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Homesimple field extension

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# simple field extension

Let $K(\alpha)$ be obtained from the field $K$ via the simple adjunction of the element $\alpha$, which is called the *primitive element* of the field extension $K(\alpha)/K$. We shall settle the structure types of the field $K(\alpha)$.

We consider the substitution homomorphism $\varphi:K[X]\rightarrow K[\alpha]$, where

$\sum a_{{\nu}}X^{\nu}\mapsto\sum a_{{\nu}}\alpha^{\nu}.$ |

According to the ring homomorphism theorem, the image ring $K[\alpha]$ is isomorphic with the residue class ring $K[X]/\mathfrak{p}$, where $\mathfrak{p}$ 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 $K[X]/\mathfrak{p}$ has no zero divisors, and hence $\mathfrak{p}$ is a prime ideal. It must be principal, for $K[X]$ is a principal ideal ring.

There are two possibilities:

1. $\mathfrak{p}=(p(X))$, where $p(X)$ is an irreducible polynomial with $p(\alpha)=0$. Because every non-zero prime ideal of $K[X]$ is maximal, the isomorphic image $K[X]/(p(X))$ of $K[\alpha]$ is a field, and it must give the structure of $K(\alpha)=K[\alpha]$. We say that $\alpha$ is algebraic with respect to $K$ (or over $K$). In this case, we have a finite field extension $K(\alpha)/K$.

2. $\mathfrak{p}=(0)$. This means that the homomorphism $\varphi$ 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 fields of fractions. Thus the simple extension field $K(\alpha)$ is isomorphic with the field $K(X)$ of rational functions in one indeterminate $X$. We say that $\alpha$ is transcendental with respect to $K$ (or over $K$). This time we have a simple infinite field extension $K(\alpha)/K$.

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