📂Abstract Algebra

Definition of Extension Fields and Proof of Kronecker's Theorem

Definition of an Extension Field¹

For a field $F$, a field $E$ with $F \le E$ is called an extension field of $F$.

Kronecker’s Theorem

If $f(x) \in F [ x ]$ is not a constant, then there exist an extension field $E$ of $F$ and $\alpha \in E$ such that $f ( \alpha ) = 0$.

Explanation

As an example of an extension field, $\mathbb{C}$ is an extension field of $\mathbb{R}$. Kronecker’s theorem implies that even if a root of the polynomial does not exist in $F$ for now, the domain can be enlarged to $E$, and once enlarged, a root exists. The very statement that a root will exist if we just blindly extend $F$ without even knowing what $F$ looks like is a theorem that feels quintessentially mathematical.

Proof

Part 1. Since $f(x)$ is not a constant function, it factors uniquely into irreducible elements over $F$; let $p(x)$ be one of those irreducible elements. Then the principal ideal $\left< p(x) \right>$ is a maximal ideal of $F [ x ]$, and $F [ x ] / \left< p(x) \right>$ is a field.

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Part 2. Existence of the Extension Field $E$

If we define the map $\psi : F \to F [ x ] / \left< p(x) \right>$ as follows, then $\psi$ naturally becomes a homomorphism. $$ \psi (a) := a + \left< p(x) \right> $$

If $\psi (a) = \psi (b)$ for some $a,b \in F$, then $$ a + \left< p(x) \right> = b + \left< p(x) \right> $$ so $(b-a) \in \left< p(x) \right>$, which means that $(b-a)$ is a constant multiple of $p(x)$. However, since $a, b \in F$ from the start, we have $(b-a) \in F$, and for $(b-a)$ to also be a constant multiple of $p(x)$, it must be that $(b-a) = 0$. Therefore $\psi$ is injective, and $\psi$ becomes an isomorphism sending each element of $F$ into some subfield of $F [ x ] / \left< p(x) \right>$. Here, if we concretely define $E := F [ x ] / \left< p(x) \right>$, then $E$ becomes an extension field of $F$.

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Part 3. Existence of the Root $\alpha \in E$

Let $\alpha : = x + \left< p(x) \right>$; then, to begin with, $\alpha$ $\in E$. For this $\alpha$, define the evaluation function $\phi_{\alpha} : F [ x ] \to E$. Writing the irreducible element $p(x) \in F [ x ]$ chosen above concretely as $p(x) := a_{0} + a_{1} x + \cdots + a_{n} x^{n}$, we get

$$ \begin{align*} p ( \alpha) =& \phi_{\alpha} ( p(x) ) \\ =& a_{0} + a_{1} ( x + \left< p(x) \right> ) + \cdots + a_{n} ( x + \left< p(x) \right> )^n \\ =& a_{0} + a_{1} ( x + \left< p(x) \right> ) + \cdots + a_{n} ( x^n + \left< p(x) \right> ) \\ =& \left( a_{0} + a_{1} x + \cdots + a_{n} x^n \right) + \left< p(x) \right> \\ =& p(x) + \left< p(x) \right> \\ =& 0 + \left< p(x) \right> \end{align*} $$

Therefore, in $F [ x ] / \left< p(x) \right>$, we have $p( \alpha ) = 0$, and $\alpha$ satisfies $f ( \alpha ) = 0$.

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