1

I would like to understand the definition of a simple transcendental extension and the theorem that states all such extensions are isomorphic. So for example, if $K \subseteq \mathbb{C} $ is any subfield and $L:K$ a field extension with $\alpha \in L $ then $K(\alpha) \cong K(X)$, where $K(X)$ is the field of rational functions in the indeterminate $X$. First, what makes a field extension transcendental? And second, what is the explicit isomorphism between $K(\alpha)$ and $K(X)$? I've heard this isomorphism called the "evaluation homomorphism" so could you please explain what this map is doing specifically?

My understanding of this theorem is that in this context $\alpha$ (like $X$) is purely a mathematical symbol, and is no longer considered as the root of some polynomial over $K$. Thus, it is as if we are writing rational expressions in $X$ as those in $\alpha$. However, this interpretation seems banal to me (why even write it in $\alpha$ if we have it in $X$?) and what exactly makes $K(X)$ transcendental in the first place?

Kevin Sheng
  • 2,483
  • You can easily look up the definition. A field extension is transcendental iff it is not algebraic. A field extension $L$ over $K$ is algebraic iff every $\alpha \in L$ is algebraic over $K$, i.e. $\alpha$ satisfies a monic polynomial with coefficients in $K$. – hardmath Mar 25 '15 at 21:55
  • 1
    The point is that because $\alpha$ is transcendental, then you know that if $p(X)$ and $q(X)$ are two (univariate) functions rational functions, then $p(\alpha)=q(\alpha)$ as values in $\mathbb{C}$ if and only if $p(X)=q(X)$ as functions - any nontrivial equality $p(\alpha)=q(\alpha)$ would lead to a relation for $\alpha$ that could be turned into a polynomial equation it satisfies and thus it would be algebraic. – Steven Stadnicki Mar 25 '15 at 21:56
  • Oh, so that's where the name evaluation homomorphism comes from? We take a $p(X)$ from $K(X)$ and map it to the value in $\mathbb{C}$, $p(\alpha) \in K(\alpha)$? – Kevin Sheng Mar 25 '15 at 22:02
  • Exactly - and the transcendentality is what guarantees that this mapping is an isomorphism. – Steven Stadnicki Mar 25 '15 at 22:14

0 Answers0