Ring homomorphism between fields of rational functions in many indeterminates

We let $$\mathbb{F}=\mathbb{Q}(x_1,x_2,x_3)$$ and $$\mathbb{K}=\mathbb{Q}(u_1,u_2)$$ be the field of rational functions in indeterminates $$x_1,x_2,x_3$$ and $$u_1,u_2$$, respectively. We define a map:

$$\pi: \begin{cases} \mathbb{F} \rightarrow \mathbb{K} \\ x_1,x_3 \mapsto u_1 \\ x_2 \mapsto u_2 \end{cases}$$

We want to show that $$\pi$$ is a ring homomorphism. Let $$f,g \in \mathbb{F}$$ be expressed as $$f=\frac{p_1(x_1,x_2,x_3)}{q_1(x_1,x_2,x_3)}$$ and $$g=\frac{p_2(x_1,x_2,x_3)}{q_2(x_1,x_2,x_3)}$$ ($$p_1,q_1,p_2,q_2$$ are over $$\mathbb{Z}$$). Then $$\pi(f)=\frac{p^{'}_1(u_1,u_2)}{q^{'}_1(u_1,u_2)}\in \mathbb{K}$$ and $$\pi(g)=\frac{p^{'}_2(u_1,u_2)}{q^{'}_2(u_1,u_2)}\in \mathbb{K}$$. We will often write $$p$$ for a polynomial $$p(x_1,\dots,x_n)$$. We then have that: $$$$\pi(f+g)=\pi(\frac{p_1q_2 + p_2q_1}{q_1q_2})=\frac{p_1^{'}q_2^{'}+p_2^{'}q_1^{'}}{q_1^{'}q_2^{'}}=\frac{p_1^{'}}{q_1^{'}} + \frac{p_2^{'}}{q_2^{'}} = \pi(f)+\pi(g),$$$$ and that $$$$\pi(fg) = \frac{p_1^{'}}{q_1^{'}}\frac{p_2^{'}}{q_2^{'}}=\pi(f)\pi(g).$$$$ Moreover, $$\pi(1_\mathbb{F})=\pi(\frac{p}{p})=\frac{p^{'}}{p^{'}}=1_{\mathbb{K}}$$ and so $$\pi$$ is a ring homomorphism, as required. Is this correct?

• What is $\pi\left(\frac{1}{x_1-x_3}\right)$? – Arthur Feb 25 '20 at 12:21
• Ohh, so it is not even well-defined... – billy192 Feb 25 '20 at 12:27
• You can only get ring homomorphisms from $\Bbb{Q}[x_1,x_2,x_3]$. A ring homomorphism from a field is always injective. Looking at the transcendence degrees reveals right away that $\Bbb{K}$ has no subfields isomorphic to $\Bbb{F}$. Hence there are no ring homomorphisms in this direction. – Jyrki Lahtonen Feb 25 '20 at 12:33

Given a field $$K$$ and a ring $$R$$, it known that any well-defined homomorphism $$f:K\to R$$ is either trivial (if you allow that kind of homomorphism), or injective. Indeed, if $$f(a) = 0$$ for some $$a\neq 0$$, then $$f(1) = f\left(a\cdot \frac1a\right) = f(a)\cdot f\left(\frac1a\right) = 0$$In this case, however, you have $$\pi(x_1-x_3) = 0$$, yet $$\pi$$ is not trivial. So it cannot be a homomorphism.