Show that the field \[ \mathbb Q(t_1,\ldots,t_n) \] of rational functions in \(n\) variables over \(\mathbb Q\) is isomorphic to a subfield of \(\mathbb R\).
Solution
1. If \(K\subseteq\RR\) is countable, then the set of real numbersalgebraic over \(K\) is countable.
The polynomial ring \(K[x]\) is countable, since each polynomial is a finite sequence of elements of the countable set \(K\). Every nonzero polynomial has only finitely many real roots. Hence the set of real numbers algebraic over \(K\) is a countable union of finite sets and is therefore countable.
2. There exist \(\alpha_1,\ldots,\alpha_n\in\RR\) that are algebraicallyindependent over \(\QQ\).
Choose the numbers inductively. Since \(\QQ\) is countable, step 1 shows that only countably many real numbers are algebraic over \(\QQ\). As \(\RR\) is uncountable, choose \(\alpha_1\) transcendental over \(\QQ\).
Suppose \(\alpha_1,\ldots,\alpha_k\) have been chosen algebraically independent over \(\QQ\). The field \[ K_k=\QQ(\alpha_1,\ldots,\alpha_k) \] is countable: it consists of quotients of values of polynomials with rational coefficients, and there are only countably many such pairs of polynomials. By step 1, only countably many reals are algebraic over \(K_k\). Choose \(\alpha_{k+1}\in\RR\) outside that set. Then \(\alpha_{k+1}\) is transcendental over \(K_k\), so \(\alpha_1,\ldots,\alpha_{k+1}\) are algebraically independent over \(\QQ\). Continuing finitely many times gives the required \(n\) numbers.
3. Evaluation at \((\alpha_1,\ldots,\alpha_n)\) defines an injective ringhomomorphism\[ \varphi:\QQ[t_1,\ldots,t_n]\longrightarrow\RR. \]
Define \[ \varphi(f)=f(\alpha_1,\ldots,\alpha_n). \] This is a ring homomorphism. By the algebraic independence obtained in step 2, no nonzero polynomial in \(\QQ[t_1,\ldots,t_n]\) vanishes at \((\alpha_1,\ldots,\alpha_n)\). Thus \(\ker\varphi=0\), so \(\varphi\) is injective.
4. The map in step 3 extends to an injective field homomorphism\[ \Phi:\QQ(t_1,\ldots,t_n)\longrightarrow\RR. \]
For \(f,g\in\QQ[t_1,\ldots,t_n]\) with \(g\neq0\), set \[ \Phi\left(\frac fg\right) = \frac{\varphi(f)}{\varphi(g)}. \] Step 3 gives \(\varphi(g)\neq0\), so this quotient is defined. The universal property of the fraction field shows that \(\Phi\) is a field homomorphism extending \(\varphi\). Since a field homomorphism sending \(1\) to \(1\) has zero kernel, \(\Phi\) is injective.
5. The field \(\QQ(t_1,\ldots,t_n)\) is isomorphic to a subfield of\(\RR\).
By step 4, \(\Phi\) is an injective field homomorphism. Therefore it identifies \(\QQ(t_1,\ldots,t_n)\) with the subfield \(\Phi(\QQ(t_1,\ldots,t_n))\) of \(\RR\).
6. Q.E.D.
Step 5 is the required conclusion.