Difference between revisions of "Field extension"

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If <math>K</math> and <math>L</math> are [[field]]s and <math>K\subseteq L</math>, then <math>L/K</math> is said to be a '''field extension'''. We sometimes say that <math>L</math> is a field extension of <math>K</math>.
 
If <math>K</math> and <math>L</math> are [[field]]s and <math>K\subseteq L</math>, then <math>L/K</math> is said to be a '''field extension'''. We sometimes say that <math>L</math> is a field extension of <math>K</math>.
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If <math>L/K</math> is a field extension, then <math>L</math> may be thought of as a [[vector space]] over <math>K</math>. The dimension of this vector space is called the ''degree'' of the extension, and is denoted by <math>[L:K]</math>.
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Given three fields <math>K\subseteq L\subseteq M</math>, then, if the degrees of the extensions <math>M/L</math>, <math>L/K</math> and <math>M/K</math>, are finite, then are related by the [[tower law]]: <cmath>[M:K] = [M:L]\cdot[L:M]</cmath>
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One common way to construct an extension of a given field <math>K</math> is to consider an [[irreducible polynomial]] <math>g(x)</math> in the [[polynomial ring]] <math>K[x]</math>, and then to form the quotient ring <math>K(\alpha) = K[x]/\langle g(x)\rangle</math>. Since <math>g(x)</math> is irreducible, <math>\langle g(x)\rangle</math> is a [[maximal ideal]] and so <math>K(\alpha)</math> is actually a field. We can embed <math>K</math> into this field by <math>a\mapsto [a]</math>, and so we can view <math>K(\alpha)</math> as an extension of <math>K</math>. Now if we define <math>\alpha</math> as <math>[x]</math>, then we can show that in <math>K(\alpha)</math>, <math>g(\alpha) = 0</math>, and every element of <math>K(\alpha)</math> can be expressed as a polynomial in <math>\alpha</math>. We can thus think of <math>K(\alpha)</math> as the field obtained by 'adding' a root of <math>g(x)</math> to <math>K</math>.
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It can be shown that <math>[K(\alpha):K] = \deg g</math>.
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As an example of this, we can now ''define'' the [[complex numbers]], <math>\mathbb{C}</math> by <math>\mathbb{C} = \mathbb{R}[i] = \mathbb{R}[x]/\langle x^2+1\rangle</math>.
  
 
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[[Category:Advanced Mathematics Topics]]
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[[Category:Definition]]
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[[Category:Field theory]]

Latest revision as of 23:46, 24 March 2009

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 $K(\alpha) = K[x]/\langle g(x)\rangle$. Since $g(x)$ is irreducible, $\langle g(x)\rangle$ 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 $\mathbb{C} = \mathbb{R}[i] = \mathbb{R}[x]/\langle x^2+1\rangle$.

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