3.1 Field Extension

Definition

A field $F$ is said to be an extension of a field $K$, if $K\subseteq F$ is a subfield, denoted by $F/K$.

Remark.

• $1_K=1_F$. Because $1_K^{-1}=1_K$ inside $K$ and $1_K^{-1}=1_K$ insider $F$, we have that $1_K\cdot 1_K=1_K$ with multiplication in $K$ and $1_K\cdot 1_K=1_F$ with multiplication in $F$.
• $F$ is a $K$ vector space. Denote $[F:K]={\dim }_{K}F$, called degree of the extension.
• We say $F/K$ is a finite extension if $[F:K]<\infty$.

Theorem

Suppose $K\stackrel{m}{\subseteq }E\stackrel{n}{\subseteq }F$ with finite $m,n$. Then $[F:K]=[F:E][E:K]$.

\begin{proof} Suppose $[F:E]=n$, then $F={\oplus }_{i=1}^{n}E{x}_i$ with $x_i\in F$. Suppose $[E:K]=n$, then $E={\oplus }_{j=1}^{m}K{y}_j$ with $y_j\in E$. Note that $\{x_i{y}_j:1⩽i⩽n,1⩽j⩽m\}$ are linear independent and it is a basis of $F$ over $K$, then we finish the proof. \end{proof}

Remark. For $F/E/K$, i.e., $K\subseteq E\subseteq F$, we say $E$ is an intermediate extension.

Definition

The followings hold.

• Let $F$ be a field and let $X\subseteq F$ be a subset. Define the subfield (rep. subring) generated by $X$ as the intersection of all subfields (rep. subring) containing $X$.
• Let $F/K$ be a field extension and let $X\subseteq F$ be a subset. Denote $K[x]$ be the subring of $F$ generated by $K\cup X$, and denote $K(x)$ be the subfield generated by $K\cup X$.
• If $X=\{u_1,\cdots ,u_n\}$ is finite, then we say $K(u_1,\cdots ,u_n)$ is a finitely generated extension of $K$. If $X=\{n\}$, then we call $K(n)/K$ is a simple extension.

Theorem

Let $F/K$ be a field extension. Assume that $u\in F$ and $u_1,\cdots ,u_n\in F$. Then:

• $K[u]=\{f(u):f\in K[x]\text{mbox}isapolynormial\}$;
• $K[u_1,\cdots ,u_n]=\{f(u_1,\cdots ,u_n):f\in K[x_1,\cdots ,x_n]\}$;
• $K(u)=\{f(u)/g(u):f,g\in K[u],g(u)\ne 0\}$;
• $K(u_1,\cdots ,u_n)=\{f(u_1,\cdots ,u_n)/g(u_1,\cdots ,u_n):f,g\in K[x_1,\cdots ,x_n],g(u_1,\cdots ,u_n)\ne 0\}$.

\begin{proof} Easy. \end{proof}

Definition

Suppose $L,M\subseteq F$, then denote the composite $LM$ as the subfield generated by $L\cup M$.

Remark. If $K\subseteq L\subseteq F$ and $K\subseteq M\subseteq F$, then $[LM:K]⩽[M:K][L:K]$.

Definition

Let $F/K$ be a field extension.

• We say $u\in F$ is algebraic over $K$ if there exists non-zero $f\in K[x]$ such that $f(u)=0$.
• We say $u\in F$ is transcendental over $K$ if for all non-zero $f(x)\in K[x]$, $f(u)\ne 0$.
• We say $F/K$ is an algebraic extension if all elements in $F$ is algebra over $K$.
• We say $F/K$ is an transcendental extension if there exists elements in $F$ is transcendental over $K$.

Example. Consider $\mathbb{Q}\subseteq \mathbb{R}\subseteq \mathbb{C}$, then:

• $\mathbb{C}/\mathbb{R}$ is algebraic;
• $\mathbb{R}/\mathbb{Q}$ is transcendental.

Example. Let $K$ be a field. Then $K(x)=\mathrm{Frac}(K[x])$ is transcendental over $K$.

Theorem

Let $F/K$ be a field extension, then there exists an isomorphism of fields $K(u)\stackrel{\phi }{\simeq }K(x)$ such that $\phi {|}_K=\mathrm{id}$.

\begin{proof} Define

$$\phi :K(x)\to K(u),\frac{f(x)}{g(x)}\mapsto \frac{f(u)}{g(u)}.$$

It is obvious that $\phi$ is a field homomorphism, which is injective. Thus $K(u)\simeq K(x)$ by ^248a7b. \end{proof}

Theorem

Let $F/K$ be a field extension and let $u\in F$ be algebraic over $K$. Then:

• $K(u)=K[u]$;
• $K(u)\simeq K[x]/(f)$ where $f\in K[x]$ is an irreducible monic polynomial with $\mathrm{deg}f⩾1$ uniquely determined by $f(u)=0$, and $g(u)=0$ iff $f\mid g$;
• $[K[u]:K]=n=\mathrm{deg}f$;
• $\{1,u,\cdots ,u^{n-1}\}$ is a basis of $K(u)$ as a $K$-vector space;
• any $b\in K(u)$ can be uniquely written as $b={\sum }_{i=0}^{n-1}{k}_i{u}^i$ with $k_i\in K$.

\begin{proof} Most details are skipped. Define $\phi :K[x]\to K[u],f(x)\mapsto f(u)$, then $\ker \phi =(f)$ by $K[x]$ PID. It follows that $K[u]\simeq K[x]/(f)$. Since $K[u]$ is an integral domain, $(f)$ is a prime ideal and so $f$ is irreducible. Note that in PID each non-zero prime ideal is a maximal ideal, then $K[u]$ is a field and so $K(u)=K[u]$. \end{proof}

Theorem

Suppose $F/K$ with $[F:K]<\infty$, then $F$ is finitely generated extension and it is an algebraic extension.

\begin{proof} For any $u\in F\setminus K$, there exists $n\in {\mathbb{N}}_{+}$ such that $\{1,u,\cdots ,u^{n-1}\}$ are linear independent, then ${\sum }_{i=0}^n{\alpha }_i{u}^i=0$ where ${\alpha }_i$ are not all zero. Hence $u$ is algebraic. Also, since $F={\oplus }_{i=1}^{n}K{\alpha }_i$ for some ${\alpha }_i\in F$, then $K({\alpha }_1,\cdots ,{\alpha }_n)=F$. \end{proof}

Theorem

Let $F/K$ be a field extension. Suppose there exists finite subset $X\subseteq F$ such that $F=K(X)$, and each element of $X$ is algebraic over $K$, then $[F:K]<\infty$.

\begin{proof} By ^ab12ed and ^71f159. \end{proof}

Theorem

Let $F/K$ be a field extension. Define

$$E=\{x\in F:x\text{mbox}isalgebricoverK\}.$$

Then $E$ is a subfield.

Remark. It is a corollary of ^48e5f4. 高辉说他在厕所证明了这个。

Example. Consider $\mathbb{C}/\mathbb{Q}$. If $x\in \mathbb{C}$ is algebraic over $\mathbb{Q}$, it is called an algebraic number. Then $\overline{\mathbb{Q}}=\{\text{mbox}allalgebraicnumbers\}\subseteq \mathbb{C}$ is a subfield, called algebraic closure of $\mathbb{Q}$.