📂Abstract Algebra

Proof of the Conjugate Isomorphism Theorem

Definition¹

Let $\alpha$ be algebraic over a field $F$.

1. The polynomial $p(x) \in F [ x ]$ whose leading coefficient is $1$ and which satisfies $p( \alpha ) = 0$ is called the irreducible polynomial for $\alpha$ over $F$, and is denoted by $\text{irr} ( \alpha , F) = p(x)$.
2. The degree of the leading term of $\text{irr} ( \alpha , F)$ is called the degree of $\alpha$ over $F$, and is denoted by $\deg ( \alpha , F )$.
3. For an algebraic extension field $E$ of $F$, if $\text{irr} ( \alpha , F) = \text{irr} ( \beta , F)$, then the two elements $\alpha , \beta \in E$ are said to be conjugate over $F$.

Example

As a natural example, consider the complex conjugate $\overline{ a + ib } = a - ib$. If we define the irreducible polynomial $$ p(x) := (x^2 - 2ax + a^2 + b^2) \in \mathbb{R}[x] $$ then $$ \begin{align*} & p\left( a + ib \right) \\ =& a^2 + i2ab - b^2 - 2a^2 - i2ab + a^2 + b^2 \\ =& 0 \\ =& a^2 - i2ab - b^2 - 2 a^2 + i2ab + a^2 + b^2 \\ =& p \left( a - ib \right) \end{align*} $$ so, even in the language of algebra, we can see that $\left( a + ib \right) , \left( a - ib \right) \in \mathbb{C}$ are conjugate over $\mathbb{R}$.

As an example for a more intuitive understanding of conjugates, consider the quintic equation $$ x^5 + x^3 + x^2 + 1 = 0 $$

Factoring it into irreducible elements of $\mathbb{R} [ x ]$ gives $$ x^5 + x^3 + x^2 + 1 = (x^3+1)(x^2+1) = (x+1) ( x^2 + x + 1 ) ( x^2+ 1 ) $$ so we can obtain the solutions $x=-1$, $x= \pm 1$, and $\displaystyle x= {{-1 \pm \sqrt{ - 3} } \over {2}}$. All of these satisfy the given equation, but $x=-1$ is a solution of a linear equation and therefore cannot be conjugate to the other solutions. Also, $x=i$ and $\displaystyle x= {{-1 + \sqrt{ - 3} } \over {2}}$ are zeros of different irreducible polynomials, so they are not conjugate.

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 conjugate over $\mathbb{Q}$.

Theorem

Let $\alpha$ be algebraic over a field $F$ with $\deg ( \alpha , F) = n$. If we define the map $\psi_{\alpha , \beta} : F( \alpha ) \to F ( \beta )$ by $$ \psi_{ \alpha , \beta } ( c_{0} + c_{1} \alpha + \cdots + c_{n-1} \alpha^{n-1} ) := c_{0} + c_{1} \beta + \cdots + c_{n-1} \beta^{n-1} $$ then

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

Proof

$( \implies )$

If we set $$ \text{irr} ( \alpha , F) := a_{0} + a_{1} x + \cdots + a_{n} x^{n} $$ then $$ a_{0} + a_{1} \alpha + \cdots + a_{n} \alpha^{n} = 0 $$ holds. Since $\psi_{ \alpha , \beta }$ is an isomorphism, $$ \begin{align*} \psi_{ \alpha , \beta } ( 0) =& \psi_{ \alpha , \beta } ( c_{0} + c_{1} \alpha + \cdots + c_{n} \alpha^{n} ) \\ =& c_{0} + c_{1} \beta + \cdots + c_{n} \beta^{n} \\ =& 0 \end{align*} $$ holds. This means that $\text{irr} ( \beta , F)$ divides $\text{irr} ( \alpha , F)$, and since the same applies to $( \psi_{ \alpha , \beta } )^{-1} = \psi_{ \beta , \alpha }$, 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 evaluation homomorphisms $\phi_{\alpha} : F [ x ] \to F(\alpha)$ and $\phi_{\beta} : F [ x ] \to F(\beta)$. Then, since $p( \alpha ) = p( \beta ) = 0$, $\phi_{\alpha}$ and $\phi_{\beta}$ have the same kernel $\left< p(x) \right> \subset F [ x ]$.

Fundamental theorem of homomorphisms: For rings $R$, $r '$, if a homomorphism $\phi : R \to r '$ exists, then $R / \ker ( \phi ) \simeq \phi (R)$

By the fundamental theorem of homomorphisms, there exist two isomorphisms $\psi_{\alpha} : F / \left< p(x) \right> \to F ( \alpha )$ and $\psi_{\beta} : F / \left< p(x) \right> \to F (\beta )$. Thus, if we set $$ \psi_{\alpha , \beta } := \psi_{\alpha} \circ ( \psi_{\alpha} )^{-1} $$ then $\psi_{\alpha , \beta } : F ( \alpha ) \to F ( \beta )$ is also an isomorphism. Therefore, for $( c_{0} + c_{1} \alpha + \cdots + c_{n-1} \alpha^{n-1} ) \in F ( \alpha )$, the following holds. $$ \begin{align*} & \psi_{ \alpha , \beta } ( c_{0} + c_{1} \alpha + \cdots + c_{n-1} \alpha^{n-1} ) \\ =& \left( \psi_{\alpha} \circ ( \psi_{\alpha} )^{-1} \right) ( c_{0} + c_{1} \alpha + \cdots + c_{n-1} \alpha^{n-1} ) \\ =& \psi_{\beta} \left( ( c_{0} + c_{1} x + \cdots + c_{n-1} x^{n-1} ) + \left< p(x) \right> \right) \\ =& c_{0} + c_{1} \beta + \cdots + c_{n-1} \beta^{n-1} \end{align*} $$

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

Corollary

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

Proof of the Corollary

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

Since $f ( a + ib) = 0$, $$ f( a + ib ) := c_{0} + c_{1} ( a + ib ) + \cdots + c_{n} ( a + ib )^{n} = 0 $$ Since $i$ and $-i$ are conjugate over $\mathbb{R}$, 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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