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I'm reading a book about Algebraic Geometry and I'm wondering this problem about homomorphism theorem.

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?

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  • $\begingroup$ Do you know the division algorithm? $\endgroup$ Commented Apr 4, 2022 at 3:06
  • $\begingroup$ 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. $\endgroup$
    – uiui
    Commented Apr 4, 2022 at 3:15
  • $\begingroup$ So it looks like you kind of know it. Please go look up the polynomial division algorithm, and pay attention to the whole thing. $\endgroup$ Commented Apr 4, 2022 at 3:20
  • $\begingroup$ 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. $ $\endgroup$
    – uiui
    Commented Apr 4, 2022 at 4:11
  • $\begingroup$ @uiui You don't need the division algorithm for multivariable polynomials. $\endgroup$
    – user26857
    Commented Apr 4, 2022 at 5:39

1 Answer 1

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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 $g\in I$.

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    $\begingroup$ Certainly, I didn't need the division algorithm for multivariable polynomials. Thank you for your clear answer. $\endgroup$
    – uiui
    Commented Apr 4, 2022 at 10:17

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