# Show that kernel of homomorphism is equal to an ideal defining rational space curve

Let ideal $$I = (X_2-f_2(X_1),...,X_n-f_n(X_1))\subset \mathbb{C}[X_1,...,X_n]$$, $$f_2,...,f_n \in \mathbb{C}[x_1]$$ and homomorphism $$\phi : \mathbb{C}[X_1,...,X_n] \longrightarrow \mathbb{C}[X_1]$$ s.t. $$X_1 \mapsto X_1,X_i \mapsto f_i(X_1)$$, $$i>1$$. I want to show $$\ker\phi = I$$.

Although not explicitly stated in that book, I think it is also assumed that $$\phi(a)= a \quad (a\in \mathbb{C})$$.

I proved easily $$I \subseteq \ker\phi$$ using definition of $$\phi$$. But I couldn't show the other side.

I even tried to represent $$g \in \ker\phi$$ in concrete terms, but I could not show that it is in $$I$$.

Can anyone help me solve this?

• Do you know the division algorithm? Commented Apr 4, 2022 at 3:06
• So, $g = P*h + R$ ($g \in Ker\phi, h\in I)$ and $\phi(g) =0$ then $\phi(R) =0$, therefore $g \in I$...?I am embarrassed now, but I think I can show by this.
– uiui
Commented Apr 4, 2022 at 3:15
• So it looks like you kind of know it. Please go look up the polynomial division algorithm, and pay attention to the whole thing. Commented Apr 4, 2022 at 3:20
• On second thought, I didn't know about division algorithm about multivariable polynomial. I looked up and I wrote proof: by division, $g = q_2G_2+,...,+q_nG_n + r\quad$ $(G_i = (X_i - f_i(X_1))$ then $r$ has no $X_2,...,X_n$. $\phi(g) = 0$ so $\phi(r(X_1)) = 0$, by def of $\phi, r = 0.$
– uiui
Commented Apr 4, 2022 at 4:11
• @uiui You don't need the division algorithm for multivariable polynomials. Commented Apr 4, 2022 at 5:39

Let $$g\in k[x_1,\dots,x_n]$$ such that $$g(x_1,f_2(x_1),\dots,f_n(x_1))=0$$. Write $$g(x_1,\dots,x_n)=(x_2-f_2(x_1))q_2(x_1,\dots,x_n)+r_2(x_1,x_3,\dots,x_n).$$ (Here we considered $$g$$ as a polynomial in $$x_2$$ with coefficients in $$k[x_1,x_3,\dots,x_n]$$.) Since $$g(x_1,f_2(x_1),\dots,f_n(x_1))=0$$ we get $$r_2(x_1,f_3(x_1),\dots,f_n(x_1))=0$$. Continuing this way we get $$g=(x_2-f_2(x_1))q_2+\dots+(x_n-f_n(x_1))q_n+r_n(x_1)$$ and using once again that $$g(x_1,f_2(x_1),\dots,f_n(x_1))=0$$ we obtain $$r_n(x_1)=0$$, and this shows that $$g\in I$$.