Field extension

Revision as of 23:07, 1 September 2008 by Jam (talk | contribs) (Talked about K(a))

If $K$ and $L$ are fields and $K\subseteq L$, then $L/K$ is said to be a field extension. We sometimes say that $L$ is a field extension of $K$.

If $L/K$ is a field extension, then $L$ may be thought of as a vector space over $K$. The dimension of this vector space is called the degree of the extension, and is denoted by $[L:K]$.

Given three fields $K\subseteq L\subseteq M$, then, if the degrees of the extensions $M/L$, $L/K$ and $M/K$, are finite, then are related by the tower law: \[[M:K] = [M:L]\cdot[L:M]\]

One common way to construct an extension of a given field $K$ is to consider an irreducible polynomial $g(x)$ in the polynomial ring $K[x]$, and then to form the quotient ring $K(\alpha) = K[x]/<g(x)>$. Since $g(x)$ is irreducible, $<g(x)>$ is a maximal ideal and so $K(\alpha)$ is actually a field. We can embed $K$ into this field by $a\mapsto [a]$, and so we can view $K(\alpha)$ as an extension of $K$. Now if we define $\alpha$ as $[x]$, then we can show that in $K(\alpha)$, $g(\alpha) = 0$, and every element of $K(\alpha)$ can be expressed as a polynomial in $\alpha$. We can thus think of $K(\alpha)$ as the field obtained by 'adding' a root of $g(x)$ to $K$.

It can be shown that $[K(\alpha):K] = \deg g$.

As an example of this, we can now define the complex numbers, $\mathbb{C}$ by $\mathbb{C} = \mathbb{R}[i] = \mathbb{R}[x]/<x^2+1>$.

This article is a stub. Help us out by expanding it.

Invalid username
Login to AoPS