I was reading the accepted answer for this question, and trying to see why there is no ring map(sending $1$ to $1$) from $\mathbb{Q}(x,y)$ to $\mathbb{Q}(t)$.

Here is my approach:

Suppose there exists a field homomorphism $\phi : \mathbb{Q}(x, y) \rightarrow \mathbb{Q}(t)$. I suspect that for given $a, b \in \mathbb{Q}(t)$ there exists a nonzero polynomial $h \in \mathbb{Q}[t_1, t_2]$ such that $h(a, b)=0$. If this is true, there is $0\neq h \in \mathbb{Q}[t_1, t_2]$ sastisfying $h(\phi(x), \phi(y))=0$. It follows that $0 \neq h(x, y) \in \ker \phi$, which is a contradiction. But I'm not able to verify the gap.

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    $\begingroup$ Write $a=\frac{a_1}{a_2}$, $b=\frac{b_1}{b_2}$ as a quotient of coprime polynomials. Define $h(X,Y)$ to be the resultant ( en.m.wikipedia.org/wiki/Resultant ) of the polynomials $a_1(t)-Xa_2(t)$ and $b_1(t)-Yb_2(t)$ where the polynomials are seen as elements of $A[t]$, $A=\mathbb{Q}[X,Y]$. Then, for all but finitely many $t \in \mathbb{Q}$, $h(a(t),b(t))=0$, so formally $h(a,b)=0$. $\endgroup$ – Mindlack Aug 25 '19 at 21:30
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    $\begingroup$ Actually, your idea is on the money. You work it out this way: supposing that $a(t)$ is non constant, you show that $\Bbb Q(t)$ is algebraic over $\Bbb Q(a)$. Then $b$ will be algebraic over $\Bbb Q(a)$ as well, so there’s a monic polynomial $G(X)\in\Bbb Q(a)[X]$ such that $G(b)=0$. Now clear of all the denominators among the coefficients of $G$, and you’ll get a polynomial $h(t_1,t_2)$, just as you were hoping to do. If you get bogged down in the details and don’t see your way to the end, I can write it up as an answer for you. $\endgroup$ – Lubin Aug 26 '19 at 4:08
  • $\begingroup$ @Lubin It seems nice to me. I'll accept it if you post your comment as an answer. Thank you. $\endgroup$ – Luxerhia Aug 27 '19 at 12:19

In a word, as I said in a comment, for there to be such a ring morphism there would have to be a transcendence-degree-two subfield of the field $\Bbb Q(t)$, which has transcendence degree one. Impossible, because the absolute transcendence degree of $\Bbb Q$ is zero. This argument fails when the base has infinite transcendence degree over $\Bbb Q$, like $\Bbb C$.

Be that as it may, giving a direct proof is not hard, though it can be tiresomely long (especially when a wordy geezer is at the keyboard). I’ll appreciate suggestions for shortening what appears below.

As you have observed, the ring morphism in question has to be one-to-one, because the domain is a field, so you take the images $a(t),b(t)$ of $x,y$ respectively in $\kappa=\Bbb Q(t)$ and hope to find a nonzero $\Bbb Q$-polynomial $F(X,Y)$ such that $F(a,b)=0$. Since $F(x,y)\ne0$, there’s your contradiction.

The proof is in two parts, the easy and then the harder (at least the longer). First part is, having chosen $a(t)$ as a starting point, to show that $t$ is algebraic over $\kappa=\Bbb Q(a)$. Second part is to take $b(t)$, now known to be algebraic over $\kappa$ (because everything in the big field is now algebraic over $\kappa$), and take its minimal $\kappa$-polynomial and convert this to a $\Bbb Q$-polynomial $F$ of the desired type.

So much for the program. Now to expand it to the tiresome totality.

Let $a(t)=g(t)/h(t)$ where $g$ and $h$ are $\Bbb Q$-polynomials. Now form $a\!\cdot\! h(T)-g(T)\in\Bbb Q(a)[T]$, which you see is a polynomial over $\kappa$ that vanishes at $T=t$, so that $t$ is algebraic over $\kappa$. (This is the appearance of transcendence-degree one in the argument.)

That was the quick and easy part. Now $b$ is also algebraic over $\kappa$, so that it satisfies a monic $\kappa$-polynomial $$ \Phi(Z)=Z^m+c_{m-1}Z^{m-1} +\cdots+c_1Z+c_0=Z^m+\frac{g_{m-1}}{h_{m-1}}Z^{m-1} +\cdots+\frac{g_1}{h_1}Z+\frac{g_0}{h_0}\,. $$ Here, the $g_i$ and the $h_i$ are in $\Bbb Q[a]$. When you multiply the displayed minimal polynomial for $b$ by the product of all the $h_i$, call it $H(a)$, you get the polynomial $$ H(a)\Phi(Z)=H(a)Z^m+\gamma_{m-1}(a)Z^{m-1}+\cdots+\gamma_1(a)Z+\gamma_0(a)\,, $$ where each $\gamma_i=g_i(a)\!\cdot\!\bigl(H(a)/h_i(a)\bigr)$, an element of $\Bbb Q[a]$. Make the substitution $Z\mapsto b(t)$ and get zero. This is your $\Bbb Q$-polynomial in two variables vanishing at $(a,b)$.


The resultant stuff can be obtained from the symmetric polynomials stuffs

$$\phi(x)= \frac{a(t)}{b(t)},\qquad \phi(y)= \frac{c(t)}{d(t)}, \qquad a(t),b(t),c(t),d(t)\in \Bbb{Q}[t]$$

$$ c(S) - d(S)\frac{c(t)}{d(t)} \in \Bbb{Q}(\frac{c(t)}{d(t)})[S], \qquad c(S) - d(S)\frac{c(t)}{d(t)} = \alpha(t) \prod_j (S-g_j(t))$$

The coefficients of the polynomial on the left are the elementary symmetric polynomials in the roots $g_j(t)$, from which we know that any symmetric polynomial in the $g_j(t)$ will be in $\Bbb{Q}(\frac{c(t)}{d(t)})$.


$$\prod_j (S-\frac{a(g_j(t))}{b(g_j(t))})= P(\frac{c(t)}{d(t)} ,S) \in \Bbb{Q}(\frac{c(t)}{d(t)})[S], \qquad P(U,V) \in \Bbb{Q}(U,V)^*, \qquad P(\frac{c(t)}{d(t)},\frac{a(t)}{b(t)} )=0 $$

And $\phi(P(\phi(x),\phi(y))) = 0$ contradicting that $\phi(\frac1{P(\phi(x),\phi(y))})\phi(P(\phi(x),\phi(y)))=1$

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    $\begingroup$ Am I missing something? Why didn’t you base your argument on transcendence degree? After all, $\Bbb Q(t)$ can’t contain a subfield of tr.deg. two. $\endgroup$ – Lubin Aug 26 '19 at 2:37

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