# Field extension $\mathbb{R/Q}$ is transcedental?

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?

• 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 $\Bbb Q$. – Gregory Grant May 31 '15 at 19:40
• @GregoryGrant Related to the "not obvious" note: showing that any given number is transcendental (or even rational) is often very difficult. – Arthur May 31 '15 at 19:44
• 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 $\Bbb Q$. – Lubin May 31 '15 at 19:51
• 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. – Mark Bennet May 31 '15 at 20:01
• @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. – Mark Bennet May 31 '15 at 20:03

$K$ 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.
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.