x] / \ker \varphi$ is a field and $$ F[x] / \ker\varphi = F + F\overline{x} + \cdots + F\overline{x}^{n-1}. $$ With $x \mapsto \alpha$ and $p(\alpha) = 0$ for $p(x) \in \ker\varphi$, we obtain $$ \mathrm{Im}\varphi = F + F\alpha + \cdots + F\alpha^{n-1}. $$ So $\mathrm{Im}\varphi$ is field extension of $F$ and $[\mathrm{Im}\varphi : F] < \infty$. By definition of $\varphi$, we have $F[\alpha] \subseteq \mathrm{Im}\varphi$. The above implies $\mathrm{Im}\varphi \subseteq F[\alpha]$. So $\mathrm{Im}\varphi = F[\alpha]$. Now $\mathrm{Im}\varphi$ is a field $\implies$ $F[\alpha] = F(\alpha)$. Therefore, $[F(\alpha) : F] < \infty$ and $[F[\alpha] : F] < \infty$.

(4) $\implies$ (1): $[F[\alpha] : F] < \infty$ implies that $F[\alpha] = F + F\alpha + \dots + F \alpha^{n-1}$ for some $n \in \mathbb{Z}_{\ge 0}$. Hence $1 \in F[\alpha]$ can be written as $1 = c_{0} + c_{1}\alpha + \cdots + c_{n-1}\alpha^{n-1}$, i.e., $c_{0} - 1 + c_{1}\alpha + \dots + c_{n-1}\alpha^{n-1} = 0$. So $\alpha$ is algebraic over $F$.

(3) $\implies$ (4): $F(\alpha) \subseteq F[\alpha] \subseteq F$. Then use Theorem 1207.