02 - Fields

Field

Fields are rings with multiplicative inverses for every non-zero element

Characteristic of Field

Characteristic of Field is always Zero or Prime

Field of Fractions

Let $R$ be a ring then

$F(R)$ is the field of fractions of $R$

where ring $R$ embeds into $F(R)$ through map

$$r\mapsto \frac{r}{1}$$

with elements of $F(R)$ in form

$$\frac{a}{b}\text{ where }a,b\in R$$

and for $a,b,c,d\in R$

$$\frac{a}{b}=\frac{c}{d}⟺ad=bc$$

Unique Injective Homomorphism

Let $k$ be a field
Let $\theta :R\to K$ be an embedding (injective homomorphism)

Then there exists unique injective homomorphism

$$\tilde{\theta }:F(R)\to K$$

that extends $\theta$ (${\tilde{\theta }}_R=\theta$ where $R$ is a subring of $F(R)$)

Field Extension

Let $E,F$ be fields

$E$ is a field extension of $F$ if

$$F\subseteq E$$

written as $E/F$

Note the inclusion of $F$ into $E$ gives $E$ the structure of an $F$-vector space

Degree of Field Extension

Let $E$ be a field extension of $F$ so $E/F$

If $E$ is finite-dimensional as a $F-$vector space then

$$[E:F]={\dim }_F(E)$$

is the degree of field extension $E/F$

Dimension Formula under Field Extension lemma

Let $E/F$ be a field extension
Let $d=[E:F]<\infty$

If $V$ is an $E$-vector space then

1. $V$ can be viewed as a $F$-vector space

2. $V$ is finite dimensional as a $F$-vector space if and only if $V$ is finite dimensional as an $E$-vector space

3. ${\dim }_F(V)=[E:F]{\dim }_E(V)$

Proof

If $V$ is an $E$-vector space then restrict scalar multiplication map to subfield $F$ then
$V$ is an $F$-vector space

Moreover if $V$ is finite dimensional as a $F$-vector space then it is also as a $E$-vector space

Conversely suppose $V$ is finite-dimensional $E$-vector space
Let $\{x_1,x_2,\cdots ,x_d\}$ be a $F$-basis of $E$
Let $\{e_1,\cdots ,e_n\}$ be a $E$-basis of $V$

Check that $\{x_i{e}_j:1\le i\le d,1\le j\le n\}$ is an $F$-basis of $V$
If $v\in V$ then as $E$-basis of $V$ then

$$v=\sum _{i=1}^n{\lambda }_i{e}_i\text{ for }{\lambda }_i\in E$$

As $\{x_1,\cdots ,x_d\}$ is an $F$-basis of $E$ then

$$\exists {\mu }_j^i\text{ such that }{\lambda }_i=\sum _{j=1}^d{\mu }_j^i{x}_j$$

Hence

$$v=\sum _{i=1}^n{\lambda }_i{e}_i=\sum _{i=1}^n\left(\sum _{j=1}^d{\mu }_j^i{x}_j\right)e_i=\sum _{1\le i\le n,1\le j\le d}{\mu }_j^i(x_j{e}_i)$$

Thus set $\{x_i{e}_j:1\le i\le d,1\le j\le n\}$ spans $V$ as a $F$-vector space
Set is linearly independent as if $v=0$ then ${\lambda }_i=0$ hence ${\mu }_j^i=0$

Hence dimension formula follows

Tower Law lemma

Let $F\subset E\subset K$ be fields

$$[K:F]\text{ is finite }⟺[E:F],[K:E]\text{ are both finite }$$

If all are finite then

$$[K:F]=[E:F][K:E]$$

Proof

Apply Dimension formula under field extension to $E$-vector space $K$

Algebraic

Let $\alpha \in \mathbb{C}$ then

$\alpha$ is algebraic over $\mathbb{Q}$ if

Exists field $E$ which is a finite extension of $\mathbb{Q}$ containing $\alpha$

Transcendental

Let $\alpha \in \mathbb{C}$ then

$\alpha$ is transcendental if $\alpha$ is not algebraic

Sub-Field Generated by a Subset

Let $T\subseteq \mathbb{C}$

Let the field generated by $T$ denoted as $\mathbb{Q}(T)$
where $\mathbb{Q}(T)$ is the smallest subfield that contains $T$

If $T=\{\alpha \}$ is single element then $\mathbb{Q}(T)\equiv \mathbb{Q}(\alpha )$ and field extension is called simple

Note that any subfield of $\mathbb{C}$ contains $\mathbb{Q}$, since it contains $\mathbb{Z}$ (as $1\in T$) and hence $\mathbb{Q}$

Relation between Algebraic and Simple Field Extension

Let $\alpha \in \mathbb{C}$ then

$$\alpha \text{ is algebraic }⟺\mathbb{Q}(\alpha )\text{ is a finite extension of }\mathbb{Q}$$

Generally

Let $F$ be any subfield of $\mathbb{C}$
Let $F(\alpha )=\mathbb{Q}(F\cup \{a\})$ then

\alpha \text{ is algebraic over } F \quad \iff \quad F(\alpha) / F \text{ is a finite extension of } F $$

Existence of Monic Irreducible Polynomial for Isomorphism of Field Extension lemma

Let $E/F$ be a finite extension of fields (both say subfields of $\mathbb{C}$)
Let $\alpha \in E$ then

Exists unique monic irreducible polynomial $f\in F[t]$ such that
Evaluation homomorphism $\varphi :F[t]\to E$ sends $t$ to $\alpha$ induces isomorphism

$$F(\alpha )\cong F[t]/⟨f⟩$$

Proof

Field $K=F(\alpha )$ is a finite extension of $F$
since subfield of $E$ so sub-$F$-vector space of finite dimensional $F$-vector space $E$

Let $d=[K:F]={\dim }_F(K)$
As $\{{\alpha }^i:0\le i\le d\}$ has $d+1$ elements then must be linearly dependent
Hence exists ${\lambda }_i\in F$ for $0\le i\le d$ (not all zero) such that

$$\sum _{i=0}^d{\lambda }_i{\alpha }^i=0$$

If

$$g=\sum _{i=0}^d{\lambda }_i{t}^i\in F[t]\setminus \{0\}$$

then

$$g(a)=0$$

Hence kernel $I$ of homomorphism $\varphi :F[t]\to E$ defined by

$$\varphi \left(\sum _{j=0}^m{c}_j{t}^j\right)=\sum _{j=0}^m{c}_j{\alpha }_j$$

has non-zero kernel

As non-zero ideal in $F[t]$ is generated by unique monic polynomial then

$$I=⟨f⟩$$

where $f$ is monic and uniquely determined by $\varphi$ (and hence $\alpha$)

By First Isomorphism Theorem then
Image $S$ of $\varphi$ is isomorphic to $F[t]/I$

As $S$ is a subring of field then it is an integral domain so

$$⟨f⟩\text{ is a prime ideal}$$

Hence by definition of prime ideals in $F[t]$ then it is maximal
Hence $S$ is a field

Thus any subfield of $\mathbb{C}$ containing $F$ and $\alpha$ must contain $S$ hence

$$S=F(\alpha )$$

Minimal Polynomial

Let $a\in \mathbb{C}$ then

Minimal Polynomial $f$ is the polynomial associated by $\alpha$ from
Existence of monic irreducible polynomial for isomorphism of field extension
where $f$ satisfies

$$\mathbb{Q}(a)\cong \mathbb{Q}[t]/⟨f⟩$$

with

$$[\mathbb{Q}(\alpha ):\mathbb{Q}]=\mathrm{deg}(f)$$