1

Working through Stewart's "Galois Theory" at the moment, want to understand simple transcendental field extensions and the theorem that states that the field of rational expressions in a field $K$ is one (and vice versa). I think I understand the logic behind it, it being that if an element is transcendtal over $K$ you basically have no simpler way of creating a working field containing the element other than taking all possible rational expressions of it, while if an element is algebraic, the possible "size" of the field shrinks somewhat, because it does not need to contain "as much". What I'm having trouble with is the mechanics of the proof, specifically:

Theorem 5.3. The set of rational expressions $K(t)$ is a simple transcendental extension of the subfield $K$ of $\Bbb C$.

Proof. Clearly $K(t) : K$ is a simple extension, generated by $t$. If $p$ is a polynomial over $K$ such that $p(t) = 0$ then $p = 0$ by definition of $K(t)$, so the extension is transcendental.

The part I don't understand is in bold italics. I understand that if the only polynomial that $t$ satisfies is the zero polynomial, then $t$ is transcendental, but how does the definition of $K(t)$ imply that $p$ even is zero in the first place?

Edit: Just to clarify, in the theorem $t$ is a new variable and $K(t)$ is the field of rational expressions in $t$ over $K$.

Community
  • 1
Hilbert Jr.
  • 1,465
  • $p(t)$ is an element of $K(t)$. For it to equal zero is for it to be the zero element, that is, for $p$ to be the zero polynomial. – Gerry Myerson Dec 14 '19 at 23:14
  • The idea here is (I suspect) that $t$ is a fresh variable in the definition of $K(t)$ and won't be subject to any algebraic relations that don't just follow from the field axioms. If $K(t)$ denotes the field generated by an element $t$ in some given superfield of $K$, then $t$ might be a root of $p$. Without seeing the definition of $K(t)$, I can't comment in more detail. – Rob Arthan Dec 14 '19 at 23:37
  • My bad, $t$ is a fresh variable and $K(t)$ is the field of rational expressions in it over $K$, will add it to the original post now. – Hilbert Jr. Dec 14 '19 at 23:40

1 Answers1

0

Either $t$ satisfies a nonzero polynomial equation $p(t) = 0$ for some polynomial, $p$, with coefficients in $K$ or it does not.

  • If there is such a $p$, $K(t)$ is an algebraic extension.
  • Otherwise, $K(t)$ is a transcendental extension.

The theorem hypothesizes that $K(t)$ is a transcendental extension. Therefore, there is no nonzero polynomial, $p$, with coefficients in $K$ such that $p(t) = 0$.

Eric Towers
  • 67,037
  • I think the question is about the field of rational functions in an indeterminate $t$ over the field $K$, not the field generated by $t$ in some given extension of $K$. I've raised a comment on the question asking the OP to clarify. – Rob Arthan Dec 14 '19 at 23:34