simple field extension

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

We consider the substitution homomorphism  $\phi :K[X]\to K[\alpha ]$, 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 $K[X]/𝔭$, where $𝔭$ 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 $K[X]/𝔭$ has no zero divisors, and hence $𝔭$ is a prime ideal.  It must be principal, for $K[X]$ is a principal ideal ring.

There are two possibilities:

1. $𝔭=(p(X))$, 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 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. $𝔭=(0)$.  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 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 (http://planetmath.org/Algebraic) with respect to $K$ (or over $K$).  This time we have a simple infinite field extension  $K(\alpha )/K$.