# 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.

• 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 $h(a,b)=0$. – Mindlack Aug 25 '19 at 21:30
• 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. – Lubin Aug 26 '19 at 4:08
• @Lubin It seems nice to me. I'll accept it if you post your comment as an answer. Thank you. – 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 $$\Bbb{Q}(\frac{c(t)}{d(t)})$$.

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 $$\phi(\frac1{P(\phi(x),\phi(y))})\phi(P(\phi(x),\phi(y)))=1$$

• 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. – Lubin Aug 26 '19 at 2:37