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I am not too familiar with infinites and I don't trust myself with them, I am not sure if any step I took supposed something that I shouldn't have. Below I am presenting an exercise in the book "Mathematics++" and my proof.

The statement

We recall that real numbers $\xi_1,\dots, \xi_n$ are algebraically independent (over the rational) if there is no nonzero polynomial $f\in\mathbb{Q}[x_1,\dots,x_n]$ with $f(\xi_1,\dots, \xi_n)=0$. Prove that for every $n$ there exist $n$ algebraically independent real numbers. Hint: one can use a cardinality argument or a measure argument, for example.

My proof

Suppose, for the sake of contradiction, that for a number $n$, all $n$-tuples $\xi\in\mathbb{R}^n$ are algebraically dependent.

Let $F:\mathbb{R}^n\to \mathbb{Q}[x_1,\dots,x_n]$ be a function which maps a tuple $\xi$ to a nonzero polynomial $f\in\mathbb{Q}[x_1,\dots,x_n]$ such that $f(\xi) = 0$ (for example we can choose the polynomial of smallest degree and of smallest lexilographic order of coefficients, but I don't think those detail need to be clarified).

Given $F$, and since $\mathbb{Q}[x_1,\dots,x_n]$ is a countable set, we can find a way to count $\mathbb{R}^n$. Indeed, let's enumerate the polynomials in $\mathbb{Q}[x_1,\dots,x_n]$ as $Q =\{q_1, q_2\dots \}$. Let $i$ iterate through $\{1, 2\dots\}$, for a polynomial $q_i$ consider the set $F^{-1}(q_i)=\{\xi\in\mathbb{R}^n\ |\ F(\xi) = q_i\}$. Since $q_i$ can only have a finite amount of roots, $F^{-1}(q_i)$ is finite and so is countable. That way for a specific $i$, we can enumerate the tuples $\xi\in F^{-1}(q_i)$. Since every $\xi\in\mathbb{R}^n$ is assiged to a polynomial $F(\xi)$, we will eventually count all of $\mathbb{R}^n$, which is a contradiction since $\mathbb{R}$ is uncountable.

Asaf Karagila
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Yuumita
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  • @ShyamalSayak The negative of "for all ..., there exists..." is "there exists ..., such that for all ...", so OP is correct here. Also, you should not (propose an) edit in a way that changes the meaning unless it's an actual typo/slip of mind. Commenting like you did to let the author know is the way to go, but be mindful to avoid going against the poster's intent when editing in the future. – Bruno B Jul 24 '23 at 18:14
  • I now notice that $F^{-1}(q_i)$ is not finite necessarily (I'm not used to multivariate polynomials...), however I believe it is still countable. Not sure how to prove it thought right now. – Yuumita Jul 24 '23 at 18:19
  • @Yuumita Can you show that $F$ is well-defined? –  Jul 24 '23 at 18:21
  • @ShyamalSayak Perhaps. However as I mentioned in a previous comment I make a bigger mistake. In fact a polynomial $q_i$ can have uncountably many roots. So my proof is not correct. – Yuumita Jul 24 '23 at 18:23
  • As long as you use the axiom of choice, you can define such an $F$ by assumption with no problem. – Bruno B Jul 24 '23 at 18:23
  • @Yuumita Do you think using the idea that $\mathbb{R}/\mathbb{Q}$ is not an algebraic extension would help? –  Jul 24 '23 at 18:33
  • @RobArthan Polynomials in multiple variables are different from the one variable case as almost all of them have uncountably many roots, see here: Polynomial with infinitely many zeros. – Bruno B Jul 24 '23 at 21:00
  • Bruno B: of course, mental aberration on my part. – Rob Arthan Jul 24 '23 at 21:06

1 Answers1

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Your idea works if you use induction. Once you have $\xi_1, \dots, \xi_n$ that are algebraically independent, observe that $\{x \mid \xi_1, \dots, \xi_n, x$ are not algebraically independent$\}$ is countable. That's because there are only countably many polynomials with rational coefficients, and each polynomial with a non-zero coefficient for at least one term containing $x$ corresponds to only finitely many choices of $x$ (since $\xi_1, \dots, \xi_n$ are fixed).

By the way, the axiom of choice isn't needed for this argument since the finitely many $x\text{'s}$ for each polynomial can be ordered linearly using the usual ordering on $\mathbb R.$

Mitchell Spector
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