📂Abstract Algebra

Proof of the Conjugate Isomorphism Theorem

Definition ¹

Let’s say $\alpha$ is algebraic over the field $F$.

1. A reduced polynomial for $\alpha$ over $F$, whose highest coefficient is $1$ and satisfies $p( \alpha ) = 0$, is represented as $\text{irr} ( \alpha , F) = p(x)$.
2. The highest degree of $\text{irr} ( \alpha , F)$ is called the degree of $\alpha$ over $F$, and is represented as $\deg ( \alpha , F )$.
3. For the algebraic extension field $E$ of $F$, if we say $\text{irr} ( \alpha , F) = \text{irr} ( \beta , F)$, then two elements $\alpha , \beta \in E$ are called conjugates over $F$.

Example

As a natural example, consider the conjugate complex number $\overline{ a + ib } = a - ib$. For the reduced polynomial $$p(x) := (x^2 - 2ax + a^2 + b^2) \in \mathbb{R}[x]$$, defining it as $$\begin{align*} & p\left( a + ib \right) \\ =& a^2 + i2ab - b^2 - 2a^2 - i2ab + a^2 + b^2 \\ =& 0 \\ =& a^2 - i2ab - b^2 - 2 a^2 + i2ab + a^2 + b^2 \\ =& p \left( a - ib \right) \end{align*}$$, using the expression of algebra, we can see that $\left( a + ib \right) , \left( a - ib \right) \in \mathbb{C}$ is a conjugate over $\mathbb{R}$.

For a more intuitive understanding of conjugates, consider the error equation $$x^5 + x^3 + x^2 + 1 = 0$$.

Factoring it into irreducibles $\mathbb{R} [ x ]$, $$x^5 + x^3 + x^2 + 1 = (x^3+1)(x^2+1) = (x+1) ( x^2 + x + 1 ) ( x^2+ 1 )$$, therefore, solutions $x=-1$, $x= \pm 1$, and $\displaystyle x= {{-1 \pm \sqrt{ - 3} } \over {2}}$ can be found. Although all of these satisfy the given equation, $x=-1$ is a solution to a linear equation and, therefore, cannot be a conjugate of the other solutions. Moreover, $x=i$ and $\displaystyle x= {{-1 + \sqrt{ - 3} } \over {2}}$ are not conjugates since they are zeros of a different reduced polynomial.

As an example of conjugates without complex numbers, $\sqrt{2}$ and $(- \sqrt{2})$, which are zeros of $(x^2 - 2 ) \in \mathbb{Q} [ x ]$, are conjugates over $\mathbb{Q}$.

Theorem

Let’s say $\deg ( \alpha , F) = n$ for algebraic $\alpha$ over the field $F$. By defining the mapping $\psi_{\alpha , \beta} : F( \alpha ) \to F ( \beta )$ as $$\psi_{ \alpha , \beta } ( c_{0} + c_{1} \alpha + \cdots + c_{n-1} \alpha^{n-1} ) := c_{0} + c_{1} \beta + \cdots + c_{n-1} \beta^{n-1}$$,

• $\psi_{ \alpha , \beta }$ is a homomorphism $\iff$ $\text{irr} ( \alpha , F) = \text{irr} ( \beta , F)$

Proof

Assuming $( \implies )$,

$$\text{irr} ( \alpha , F) := a_{0} + a_{1} x + \cdots + a_{n} x^{n}$$, then $$a_{0} + a_{1} \alpha + \cdots + a_{n} \alpha^{n} = 0$$. Since $\psi_{ \alpha , \beta }$ is a homomorphism, $$\begin{align*} \psi_{ \alpha , \beta } ( 0) =& \psi_{ \alpha , \beta } ( c_{0} + c_{1} \alpha + \cdots + c_{n} \alpha^{n} ) \\ =& c_{0} + c_{1} \beta + \cdots + c_{n} \beta^{n} \\ =& 0 \end{align*}$$. This means that $\text{irr} ( \beta , F)$ divides $\text{irr} ( \alpha , F)$, and the same applies to $( \psi_{ \alpha , \beta } )^{-1} = \psi_{ \beta , \alpha }$, therefore the following holds: $$\text{irr} ( \beta , F) = \text{irr} ( \alpha , F)$$.

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$( \impliedby )$

Let $$p(x) := \text{irr} ( \beta , F) = \text{irr} ( \alpha , F)$$ and define the substitution functions $\phi_{\alpha} : F [ x ] \to F(\alpha)$ and $\phi_{\beta} : F [ x ] \to F(\beta)$. Then, because of $p( \alpha ) = p( \beta ) = 0$, $\phi_{\alpha}$, and $\phi_{\beta}$ have the same kernel $\left< p(x) \right> \subset F [ x ]$.

Fundamental Theorem of Homomorphism: For rings $R$, $r '$, if there exists a homomorphism $\phi : R \to r '$, then $R / \ker ( \phi ) \simeq \phi (R)$.

By the Fundamental Theorem of Homomorphism, two homomorphisms $\psi_{\alpha} : F / \left< p(x) \right> \to F ( \alpha )$ and $\psi_{\beta} : F / \left< p(x) \right> \to F (\beta )$ exist. By setting $$\psi_{\alpha , \beta } := \psi_{\alpha} \circ ( \psi_{\alpha} )^{-1}$$, $\psi_{\alpha , \beta } : F ( \alpha ) \to F ( \beta )$ is also a homomorphism. Therefore, the following holds for $( c_{0} + c_{1} \alpha + \cdots + c_{n-1} \alpha^{n-1} ) \in F ( \alpha )$: $$\begin{align*} & \psi_{ \alpha , \beta } ( c_{0} + c_{1} \alpha + \cdots + c_{n-1} \alpha^{n-1} ) \\ =& \left( \psi_{\alpha} \circ ( \psi_{\alpha} )^{-1} \right) ( c_{0} + c_{1} \alpha + \cdots + c_{n-1} \alpha^{n-1} ) \\ =& \psi_{\beta} \left( ( c_{0} + c_{1} x + \cdots + c_{n-1} x^{n-1} ) + \left< p(x) \right> \right) \\ =& c_{0} + c_{1} \beta + \cdots + c_{n-1} \beta^{n-1} \end{align*}$$.

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Meanwhile, we can derive the following useful corollary for equations with real coefficients.

Corollary

For $f(x) \in \mathbb{R} [ x ]$, if $f ( a + ib) = 0$, then $f ( a - ib) = 0$.

Proof of the corollary

Let’s say $f(x) := c_{0} + c_{1} x + \cdots + c_{n} x^{n}$.

Since $f ( a + ib) = 0$, $$f( a + ib ) := c_{0} + c_{1} ( a + ib ) + \cdots + c_{n} ( a + ib )^{n} = 0$$, $i$, and $-i$ are conjugates over $\mathbb{R}$, therefore, the following holds: $$0 = \psi_{i , -i} \left( 0 \right) = \psi_{i , -i} \left( f ( a + ib) \right)= f ( a - ib)$$.

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