Why there is no nonzero ring homomorphism from $\mathbb{Q}(x, y)$ to $\mathbb{Q}(t)$? 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.
 A: 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 $(a,b)$.
A: 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 $\Bbb{Q}(\frac{c(t)}{d(t)})$.
Whence
$$\prod_j (S-\frac{a(g_j(t))}{b(g_j(t))})= P(\frac{c(t)}{d(t)} ,S) \in \Bbb{Q}(\frac{c(t)}{d(t)})[S], \qquad P(U,V) \in \Bbb{Q}(U,V)^*, \qquad P(\frac{c(t)}{d(t)},\frac{a(t)}{b(t)} )=0 $$
And $\phi(P(\phi(x),\phi(y))) = 0$ contradicting that $\phi(\frac1{P(\phi(x),\phi(y))})\phi(P(\phi(x),\phi(y)))=1$
