2.8 Field extensions

def: Subfield

$$\begin{align*} &\sphericalangle \\ &E \in \mathcal F \\ &L \subseteq_{(G, +)} E \\ &L^* \subseteq_{(G, \cdot)} E^* \\ \hline \\ &L \subseteq_F E \end{align*}$$

def: Field homomorphism

$$\begin{align*} &\sphericalangle \\ &F, E \in \mathcal F \\ &\phi: F \rightsquigarrow_R E \\ &\phi(1)=1 \\ \hline \\ &\phi: F \rightsquigarrow_F E \\ \end{align*}$$

def: Field extension

$$\begin{align*} &\sphericalangle \\ &E \in \mathcal F \\ &F \subseteq_F E \\ \hline \\ & E/F \in \mathcal F,E - \text{extension of }F, F - \text{base field} \end{align*}$$

note

Do not confuse the notation $E/F$ with factor group and factor rings. These are different concepts. Overall the meaning of the notation $X/Y$ depends on $X$ and $Y$ being groups, rings or fields.

def: Extension degree

$$\begin{align*} &\sphericalangle \\ &E / F \in \mathcal F \implies E \in \mathcal M_F \\ \hline \\ &[E:F] := \dim_F E \end{align*}$$

Proposition 2.8.1: Evaluation homomorphism

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ &\alpha \in E \\ &\varepsilon_{\alpha}: F[x] \to E, f \mapsto f(\alpha) \\ \hline \\ &\varepsilon_{\alpha}: F[x] \rightsquigarrow_F E \hspace{1cm} \end{align*}$$

Proof

$$\forall p, q \in F[x]: \\ \varepsilon_\alpha(p+q) = (p+q)(\alpha)=p(\alpha)+q(\alpha)=\varepsilon_\alpha(p)+\varepsilon_\alpha(q) \\ \varepsilon_\alpha(pq)=(pq)(\alpha)=p(\alpha)q(\alpha)=\varepsilon_\alpha(p)\varepsilon_\alpha(q)$$

$\square$

Extensions taxonomy

Finitely generated extensions

def: Finitely generated extension

$$\begin{align*} &\sphericalangle \\ & E/F \in \mathcal F \\ & A \subseteq E, |A| \lt \infty \\ & F(A): F \subseteq_F F(A) \subseteq_F E, A \subseteq F(A) \\ &\forall M: F \subseteq_F M \subseteq_F E, A \subseteq M \implies F(A) \subseteq M\\ \hline \\ &F(A)- \text{finitely generated extension} \end{align*}$$

note

If $A=\{\alpha_1, \ldots, \alpha_n\}$ then we just write $F(\alpha_1, \ldots, \alpha_n)$

def: Simple extension

$$\begin{align*} &\sphericalangle \\ & E/F \in \mathcal F \\ & \alpha \in E\\ \hline \\ &F(\alpha)- \text{simple extension} \end{align*}$$

Algebraic and transcendental elements

def: Algebraic closure, algebraic and transcendental elements

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ &\overline{F}_E:=\{\alpha \in E: \exists p(x) \in F[x] \setminus F: p(\alpha)=0\} \\ \hline \\ &\overline{F}_E - \text{algebraic closure of } F \text{ in } E \\ & \alpha \in \overline{F}_E - \text{algebraic element over } F \text{ in the field } E \\ &\beta \in E \setminus \overline{F}_E - \text{transcendental element over } F \text{ in the field } E \\ \end{align*}$$

def: Minimal polynomial

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ & \alpha \in \overline{F}_E \\ &p \in F[x]^-, LC(p)=1, p(\alpha) = 0 \\ &\forall q \in F[x], q(\alpha) = 0: p \mid q \\ \hline \\ &\mathfrak{pol}_F(\alpha):= p - \text{minimal polynomial for } \alpha \text{ over }F\\ \end{align*}$$

Proposition 2.8.2: Minimal polynomial is unique

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ & \alpha \in \overline{F}_E \\ \hline \\ &\begin{align*} &\exists! \mathfrak{pol}_F(\alpha) \hspace{0.5cm} \tag{a}\\ &p, q \in F[x]^-, p(\alpha)=q(\alpha)=0 \implies p(x)=uq(x), u \in F^* \hspace{0.5cm} \tag{b}\\ \end{align*} & \end{align*}$$

Proof

a.

$\varepsilon_\alpha$ - evaluation homomorphism. Note that all polynomials which has $\alpha$ as a root are exactly $\ker \phi$.

$$(2.5.13) \implies F[x] \in \mathcal R^{\mathcal {PID}} \overset{(2.3.8)}\implies \\ \exists p(x) \in F[x] \setminus F: \ker \varepsilon_\alpha = \lang p(x) \rang_I$$

Thus $q \in F[x], q(\alpha) = 0 \implies p \mid q$. Any polynomial of the same degree must be of the form $ap(x)$, $a \in F$. Thus adding the requirement $LC(p)=1$, we finish the proof.

b.

$$\mathfrak{pol}_F(\alpha) \mid p, \mathfrak{pol}_F(\alpha) \mid q \implies p = u \cdot \mathfrak{pol}_F(\alpha), q = v \cdot \mathfrak{pol}_F(\alpha) \\ p, q, \mathfrak{pol}_F(\alpha) \in F[x]^- \implies u,v \in F^* \implies p=uv^{-1}q=mq, m \in F^*$$

$\square$

def: Algebraic element degree

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ & \alpha \in \overline{F}_E \\ \hline \\ &\deg \alpha := \deg \mathfrak{pol}_F(\alpha) \end{align*}$$

Algebraic extensions

def: Algebraic extension

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ &E = \overline{F}_E \\ \hline \\ &E/_AF, E - \text{algebraic extension of } F \end{align*}$$

Proposition 2.8.3: Simple algebraic extension is a polynomials factor field

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ & \alpha \in \overline{F}_E \\ \hline \\ &\begin{align*} & \exists \eta: F[x] / \lang \mathfrak{pol}_F(\alpha) \rang_I \cong F(\alpha) \hspace{1cm} \tag{a}\\ & \forall a \in F: \eta(a+\lang \mathfrak{pol}_F(\alpha) \rang_I)=a \hspace{1cm} \tag{b}\\ & \eta(x + \lang \mathfrak{pol}_F(\alpha) \rang_I) = \alpha \tag{c} \end{align*} \end{align*}$$

Proof

a.

$\varepsilon_\alpha$ - evaluation homomorphism from $(2.8.1)$. It's easy to see that $\ker \varepsilon_{\alpha}=\lang \mathfrak{pol}_F(\alpha) \rang_I, \varepsilon_\alpha(F[x])=F[\alpha]$.

$$(2.3.9) \implies F[\alpha] \cong F[x] / \lang \mathfrak{pol}_F(\alpha)\rang_I \\ (2.5.13) \implies F[x] \in \mathcal R^{\mathcal{PID}}\\ F[x] \in \mathcal R^{\mathcal{PID}}, \mathfrak{pol}_F(\alpha) \in F[x]^- \overset{(2.4.14)}\implies \\\lang \mathfrak{pol}_F(\alpha)\rang_I \in \mathfrak M_I(F[x]) \overset{(2.4.21)}\implies F[x] / \lang \mathfrak{pol}_F(\alpha)\rang_I \in \mathcal F$$

So $F[\alpha] \cong F[x] / \lang \mathfrak{pol}_F(\alpha)\rang_I \implies F[\alpha] \in \mathcal F$.

$$F, \alpha \in F[\alpha], F[\alpha] \in \mathcal F \implies F(\alpha) \subseteq F[\alpha]$$

But it's also obvious that every element of $F[\alpha]$ will be necessary in $F(\alpha)$. Thus $F(\alpha)=F[\alpha] \cong F[x] / \lang \mathfrak{pol}_F(\alpha)\rang_I$

b., c.

By construction of the isomorphism in the first isomorphism theorem we must have

$$\forall a \in F: \eta(a+\lang \mathfrak{pol}_F(\alpha) \rang_I)=\varepsilon_\alpha(a)=a \\ \eta(x+\lang \mathfrak{pol}_F(\alpha) \rang_I)=\varepsilon_\alpha(x)=\alpha$$

$\square$

note

From the proposition it's also clear that no matter what root of the polynomial we take to build an extensions, all such extensions are isomorphic

Example: $\alpha=\sqrt 2$

Consider $\mathbb Q(\sqrt 2)$. The minimal polynomial over $\mathbb Q$ will be $p(x):=x^2-2$. Why it's irreducible? The only way for it to be reducible is to be $p(x)=(x-a)(x-b)$ which would imply $a, b \in \mathbb Q: ab=\sqrt 2$ which is impossible in rationals. So we'll have $\mathbb Q(\sqrt 2)\cong \mathbb Q[x]/\lang x^2-2 \rang_I$. Note that $\mathbb Q(\sqrt 2)=\{a+b\sqrt 2, a, b \in \mathbb Q\}$, while $\mathbb Q[x]/\lang x^2-2 \rang_I = \{a+bx, a, b, \in \mathbb Q\}$. It's pretty clear that these two fields are isomorphic and each $a\in \mathbb Q \mapsto a$ and $\sqrt 2 \mapsto x$. Exactly as $(2.8.3)$ states.

note

It may come as a surprise that a field $F(\alpha)$ can be isomorphic to a quotient field of polynomials. One might wonder, should the field not also include fractions like $\frac{1}{\alpha}$? To clarify, let's look at a straightforward example: consider the field $\mathbb Q(\sqrt{2})$, which is an extension of $\mathbb Q$ where $\alpha = \sqrt{2}$. Within this field, $\frac{1}{\sqrt{2}}$ simplifies to $\frac{\sqrt{2}}{2}$, showing that it can be represented as a polynomial of degree one in $\alpha$.

Finite extensions

def: Finite extension

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ &[E: F] < \infty \\ \hline \\ &E - \text{finite extension of } F \\ \end{align*}$$

Proposition 2.8.4: Simple algebraic extension is finite

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ & \alpha \in \overline{F}_E \\ \hline \\ &\begin{align*} &[F(\alpha):F] = \deg \alpha \tag{a}\\ &n:=\deg \alpha, \{1, \alpha, \ldots, \alpha^{n-1} \} \in \mathfrak B_F(F(\alpha)) \hspace{1cm} \tag{b}\\ \end{align*} \end{align*}$$

Proof

From $(2.8.3)$ we know that $F(\alpha) \cong F[x]/\lang \mathfrak{pol}_F(\alpha) \rang_I$. Assume $\mathfrak{pol}_F(\alpha) = x^n+a_{n-1}x^{n-1} + \ldots+a_0$. Consider $bx^n + \lang \mathfrak{pol}_F(\alpha) \rang_I=bx^n - b(x^n+a_{n-1}x^{n-1} + \ldots+a_0) + \lang \mathfrak{pol}_F(\alpha) \rang_I=ba_{n-1}x^{n-1} + \ldots+ba_0 + \lang \mathfrak{pol}_F(\alpha) \rang_I$. Using the same techinque we can reduce any plolynomial of degree $\ge n$ to polynomial of degree $\leq n-1$. So $\lang 1 + \lang \mathfrak{pol}_F(\alpha) \rang_I, x+\lang \mathfrak{pol}_F(\alpha) \rang_I, \ldots, x^{n-1} + \lang \mathfrak{pol}_F(\alpha) \rang_I \rang_M=F[x]/\lang \mathfrak{pol}_F(\alpha) \rang_I$. Or using isomopshism properties from $(2.8.3)$:

$$\lang 1, \alpha, \ldots, \alpha^{n-1} \rang_I = F(\alpha)$$

Because $n=\deg \alpha = \deg \mathfrak {pol}_F(\alpha)$, $\alpha$ cannot be a root of a polynomial of degree less than $n$, so $a_{n-1}\alpha^{n-1} + \ldots+a_0 \implies a_{n-1} = \ldots = a_0$.

Thus we proved $\{1, \alpha, \ldots, \alpha^{n-1} \} \in \mathfrak B_F(F(\alpha))$

$\square$

Proposition 2.8.5: Finite extension is algebraic

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ &[E:F] \lt \infty \\ \hline \\ &E /_A F \end{align*}$$

Proof

$$n:=[E:F] \\ \forall \alpha \in E \implies \{1, \alpha, \ldots , \alpha^n\} - \text{linearly dependent in } \mathcal M_F \implies\\ \exist a_1, \ldots, a_n \in F, \exists a_i \ne 0: a_0+a_1 \alpha + \ldots + a_n \alpha^n=0 \implies \\ \exists p \in F[x] \setminus F: p(\alpha) = 0$$

$\square$

Proposition 2.8.6: Finite extension chain: dimension and basis

$$\begin{align*} &\sphericalangle \\ &E/F/K \in \mathcal F \\ &[E:F] \lt \infty, [F:K] \lt \infty \\ \hline \\ &\begin{align*} &\mathfrak B_K(E)=\mathfrak B_F(E)\mathfrak B_K(F) \tag{a}\\ &[E:K]=[E:F][F:K] \hspace{1cm} \tag{b}\\ \end{align*} \end{align*}$$

Proof

a.

$$\mathfrak B_K(F)=\{x_1, \ldots, x_n\}, \mathfrak B_F(E)=\{y_1, \ldots, y_m\} \\$$ $$u \in E \implies u = \sum_i b_iy_i, b_i \in F \\ b_i \in F \implies b_i = \sum_jc_{ij}x_j \implies \\ u = \sum_i(\sum_jc_{ij}x_j)y_i =\sum_{i,j}c_{ij}(x_jy_i) \implies \\ \lang \mathfrak B_K(F)\mathfrak B_F(E) \rang_M =E$$ $$\sum_{i, j}c_{ij}x_iy_j = \sum_{j}(\sum_{i}c_{ij}x_i)y_j \implies \\ \forall j: \sum_{i}c_{ij}x_i=0 \implies \forall i,j: c_{ij}=0 \implies \\ \mathfrak B_K(E)=\mathfrak B_K(F)\mathfrak B_F(E)$$

b.

Follows from $(a)$.

$\square$

Proposition 2.8.7: Finite extension is finitely generated algebraic extension

$$\begin{align*} &\sphericalangle \\ &E \in \mathcal F \\ &F \subseteq_F E\\ \hline \\ &[E:F]<\infty \iff \exists \alpha_1, \ldots, \alpha_n \in \overline{F}_E: E = F(\alpha_1, \ldots, \alpha_n)\\ \end{align*}$$

Proof

$\implies$

$$\mathfrak B_F(E)=\{\alpha_1, \ldots, \alpha_n\} \\ \mathfrak B_F(E) \subseteq E \implies E/F(\alpha_1, \ldots, \alpha_n)\\ \lang \mathfrak B_F(E) \rang_M=E \implies E = F(\alpha_1, \ldots, \alpha_n) \\ (2.8.5) \implies \alpha_i \in \overline{F}_E$$

$\impliedby$

Since $F(\alpha_1, \ldots, \alpha_{i-1})(\alpha_i) = F(\alpha_1, \ldots, \alpha_i)$ it follows that $F(\alpha_1, \ldots, \alpha_i) /_A F(\alpha_1, \ldots, \alpha_{i-1})$ and by $(2.8.4)$ $[F(\alpha_1, \ldots, \alpha_i): F(\alpha_1, \ldots, \alpha_{i-1})]< \infty$.

$$(2.8.6.b) \implies [E : F]= \\ =[F(\alpha_1, \ldots, \alpha_n) : F(\alpha_1, \ldots, \alpha_{n-1})] \cdot \ldots \cdot [F(\alpha_1) : F] \lt \infty$$

$\square$

Splitting fields

def: Polynomials split in the field

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &P \subseteq F[x] \\ &\forall p(x) \in P: p(x)= u(x-\alpha_1)\ldots(x-\alpha_n), u, \alpha_i \in F, n \ge 0\\ \hline \\ &P \parallel F, P \text{ splits in } F \end{align*}$$

note

Similarly, for $p \in F[x]$ we'll use the notation $p \parallel F$ if $\{p\} \parallel F$.

def: Splitting field

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ &P \subseteq F[x] \setminus F \\ &R:= \{\alpha \in E: \exists p \in P: p(\alpha) = 0\}\\ &E=F(R) \\ &P \parallel E \\ \hline \\ &F(\parallel P):=E - \text{splitting field for }P \text{ over } F \end{align*}$$

note

Obviously $F(\parallel P) /_A F$

note

The notation assumes that the splitting field exists and unique. We will prove that the splitting field is indeed exists and unique up to isomorphism later in this section.

Proposition 2.8.8: Splitting field exists

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &p(x) \in F[x] \setminus F \\ &P \subseteq F[x] \setminus F \\ \hline \\ &\begin{align*} &\exists E/F, \exists \alpha \in E: p(\alpha) = 0, [E:F] \leq \deg p, E = F(\alpha) \hspace{0.5cm} \tag{a}\\ &\exists E/F: p \parallel E, [E: F] \le \deg p!, E= F(\parallel p)\tag b \\ &\exists F(\parallel P) \tag c \\ \end{align*} \end{align*}$$

Proof

a.

By $(2.5.13)$ $F[x] \in \mathcal R^{\mathcal {UFD}}$ so $p(x) = p_1(x)\ldots p_n(x), p_i \in F^{-}[x]$. Take $p_1(x)$ and define

$$E:=F[x] / \lang p_1(x) \rang_I$$

By $(2.4.14)$ $\lang p_1(x) \rang_I \in \mathfrak M_I(F[x])$. And so by $(2.4.21)$ $E \in \mathcal F$.

Consider a natural homomorphism $\phi: F \to E, a \mapsto a + \lang p_1(x) \rang_I$, it's obvious that $\ker \phi = \{0\}$, so $F \cong \phi(F) \subseteq E$, so $E / F$.

Next define

$$\alpha: = x + \lang p_1(x) \rang_I$$

We have $p_1(\alpha)=p_1(x+\lang p(x) \rang_I)=p_1(x)+\lang p(x) \rang_I = 0 + \lang p_1(x) \rang_I$. Note that $0 + \lang p_1(x) \rang_I$ is just $0$ in $E$ so we can write $p_1(\alpha)=0 \implies p(\alpha) = 0$. So $\alpha \in \overline F_E$ and by $(2.8.3)$ we have $E = F(\alpha)$.

By $(2.8.2)$ $p_1(x)=u\cdot \mathfrak{pol}_F(\alpha), u \in F^*$ so by $(2.8.4)$ $[E:F]=\deg \alpha = \deg p_1(x) \leq \deg p$.

b.

By induction on $\deg p$. If $\deg p=1$ then $E:=F$, $[E:F]=1=1!$ and $p \parallel E$.

If $\deg p = n$ then by $(a)$ we can make $E/F, [E:F] \leq n$ where $\exists \alpha \in E: p(\alpha) = 0$, so $p(x) = (x-\alpha)g(x)$. Then by induction we have field $L/E: [L:E] \le (n-1)!, g \parallel L$. So we have $p \parallel L$, $[L:F]=[L:E][E:F] \le n!$.

Finally by construction on each induction step we built an extension $F(\alpha_1, \ldots, \alpha_i)$, so $E= F(\parallel p)$.

c.

This part we'll only prove with for $|P|<\infty$, however it's also true for infinite $P$.

In $(b)$ we proved that $p(x) \in F[x] \implies \exists F(\parallel p)$. If $P = \{p_1, \ldots, p_n\}$ then it's easy to check that $F(\parallel P)=F(\parallel p_1\cdot \ldots \cdot p_n)$ so we finish the proof.

$\square$

Example: Splitting field of $(x^2-2)(x^2-3)$

$\mathbb Q(\parallel (x^2-2)(x^2-3)) = \mathbb Q(\sqrt 2, \sqrt 3)$

$$[\mathbb Q(\sqrt 2, \sqrt 3):\mathbb Q]=[\mathbb Q(\sqrt 2, \sqrt 3):\mathbb Q(\sqrt 2)][\mathbb Q(\sqrt 2):\mathbb Q]=\\ 2\cdot 2 = 4$$

Example: Splitting field of $x^3-2$

The roots of $x^3-2$ are:

$$\theta_1 = \sqrt[3]{2}, \theta_2 = \sqrt[3]{2}\cdot\frac{i\sqrt{3}-1}{2}, \theta_3 = \sqrt[3]{2}\cdot\frac{-i\sqrt{3}-1}{2}$$

$\mathbb Q(\parallel x^3-2)=\mathbb Q(\theta_1, \theta_2, \theta_3)=\mathbb Q(\sqrt[3]2, \sqrt{-3})$

$$[\mathbb Q(\sqrt[3]2, \sqrt{-3}):\mathbb Q]=6$$ $$[\mathbb Q(\theta_1, \theta_2, \theta_3):\mathbb Q]=\underbrace{[\mathbb Q(\theta_1) : \mathbb Q]}_3\underbrace{[\mathbb Q(\theta_1, \theta_2):\mathbb Q(\theta_1)]}_2\underbrace{[\mathbb Q(\theta_1, \theta_2, \theta_3):\mathbb Q(\theta_1, \theta_2)]}_1$$

Proposition 2.8.9: Isomorphism of field extensions

$$\begin{align*} &\sphericalangle \\ &E_1/_AF_1, E_2/_AF_2 \in \mathcal F \\ &\sigma: F_1 \cong F_2 \\ &\alpha \in E_1: p(x):=\mathfrak{pol}_F(\alpha)\\ &\beta \in E_2: p^{\sigma}(\beta) = 0 \\ \hline \\ &\exists \phi:F_1(\alpha) \cong F_2(\beta), \phi(\alpha) = \beta, \forall x \in F_1:\phi(x)= \sigma(x) \end{align*}$$

note

Remember that $p^{\sigma}(x)$ is action on coefficients

Proof

First, $LC(p(x))=1 \implies LC(p^{\sigma}(x))=\sigma(LC(p(x))) = \sigma(1)=1$. Additionally it can be easily verified that under isomorphism irreducible elements map to irreducible elements, so $p^{\sigma}(x) \in F_2[x]^-$ and $p^{\sigma}(x)=\mathfrak{pol}_F(\beta)$

$$(2.5.1) \implies \sigma_x: F_1[x]\cong F_2[x], p(x) \mapsto p^{\sigma}(x) \\$$

Additionally it's obvious that if rings $R_1$ and $R_2$ are isomorphic and ideal $I_1$ is mapped to ideal $I_2$ under the isomorphism then $R_1/I_1 \cong R_2/I_2$

So we have

$$\sigma(\lang p(x) \rang_I) = \lang p^{\sigma}(x) \rang_I\\ (2.4.14) \implies \lang p(x) \rang_I \in \mathfrak M_I(F_1[x]),\lang p^{\sigma}(x) \rang_I \in \mathfrak M_I(F_2[x]) \\ F_1[x]/\lang p(x) \rang_I \overset{\sigma_x}\cong F_2[x]/\lang p^{\sigma}(x) \rang_I \overset{(2.8.3)}\implies \\ F_1(\alpha) \overset{\eta}\cong F_1[x]/\lang p(x) \rang_I \overset{\sigma_x}\cong F_2[x]/\lang p^{\sigma}(x) \rang_I \overset{\tau}\cong F_2(\beta)$$

So we define $\phi:=\tau\sigma_x\eta$.

Consider some $a \in F_1$, let's see how it maps under $\phi$ (refer to $(2.8.3)$):

$$a \overset{\eta}\mapsto a+\lang p(x) \rang_I \overset{\sigma_x}\mapsto \sigma(a)+\lang p^{\sigma}(x) \rang_I \overset{\tau}\mapsto \sigma(a) \\ \alpha \overset{\eta}\mapsto x+\lang p(x) \rang_I \overset{\sigma_x}\mapsto x+\lang p^{\sigma}(x) \rang_I \overset{\tau}\mapsto \beta \\$$

So $\forall a \in F_1: \phi(x)=\sigma(x)$, and $\phi(\alpha) = \beta$

$\square$

Propostion 2.8.10: Isomorphism extension theorem for one polynomial

$$\begin{align*} &\sphericalangle \\ &F_1, F_2 \in \mathcal F \\ &\sigma: F_1\cong F_2 \\ &p(x) \in F_1[x] \\ \hline \\ &\exists \phi: F_1 (\parallel p(x)) \cong F_2 (\parallel p^{\sigma}(x)), \forall x \in F_1: \phi(x)=\sigma(x) \end{align*}$$

Proof

We'll make a proof by the induction on $n:=\deg p(x)$.

If $n=1$ then $F_1( \parallel p)=F_1, F_2( \parallel p^{\sigma}) = F_2$ and we're done.

Now assume the proposition holds for any $k<n$. If $p \parallel F_1$ then we have the same case as in $n=1$. Otherwise let's take an irreducible monic factor of degree greater than $1$ and denote it by $q(x)$. It has a corresponding factor $q^{\sigma}(x)$ in $p^{\sigma}(x)$. By $(2.8.10)$ we can take $\alpha \in E_1$ and $\beta \in E_2$ such that $\exists \eta: F_1(\alpha) \cong F_2(\beta), p(x)=(x-\alpha)p_1(x), p^{\sigma}(x)=(x-\beta)p_2(x)$. We'll use the following diagram to illustrate:

$$\begin{CD} E_1 @>\phi>> E_2 \\ @AAA @AAA \\ F_1(\alpha) @>\eta>> F_2(\beta) \\ @AAA @AAA \\ F_1 @>\sigma>> F_2 \end{CD}$$

Now if we consider $E_1:=F_1(\alpha)(\parallel p_1), E_2:=F_2(\beta)(\parallel p_2)$, by induction hypothesis $\exists \phi: E_1\cong E_2: \forall x \in F_1(\alpha): \phi(x)=\eta(x)$. Since $\forall x \in F_1: \eta(x)=\sigma(x)$, we finish the proof.

$\square$

Theorem 2.8.11: Isomorphism extension theorem

$$\begin{align*} &\sphericalangle \\ &F_1, F_2 \in \mathcal F \\ &\sigma: F_1\cong F_2 \\ &P \subseteq F_1[x] \setminus F \\ &P^{\sigma}:= \{p^{\sigma}, p \in P\}\\ \hline \\ &\exists \phi: F_1 (\parallel P) \cong F_2 (\parallel P^{\sigma}), \forall x \in F_1: \phi(x)=\sigma(x) \end{align*}$$

Proof

The proof uses the fact that if $|P|<\infty$ then $F(\parallel \{p_1, \ldots, p_n\})=F(\parallel p_1 \cdot \ldots \cdot p_n)$. It's also true for $|P|=\infty$ but we'll not prove it here.

$\square$

Corrolary 2.8.12: Splitting field is unique up to isomorphism

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &P \subseteq F[x] \setminus F \\ \hline \\ &\exists! F (\parallel P) \end{align*}$$

Proof

Follows directly from $(2.8.11)$ by asuming $F_1=F_2=F$.

$\square$

Lemma 2.8.13: Splitting fields under isomopshism

$$\begin{align*} &\sphericalangle \\ &E_1/F_1, E_2/F_2 \in \mathcal F \\ &\sigma: F_1 \cong F_2 \\ &\tau: E_1 \rightsquigarrow_F E_2, \tau|_{F_1} = \sigma \\ &P \subseteq F_1[x]: E_1 = F_1(\parallel P) \\ &P^{\sigma} := \{p^{\sigma}, p \in P\} \\ \hline \\ &\tau(E_1)=F_2(\parallel P^{\sigma}) \end{align*}$$

Proof

Consider some $p_i \in P$, it splits in $E_1$ so $p_i(x)=u\prod_j(x-\alpha_j), u \in F_1^*, \alpha_j \in E_1$. By $(2.5.1)$ we know that $p_i^{\sigma}=\sigma_x(p_i)=\tau_x(p_i)=\sigma(u)\prod_j(x-\tau(\alpha_j))$, so each $p_i^{\sigma} \parallel \tau(E_1)$ and has roots $\tau(\alpha_1), \ldots, \tau(\alpha_n)$.

Now take all roots of $P$ to be $\alpha_1, \ldots, \alpha_k$. If we consider properties of homomorphism it follows that $\tau(E_1)=\tau (F_1(\alpha_1, \ldots, \alpha_k))=\tau(F_1)(\tau(\alpha_1),\ldots, \tau(\alpha_k))=\sigma(F_1)(\tau(\alpha_1),\ldots, \tau(\alpha_k)) = F_2(\tau(\alpha_1),\ldots, \tau(\alpha_k))$.

So we have $\tau(E_1)/F_2$, $P^{\sigma} \parallel \tau(E_1), \tau(E_1)$ is generated by the roots of $P^{\sigma}$ implying $\tau(E_1)=F_2(\parallel P^{\sigma})$

$\square$

Algebraic Closures

Proposition 2.8.14: Algebraic closure is a field

$$\begin{align*} &\sphericalangle \\ &E / F \in \mathcal F \\ \hline \\ &E/\overline{F}_E/F \end{align*}$$

Proof

By definition of $\overline{F}_E$ it's obvious that $F \subseteq \overline{F}_E \subseteq E$. So all we need to prove is $\overline{F}_E \in \mathcal F$.

Consider some $a, b \in \overline{F}_E$. By $(2.8.7)$ $[F(a, b): F] < \infty$, so by $(2.8.5)$ $F(a, b)/_AF$. So

$$F(a, b) \subseteq E \implies F(a, b) =\overline F_{F(a, b)} \subseteq \overline F_E \implies \\ a\pm b, ab, ab^{-1} \in F(a, b) \subseteq \overline{F}_E$$

$\square$

Proposition 2.8.15: Algebraically closed field properties

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ \hline \\ &\text{The following are equivalent:} \\ &\begin{align*} & E/_AF \implies E =F\tag{a}\\ & E/F, [E:F]\lt \infty \implies E = F \tag{b}\\ & E/F \implies \overline{F}_E=F\tag{c}\\ & p \in F[x] \setminus F \implies F(\parallel p)=F \tag{d}\\ & p \in F[x] \setminus F \implies \exists \alpha \in F: p(\alpha) =0 \hspace{1cm} \tag{e}\\ & p \in F[x]^- \implies \deg p(x)=1 \tag{f}\\ \end{align*} \end{align*}$$

Proof

$(a) \implies (b)$

By $(2.8.5): [E:F]\lt \infty \implies E /_A F \implies E = F$

$(b) \implies (c)$

Obviously $F \subseteq \overline{F}_E$

$a \in \overline{F}_E \implies [F(a) : F] < \infty \implies F(a) = F \implies a \in F$

So $F \supseteq \overline{F}_E$

$(c) \implies (d)$

Consider some $p \in F[x]$:

$$E: = F (\parallel p) \implies E /_A F \\ (a \in E \implies a \in \overline{F}_{E} =F) \implies F/E \\$$

So $F(\parallel p)=F$.

$(d) \implies (e)$

Obvoius

$(e) \implies (f)$

$p \in F[x]^- \implies \exists \alpha \in F: p(\alpha) =0$. This means that $p$ has a linear factor. But that's the only factor it can have since it's irreducible. So $\deg p = 1$.

$(f) \implies (a)$

Obviously $F \subseteq E$

$$a \in E, p:=\mathfrak{pol}_F(a) \implies \deg p = 1 \overset{(2.8.4.b)} \implies \\ [F(a):F]=1 \implies a \in F \\ E \subseteq F$$

$\square$

def: Algebraically closed field

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &F \text{ satisfies any condition of }(2.8.15) \\ \hline \\ &F - \text{algebraically closed field} \end{align*}$$

def: Algebraic closure

$$\begin{align*} &\sphericalangle \\ &\overline{F}/_AF \in \mathcal F \\ &\overline{F} - \text{algebraically closed field}\\ \hline \\ &\overline{F} - \text{algebraic closure of } F \end{align*}$$

Proposition 2.8.16: Every field has a unique algebraic closure

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ \hline \\ &\exists! \overline F \end{align*}$$

Proof

The following proof for agebraic closures use infinite extensions or extensions of infinite set. If we wanted to make it strict we would need to introduce some notions from the set theory like cardinal numbers and associated theorems. Essentially we want to prove existence and uniqueness of algebraic closures not diving deep into its structure. So we'll rather give a semi-strict proof which conveys the main idea and does not involve additional notions from the set theory.

Consider $\mathcal E$ - the set of all algebraic extensions of $F$. Let's introduce a partial order $X \le Y := X\subseteq Y$ on it.

Each chain $E_1, E_2, \ldots$ in $\mathcal E$ will have an upper bound $E = \bigcup_i E_i$. $E \in \mathcal E$ because it's a field and it's algebraic (can be verified by assuming $\forall a \in E : \exists i: a \in E_i$). So by Zorn's lemma we have a maximal $\overline{F} \in E$.

Any extension $E/_A\overline F = \overline F$ by $\overline F$ maximality. So $\overline{F}$ is indeed algebraically closed.

By $(2.8.15.d)$ every polynomial splits in $F$ so we have $F(\parallel F[x] \setminus F) = \overline{F}$. By $(2.8.12)$ $F(\parallel F[x] \setminus F)$ is unique and so is $\overline F$.

$\square$

Transcendental extensions

def: Transcendental extension

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ &\exists a \in E \setminus \overline F_E\\ \hline \\ &E/_TF, E - \text{transcendental extension of } F \end{align*}$$

Proposition 2.8.17: Simple transcendental extension is a polynomials field of fractions

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ \hline \\ &\alpha \in E \setminus \overline F_E \iff F(\alpha) \cong \mathfrak F(F[x]) \end{align*}$$

Proof

$\varepsilon_\alpha$ - evaluation homomorphism from $(2.8.1)$.

$$\alpha \in E \setminus \overline F_E \iff \forall p \in F[x] \setminus F: p(\alpha) \ne 0 \iff \\ \ker \varepsilon_{\alpha} = \{0\} \iff F[x] \cong \varepsilon_\alpha(F[x]) = F[\alpha] \iff \\ \mathfrak F(F[x]) \cong \mathfrak F(F[\alpha])$$

Finally note that obviously $F \in \mathfrak F(F[\alpha]), \alpha \in \mathfrak F(F[\alpha])$ and $\mathfrak F(F[\alpha]) \in \mathcal F$, so $F(\alpha) \subseteq \mathfrak F(F[\alpha]).$ But also every elements of $\mathfrak F(F[\alpha])$ is $\frac{f(\alpha)}{g(\alpha)}$, which obviously belong to $F(\alpha)$.

$\square$

note

This theorem also states that if $E:=\mathfrak F(F[x])$ then $F(x)=\mathfrak F(F[x])$, we'll use the notation $F(x)$ further to denote field of fractions of polynomials ring. We'll call $F(x)$ a function field.

def: Algebraically independent set

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ &T:= \{t_1, \ldots, t_k\}\\ &\forall p \in F[x_1, \ldots, x_k]: p(t_1, \ldots, t_k)=0 \implies p = 0 \\ \hline \\ &T \in \perp_A(F) \\ &T - \text{algebraically independent over } F \end{align*}$$

note

This is a generalization of linear indepence which operates only on linear polynomials to arbitrary polynomials. If we consider $F[x_1, \ldots, x_k]=\bigoplus_iF_i[x_1, \ldots, x_k]$ as a graded ring and replace $F[x_1, \ldots, x_k]$ with $F_1[x_1, \ldots, x_k]$ in the definition of Algebraically independent set, we'll get the definition of linearly independent set.

Example: Algebraically independent set

Consider function field $F(x_1, \ldots, x_n)$. Then $\{x_1, \ldots, x_n\}$ is algebraically independent set.

Example: Algebraically independece over chain of fields

$E/L/F, S \subseteq E$ if $S \bot_A(L)$ then $S \bot_A(F)$.

Lemma 2.8.18: Transcendence basis and transcendental element

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ &T \in \bot_A(F) \\ & t \in E \setminus \overline {F(T)}_E \\ \hline \\ &T \cup \{t\} \in \bot_A(F) \end{align*}$$

Proof

Assume the proposition is false:

$$T \cup \{t\} \notin \bot_A(F) \implies \\ \exists p \in F[x_1, \ldots, x_n, y] \setminus \{0\}, t_1, \ldots, t_n \in T: p(t_1, \ldots, t_n, t) = 0$$

Consider $p(y) = p(t_1, \ldots , t_n, y) = \sum_{j=0}^{m}a_jy^j, a_j \in F[t_1, \ldots, t_n], a_m \ne 0$. Note that $m \ge 1$, otherwise the algebraic independence of $T$ is violated. Since $p(t) = 0$ it follows $t \in \overline {F(t_1 , \ldots , t_n)}_E \subseteq \overline {F(T)}_E$ which is a contradiction.

$\square$

def: Transcendence basis

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ &T \subseteq E\\ &T \in \bot_A(F)\\ & E/_A F(T)\\ \hline \\ &T \in \mathfrak {B}^{\text{tr}}_F(E), T - \text{transcendence basis} \\ \end{align*}$$

Example: Transcendence basis in function field

Consder $E:=F(x_1, \ldots, x_n)$. As mentioned in the example above $\{x_1, \ldots, x_n\} \in \bot_A(F)$. So $\{x_1, \ldots, x_n\} \in \mathfrak {B}^{\text{tr}}_F(E)$.

Additionally $T:=\{x_1^{r_1}, \ldots, x_n^{r_n}\}, r_i \ge 1$ is a transcendence basis of $E$. Obviously $T \in \bot_A(F)$. Denote $L:=F(T)$ then $E/_AL$ because any element $x_i$ is a root of polynomial $p(t)=t^{r_i}-x^{r_i} \in L[t]$

Proposition 2.8.19: Transcendence basis exists

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ &S \subseteq T \subseteq E \\ &S \in \bot_A(F) \\ & E /_A F(T) \\ \hline \\ &\exists M \in \mathfrak {B}^{\text{tr}}_F(E), S \subseteq M\subseteq T \end{align*}$$

Proof

$$X:=\{Y: S \subseteq Y \subseteq T, Y \in \bot_A(F) \} \\ S \in X \implies X \ne \empty$$ $$(X, \leq_P):=(X, \subseteq) \\ C = Y_1, Y_2, \ldots \in \mathfrak C(X) \implies \forall i: Y_i \leq_P \bigcup_jY_j \in X$$

Let's prove that $\bigcup_jY_j \in X$. The only way it can be violated is $\bigcup_jY_j \notin \bot_A(F)$. In that case consider the elements $\{y_1, \ldots, y_n\}\in \bigcup_jY_j:\exists p \in F[x_1, \ldots, x_n]:p(y_1, \ldots y_n)=0$. $\exists k: \{y_1, \ldots, y_n\} \in Y_k$, which contradicts $Y_k \in \mathfrak C(X)$.

Thus $\bigcup_jY_j \in X$ and

$$(2.1.3) \implies \exists M:=\max_{\leq_P} X$$

Assume $E/_T F(M)$. If we assume $E /_A F(T) /_A F(M)$ then $E/_A F(M)$ which is a contradiction. So we have $F(T) /_T F(M)$. If $\forall t \in T: t \in \overline{F(M)}_{F(T)}$ we'll have by $(2.8.7)$ $[F(T): F(M)]<\infty$, so by $(2.8.5)$ $F(T)/_AF(M)$ which is a contradiction. So we can take $t \in T, t \in F(T) \setminus \overline{F(M)}_{F(T)}$ and in $(2.8.18)$ assume $E:=F(T), T:=M, F:=F$ to get $M \cup \{t\} \in \bot_A(F)$. But that is a contradiction to the definition of $M$, so we proved $E/_AF(M)$. But we also know that by defintion $M \in \bot_A(F)$, so $M \in \mathfrak {B}^{\text{tr}}_F(E)$.

$\square$

note

The existence of basis follows from assuming $S=\empty$, $T=E$. Then $S \in \bot_A(E), E/_AF(T)$.

Proposition 2.8.20: Transcendence bases have the same cardinality

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ &S, T \in \mathfrak {B}^{\text{tr}}_F(E) \\ &|S| \lt \infty \\ \hline \\ &|S|=|T| \end{align*}$$

Proof

$S:=\{s_1, \ldots, s_n\}$.

$E /_T F(S \setminus \{s_1\})$, otherwise $s_1$ is a root of some polynomial in $F(S \setminus \{s_1\})$ so $S \notin \bot_A(F)$.

$E/_A F(T) \implies \exists t \in T: t \in E \setminus \overline{F(S \setminus \{s_1\})}_E \implies \{s_2, \ldots, s_n, t\} \in \bot_A(F)$

Now note that $\{s_1, \ldots, s_n, t\} \notin \bot_A(F)$, so $s_1 \in \overline{F(s_2, \ldots, s_n, t)}_E$ and $\{t, s_2, \ldots, s_n\} \in \mathfrak{B}^{\text{tr}}_F(E)$.

Repeating this procedure we arrive at the basis $\{t_1, \ldots, t_n\}$, so $|T|=|S|=n$

$\square$

note

This theorem is also true for infinte bases. However for our purposes a finite case will do.

def: Transcendence degree

$$\begin{align*} &\sphericalangle \\ &E/F \in \mathcal F \\ & T \in \mathfrak {B}^{\text{tr}}_F(E) \\ \hline \\ &\text{trdeg}_F(E):=|T| \\ \end{align*}$$

Example: Transcendence basis for a function field over elliptic curve

Consider some field $F$ and $p(x, y):=y^2-x^3+x \in F[x, y]$. It can be shown directly that $p \in F[x]^-$. Consider

$$A:=F[x, y]/\lang p \rang_I, A \in \mathcal R^{\mathcal ID}$$

Below is the proof that it's indeed an integral domain:

$$(2.5.13) \implies F[x, y] \in \mathcal R^{\mathcal {UFD}} \\ (2.4.12) \implies p \in \mathfrak P(F[x, y]) \\ (2.4.1) \implies \lang p \rang_I \in \mathfrak P_I(F[x, y]) \\ (2.4.8) \implies A \in \mathcal R^{\mathcal{ID}}$$

Since $A \in \mathcal R^{\mathcal{ID}}$ we can define a field of fractions on it:

$$E:=\mathfrak F(A)$$

Note that $E/F$, since $F$ is contained in $E$ in the form of constant polynomials. If we denote:

$$u:= x + \lang p \rang_I, v:=y+\lang p \rang_I$$

Then

$E = F(u, v)$

Let's prove that $\{u\} \in \mathfrak {B}^{\text{tr}}_F(E)$. First, $E/_A F(u)$, because $v^2=u^3-u$, so $v$ is a root of polynomial $t^2 - (u^3-u) \in (F(u))[t]$. So all we need to prove is $u \in E \setminus \overline F_E$.

Assume otherwise: $u \in \overline F_E$ then we'll have $E = \overline F_E$. Below is the proof that in this case $A \in \mathcal F$:

$$\alpha \in A, \alpha \ne 0 \\ \alpha^{-1} \in E=\overline F_E \implies \\ \alpha^{-n}+a_1\alpha^{-n+1}+\ldots + a_n=0, a_i \in F \implies \alpha^{-1}=-(a_1 + a_2\alpha+\ldots + a_{n}\alpha^{n-1}) \in A$$

Since $A\in \mathcal F$ then $\lang p \rang_I \in \mathfrak M(F[x,y])$ but that cannot be true since $\lang p \rang_I \subset \lang x, y \rang_I$.

Proposition 2.8.21: Basis and transcendence degree of chain extensions

$$\begin{align*} &\sphericalangle \\ &E/L/F \in \mathcal F \\ \hline \\ &\begin{align*} & \mathfrak {B}^{\text{tr}}_F(E) = \mathfrak {B}^{\text{tr}}_F(L) + \mathfrak {B}^{\text{tr}}_L(E) \tag{a}\\ &\text{trdeg}_F(E)=\text{trdeg}_F(E)+\text{trdeg}_F(E) \tag{b} \hspace{1cm}\\ \end{align*} \end{align*}$$

Exercises

1. Which of the following is true?

   a. $\mathbb Q(\sqrt 2) \cong_{M_\mathbb Q} \mathbb Q(\sqrt 3)$

   b. $\mathbb Q(\sqrt 2) \cong_{F} \mathbb Q(\sqrt 3)$

   c. Is $\mathbb Q(\sqrt 2, \sqrt 3) \cong_F \mathbb Q(\sqrt 2 + \sqrt 3)$

2. Consider field extension $E/L/K/F \in \mathcal F$ where

$$q: = 400240955522166739341778982573590415655688281993900788533205813\\6124031650490837864442687629129015664037894272559787 \\ F: = \mathbb F_q \\ K:= F(\alpha), \alpha^2+1 = 0 \\ L:= K(\beta), \beta^3- \alpha - 1 = 0 \\ E:= L(\gamma), \gamma^2-\beta = 0 \\$$

a. What is $[E:F]$?

b. What is the basis of $E/F$?

c. Calculate $14\alpha^3 + (\beta^2+ 2\gamma)^3$

3. Consider field $\mathbb F_5$

   a. Is $\mathbb F_5[x]/\lang x^2+1\rang_I \cong \mathbb F_5[x]/\lang x^2+3\rang_I$?

   b. Build an isomopshism between extenstions $\mathbb F_5[x]/\lang x^2+2\rang_I = \mathbb F_5(a)$ and $\mathbb F_5[x]/\lang x^2+3\rang_I= \mathbb F_5(b)$. What is the value of $2+3a$ under this isomopshism?

4. If $E=\mathbb Q(\parallel x^5-x^4-2x+2)$. What is $[E:\mathbb Q]$?