0
$\begingroup$

Can somebody explain to me ( or give a proof ) why the field extension $\mathbb{R/Q}$, that is the field of real numbers as an extension of the field of rational numbers, is transcendental and not algebraic, which would mean that each element of $\mathbb R$ would be a root of some polynomial with rational coefficients only if it is transcendental?

$\endgroup$
  • 2
    $\begingroup$ Actually more precisely $\mathbb R$ has both algebraic and transcendental elements over $\mathbb Q$. $\sqrt{2}$ is algebraic over $\Bbb Q$, it's the zero of $x^2-2$. But $\pi$ is not algebraic because it is not the zero of any polynomial over $\Bbb Q$ (that's not obvious though). But most of what is in $\Bbb R$ is transcendental, because it has a higher cardinality than $\Bbb Q$. $\endgroup$ – Gregory Grant May 31 '15 at 19:40
  • $\begingroup$ @GregoryGrant Related to the "not obvious" note: showing that any given number is transcendental (or even rational) is often very difficult. $\endgroup$ – Arthur May 31 '15 at 19:44
  • $\begingroup$ To expand slightly on @GregoryGrant’s comment, an extension of $\Bbb Q$ is algebraic if all elements are algebraic over $\Bbb Q$. If that isn’t so, then the extension is transcendental. In other words, all you need for an extension to be transcendental is for one element of it to be nonalgebraic over $\Bbb Q$. $\endgroup$ – Lubin May 31 '15 at 19:51
  • $\begingroup$ The extent to which an extension is transcendental is conveniently measured by the transcendence degree. Lang's Algebra deals with this in some detail (but abstractly). The existence of a single transcendental element makes the transcendence degree at least one - choose your favourite. But the extension you are asking about is way more transcendental than that. $\endgroup$ – Mark Bennet May 31 '15 at 20:01
  • $\begingroup$ @GregoryGrant A finite extension of a finite field is algebraic, though the extension field has greater cardinality than the ground field. Your comment is valid for infinite fields though. $\endgroup$ – Mark Bennet May 31 '15 at 20:03
5
$\begingroup$

You are confusing the quantifiers when you negated the statement.

$K$ is an algebraic extension of $F$ if every element of $K$ is the root of a polynomial with coefficients in $F$. So $\forall k\exists p\in F[x]: p(k)=0$.

$K$ is a transcendental extension of $F$ if it is not algebraic. So $\lnot(\forall k\exists p\in F[x]:p(k)=0)$.

Let's parse this negation into the statement, we have $\exists k\forall p\in F[x]:p(k)\neq 0$. So $K$ is transcendental over $F$ if there is at least one element in $K$ which is not the root of any polynomial with coefficients in $F$.

Now a simple counting argument shows that $\Bbb Q[x]$ is a countable set, and every polynomial has finitely many roots. So there are only countably many algebraic numbers in any given extension of $\Bbb Q$. Since $\Bbb R$ is uncountable it means that "most" of its elements are indeed transcendental.

$\endgroup$
2
$\begingroup$

Any algebraic extension of $\mathbb{Q}$ is denumerable, since its elements are roots of polynomials with coefficients in $\mathbb{Q}$, but $\mathbb{R}$ is not denumerable so it contains elements that are not algebraic and this elements are said transcendental.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.