LEC ALG3 15

Properties of algebraic elements

Definition 119.1(Field extension).

A field extension is an injective ring homomorphism $F\to K$, where $F$ is called the base field, and $K$ is called the extension of $F$. We will denote the field extension $F\subseteq K$ by $K/F$.

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Definition 119.2(Algebraic elements).

Let $K/F$ be a field extension and $\alpha \in K$. We say that $\alpha$ is algebraic over $F$ if $\alpha$ satisfies a polynomial $f(x)\in F[x]$ ¹. $K$ is said to be algebraic if every $\alpha \in K$ is algebraic over $F$. $\alpha \in K$ is said to be transcendental over $F$ if it is not algebraic.

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Example 119.3(Example).

1. $\pi$, $e$ over $\mathbb{Q}$ are transcendental.
2. $2^{1/3}+2^{1/2}+1$ is algebraic over $\mathbb{Q}$; the minimal polynomial is $x^6-6x^5+9x^4+3x^2-42x+31$.

Proposition 119.4(Proposition).

Let $K/F$ be a field extension. Let $\alpha$ be algebraic over $F$. Then there is a unique monic irreducible polynomial $f(x)\in F[x]$ such that $f(\alpha )=0$.

Proof.

Consider the ring homomorphism $\psi :F[x]\to K$ defined by $x\mapsto \alpha$. Since $F[x]$ is a PID, $\ker \psi =(f(x))$ for some irreducible monic $f$ such that $f(\alpha )=0$. If $g(x)$ is another irreducible monic polynomial in $F[x]$ such that $g(\alpha )=0$, then $f|g$. Since $g$ is irreducible, $f$ and $g$ must be associates, and since $g$ is monic, we have $f=g$.□

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Definition 119.5(Minimal polynomial).

The unique monic polynomial in Prp 4 satisfied by $\alpha$ is denoted by $\text{Irr}(\alpha ;F)$.

Example 119.6(Example).

1. Consider $\mathbb{R}\subseteq \mathbb{C}$ and $\mathbb{Q}\subseteq \mathbb{C}$. $\text{Irr}(\sqrt{2},\mathbb{R})=x-\sqrt{2}$, $\text{Irr}(\sqrt{2},\mathbb{Q})=x^2-2$.
2. $\text{Irr}(e^{i2\pi j/n},\mathbb{R})=x^k-1$, where $k=\text{lcm}(j,n)/j$.

Monogenic algebraic extensions

Definition 119.7(Extensions generated by one element).

Let $K/F$ be a field extension and let $\alpha \in K$. We denote the smallest subfield of $K$ containing $\alpha$ and $F$ by $F(\alpha )$.

Remark 119.8(Remark).

The notation $F(\alpha )$ is suggestive. Indeed, it is easy to see that $F(\alpha )$ is the set of all rational functions in $\alpha$ with coefficients in $F$:

$$F(\alpha )=\left\{\frac{f(\alpha )}{g(\alpha )}:f,g\in F[x],g(\alpha )\ne 0\right\}.$$

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Suppose $K/F$ and $\alpha \in K$. Clearly,

$$F[\alpha ]=\left\{\sum _{i=0}^n{a}_i{\alpha }^i:a_i\in F\right\}\subseteq F(\alpha ).$$

If $\alpha$ is algebraic over $F$, the reverse inclusion holds too.

Proposition 119.9(Proposition).

Let $K/F$ be a field extension and $\alpha \in K$ be algebraic over $F$.

$$F(\alpha )=F[\alpha ]\cong F[x]/\text{Irr}(\alpha ;F).$$

Proof.

Note that for $\psi :F[x]\to F[\alpha ]$ given by $x\mapsto \alpha$, we have $\ker (\psi )=(\text{Irr}(\alpha ,F))$. Since $\psi$ is surjective, we have $F[x]/\text{Irr}(\alpha ,F)\cong F[\alpha ]$. Thus, $F[\alpha ]$ is a field, and hence $F(\alpha )=F[\alpha ]$.□

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Remark 119.10(Remark).

By Prp 119.4, for every $\alpha$ that is algebraic over $k$, there exists a unique monic irreducible polynomial $f(x)\in k[x]$ such that $f(\alpha )=0$. By Prp 119.9, $k(\alpha )\cong k[x]/\text{Irr}(\alpha ,k)$. For instance, if ${\alpha }_1,{\alpha }_2,{\alpha }_3$ are the roots of $x^3-2\in \mathbb{Q}[x]$ in $\mathbb{C}$, $x^3-2$ must be the irreducible polynomial of ${\alpha }_1,{\alpha }_2,{\alpha }_3$ since it is monic and irreducible, and we have $\mathbb{Q}({\alpha }_1)\cong \mathbb{Q}({\alpha }_2)\cong \mathbb{Q}({\alpha }_3)$.

Remark 119.11(Constructing inverses in$F[\alpha ]$).

$F[\alpha ]$ is a field by Prp 9; how can we obtain the inverse of $f(\alpha )\in F[\alpha ]$?

Let $m(x)=\text{Irr}(\alpha ,F)$. Since $m(x)$ is irreducible in $F[x]$, the gcd of $f(x)$ and $m(x)$ is $1$. Since $F[x]$ is a Euclidean domain, the euclidean algorithm supplies $g(x),h(x)\in F[x]$ such that $f(x)g(x)+m(x)h(x)=1$. Reducing modulo $m(x)$, we obtain $f(x)g(x)\equiv 1\mathrm{mod}m(x)$. Thus, $f(\alpha )g(\alpha )=1$.

Proposition 119.12(Proposition).

$F(\alpha )=F[\alpha ]$ iff $\alpha$ is algebraic over $F$.

Proof.

Let$\alpha \in K$ be transcendental over $F$. If $F(\alpha )=F[\alpha ]$, since ${\alpha }^{-1}\in F(\alpha )$, we have ${\alpha }^{-1}={\sum }_{i=0}^n{a}_i{\alpha }^i⟺0=\left({\sum }_{i=0}^n{a}_i{\alpha }^{i+1}\right)-1$. This implies $\alpha$ is algebraic over $F$, a contradiction.□

Proposition 119.13(Monogenic algebraic extensions are finite).

Let $K/F$ be a field extension and $\alpha \in K$ be algebraic over $F$. Then, ${\text{dim}}_F(F(\alpha ))=n$, where $n$ is the degree of $\text{Irr}(\alpha ;F)$. Specifically, $B=\{1,\alpha ,\dots ,{\alpha }^{n-1}\}$ is an $F$-basis for $F(\alpha )$.

Proof.

If $B$ were not linearly independent, an equation of linear dependence would yield a polynomial of degree less than $n$ of which $\alpha$ is a root, contradicting Prp 4.

Let $g(\alpha )\in F[\alpha ]$. Suppose $f=\text{Irr}(\alpha ;F)$. By the division algorithm, there exist unique $q(x),r(x)\in F[x]$ with $r(x)=0$ or $\text{deg}(r(x))<n$ such that

$$\begin{array}{rl}& g(x)=f(x)q(x)+r(x)\\ ⟹& g(\alpha )=r(\alpha ).\end{array}$$

We can now write

$$r(x)=\sum _{i=0}^{n-1}{a}_i{x}^i$$

and hence $r(\alpha )\in \text{span}(B)$. Therefore, $F[\alpha ]\subseteq \text{span}(B)$. Since $\text{span}(B)\subseteq F[\alpha ]$, we get $F[\alpha ]=\text{span}(B)$.□

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Remark 119.14(Remark).

By Prp 13 and Prp 322.4, a monogenic extension $F(\alpha )/F$ is finite iff $\alpha$ is algebraic over $F$.

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Proposition 119.15(Proposition).

Let $K/F$ be a field extension and $\alpha \in K$ be algebraic over $F$. Suppose $\text{deg}(\text{irr}(\alpha ;F))=n$. If $\beta \in F(\alpha )$, then $\text{deg}(\text{irr}(\beta ;F))\le n$.

Proof.

Clearly,$F(\beta )\subseteq F(\alpha )$¹. Since ${\text{dim}}_F(F(\alpha ))=n$, we get ${\text{dim}}_F(F(\beta ))\le n$. It follows from Prp 119.13 that $\text{deg}(\text{irr}(\beta ;F))={\text{dim}}_F(F(\beta ))\le n$.□

Proposition 119.16(Proposition).

Let $K/F$ be a field extension. Suppose $\alpha ,\beta \in K$ are algebraic over $F$. If $\text{irr}(\alpha ;F)=\text{irr}(\beta ;F)$ then there is an isomorphism $\psi :F(\alpha )\to F(\beta )$ fixing elements of $F$ such that $\psi (\alpha )=\beta$.

Proof.

Since $\text{irr}(\alpha ;F)=\text{irr}(\beta ;F)$, $F(\alpha )\cong F(\beta )$. Consider $\psi$ as in the statement. By definition, $\psi$ is a ring homomorphism (this works only because the irreducible polynomials of $\alpha$ and $\beta$ are equal). Since $\ker \psi \ne (1)$, and $F[\alpha ]$ is a field, we must have $\ker (\psi )=(0)$, that is, $\psi$ is injective. Note that the $F$-basis $\{1,\alpha ,\dots ,{\alpha }^{n-1}\}$ of $F[\alpha ]$ maps to the $F$-basis $\{1,\beta ,\dots ,{\beta }^{n-1}\}$ of $F[\beta ]$. Since $\psi$ is a $F$-linear map, we get $\psi$ is surjective.□

The converse is not true if we do not assume $\psi (\alpha )=\beta$. Consider $\mathbb{C}/\mathbb{R}$, $\alpha =i$, $\beta =2i$. $\mathbb{R}(\alpha )=\mathbb{R}(\beta )=\mathbb{C}$. The only $\mathbb{R}$-algebra isomorphisms between $\mathbb{R}(\alpha )=\mathbb{R}[x]/(x^2+1)$ and $\mathbb{R}(\beta )=\mathbb{R}[x]/(x^2+4)$ are given by

$$\begin{array}{rl}[f(x){]}_{x^2+1}& \mapsto [f(x/2){]}_{x^2+4},\\ [f(x){]}_{x^2+1}& \mapsto [f(-x/2){]}_{x^2+4},\end{array}$$

that is, by $\alpha \mapsto \beta /2$ and $\alpha \mapsto -\beta /2$.

Algebraic extensions

Definition 119.17(Algebraic extension).

A field extension $K/F$ is called an algebraic extension if every $\alpha \in K$ is algebraic over $F$.

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Proposition 119.18(Proposition).

Let $K/F$ be a field extension. Assume that $\alpha \in K$ is algebraic over $F$. Then $F(\alpha )/F$ is an algebraic extension.

Proof.

Let $n=\text{deg}(\text{Irr}(\alpha ;F))$. Let $\beta \in F(\alpha )$. Then $\{1,\beta ,\dots ,{\beta }^n\}\subseteq F(\alpha )$ is linearly dependent over $F$ by Prp 119.13, so there exist coefficients $a_0,\dots ,a_n\in F$ not all zero such that ${\sum }_{i=o}^n{a}_i{\beta }^i=0$. Thus $\beta$ is algebraic over $F$.□

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Proposition 119.19(Proposition).

Let $L/K$ be a field extension and $K/F$ be an algebraic extension. If $\alpha \in L$ is algebraic over $K$, then $\alpha$ is algebraic over $F$.

Proof.

Let $\alpha \in L$ be algebraic over $K$. Let $\text{Irr}(\alpha ;K)={\sum }_{i=0}^n{a}_n{x}^n$. The key observation is that $\alpha$ is algebraic over the finitely generated algebraic extension $F(a_0,\dots ,a_n)/F$, which is a finite extension by Prp 322.8. Also, the extension $F(a_0,\dots ,a_n,\alpha )/F(a_0,\dots ,a_n)$ is finite. By Prp 322.2, the extension $F(a_0,\dots ,a_n,\alpha )/F$ is finite, and hence algebraic by Prp 322.4. Thus, $\alpha$ is algebraic over $F$.□

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Corollary 119.20(Corollary).

Let $L/K$ and $K/F$ be algebraic extensions. Then $L/F$ is an algebraic extension.

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Misc

Proposition 119.21(Proposition).

Suppose $K/F$, $K^{\prime }/F$, and $\psi :K\to K^{\prime }$ is a ring homomorphism fixing $F$. Let $\alpha \in K$ be a root of $f(x)\in F[x]$. Then $\psi (\alpha )$ is a root of $f(x)$.

Proof.

Let$f(x)={\sum }_{i=0}^n{a}_i{x}^i$. We have $f(\alpha )=0$. Then,

$$\begin{array}{rl}f(\psi (\alpha ))& =\sum _{i=0}^n{a}_i(\psi (\alpha ){)}^i\\ & =\psi \left(\sum _{i=0}^n{a}_i{\alpha }^i\right)\\ & =\psi (0)\\ & =0.\end{array}$$ □

For instance, this shows if $\alpha \in \mathbb{C}$ is a root of $f(x)\in \mathbb{R}[x]$, then $\overline{\alpha }$ is also a root of $f(x)$.

Footnotes

1. Since $F$ is a field, we can assume that $f$ is monic. ↩ ↩²