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'simple field extension'
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| Title of object: |
simple field extension |
| Canonical Name: |
SimpleFieldExtension |
| Type: |
Definition |
| Created on: |
2004-05-31 18:19:39 |
| Modified on: |
2004-05-31 18:38:17 |
| Classification: |
msc:12F99 |
Preamble:
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Content:
Let $K(\alpha)$ be obtained from the field $K$ via the simple adjunction of the element $\alpha$. We shall settle the \PMlinkescapetext{sructure types} of $K(\alpha)$.
We consider the evaluation homomorphism $\phi: K[X] \rightarrow K[\alpha]$, where
$$\sum a_{\nu}X^\nu \rightarrow \sum a_{\nu}\alpha^\nu.$$
According to the ring homomorphism theorem, the ring $K[\alpha]$ is isomorphic with the residue class ring $K[X]/\frak{p}$, where $\frak{p}$ is the ideal of polynomials having $\alpha$ as their zero. Because $K[\alpha]$ is, as subring of the field $K(\alpha)$, an integral domain, then also $K[X]/\frak{p}$ has no zero divisors, and hence $\frak{p}$ is a prime ideal. It must be principal, for $K[X]$ is a principal ideal ring.
There are two possibilities:
1. $\frak{p} = (p(X))$, where $p(X)$ is an irreductible polynomial with $p(\alpha) = 0$. Because every non-zero prime ideal of $K[X]$ is maximal, $K[\alpha]$'s isomorphic image $K[X]/(p(X))$ is a field, and it must give the \PMlinkescapetext{structure} of $K(\alpha) = K[\alpha]$.
2. $\frak{p} = (0)$. This means that the homomorphism $\phi$ is an isomorphism between $K[X]$ and $K[\alpha]$, i.e. all expressions $\sum a_{\nu}\alpha^\nu$ behave as the polynomials $\sum a_{\nu}X^\nu$. Now, $K[\alpha]$ is no field because $K[X]$ is not such, but the isomorphy of the rings implies the isomorphy of the corresponding rings of fractions. Thus the simple extension field $K(\alpha)$ is isomorphic with the field $K(X)$ of rational functions in one indeterminate $X$. |
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