$\pi$ in $\mathbb{R}$ is transcendental over $\mathbb{Q}(\sqrt2)$

Question

We have to show The element $\pi$ in $\mathbb{R}$ is transcendental over $\mathbb{Q}(\sqrt2)$.

That is basically we have show that $\pi$ is not root of any non null polynomial in $\mathbb{Q}(\sqrt2)[x]$.

I want to prove this from the definition that is Let $K$ be a field extension of the field $F$. Then an element $\alpha$ of $K$ is said to be transcendental over $F$ iff $\alpha$ is not a root of any non-null polynomial in $F[x]$.

How to show this? Any help is appreciated. Thanks.

Answer 1

If $\pi$ were algebraic over $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{2},\pi)$ would be a finite extension of $\mathbb{Q}$. Note that every subfield of a finite field extension is finite, so $\mathbb{Q}(\pi)$ would be a finite extension, so $\pi$ is algebraic over the rationals, which is not true.

Answer 2

First, we have to assume you already know that $\pi$ is transcendental over $\Bbb Q$, because that's a difficult proof that can't be done using purely algebraic methods (because $\pi$ can't be defined using purely algebraic methods). Under those assumptions, you can prove this directly from the definition, as requested.

Suppose there were some polynomial $p(x) = \sum a_i x^i$ with $a_i \in \Bbb Q [ \sqrt 2]$ such that $p(\pi)=0$. Then for each $i, a_i = b_i+c_i \sqrt 2$, with $b_i, c_i \in \Bbb Q$. We then have $p(\pi)=\sum b_i \pi^i + \sqrt 2 \sum c_i \pi^i =0$, so $\sum b_i \pi^i = - \sqrt 2 \sum c_i \pi^i$. Squaring both sides of this equation gives you a non-zero polynomial in rational coefficients with $\pi$ as a root, which we know can't happen because $\pi$ is transcendental over $\Bbb Q$.

To see that the polynomial obtained by squaring both sides is non-zero, consider the leading coefficient $a_n$. If either $b_n=0$ or $c_n=0$, then the coefficient (in the square) associated with $\pi^{2n}$ must be nonzero because no cancellation is possible.

On the other hand, if $b_n \neq 0 \neq c_n$, then the coefficient (in the squared polynomial) associated with $\pi^{2n}$ is $b_n^2-2c_n^2$, and since $b_n, c_n \in \Bbb Q, b_n^2-2c_n^2=0 \Rightarrow 2= \left ( \frac {b_n}{c_n} \right )^2 \Rightarrow \sqrt 2 = \frac{b_n}{c_n} \in \Bbb Q$, which is a contradiction. Thus, in either case, the leading coefficient of our squared polynomial cannot be $0$ and the squared polynomial is (as required) not the zero polynomial.