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Let $R$ and $S$ be two polynomial rings $\mathbb{C}[x_1, \cdots, x_n]$ and $\mathbb{C}[y_1, \cdots, y_m]$ respectively and there is a ring homomorphism $f: R\to S$ such that

  1. $S/(f(x_1), \cdots, f(x_n))$ is a finite dimensional $\mathbb{C}$-vector space, and

  2. $S/(f(x_1), \cdots, f(x_n))$ is a complete intersection ring, i.e. $(f(x_1), \cdots, f(x_n))$ is a regular ideal or, equivalently, $(f(x_1), \cdots, f(x_n))$ equals $(u_1, \cdots, u_m)$ for some regular sequence $u_1, \cdots, u_m$.

Is it true that the ideal $\text{ker}(f)$ of $R$ can be generated by $n-m$ elements?

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  • $\begingroup$ Makes no sense. $Im f$ contains 1 and so if it is an ideal, it must be all of $S$. $\endgroup$
    – Mohan
    Sep 30 '15 at 14:07
  • $\begingroup$ I corrected my question. $\endgroup$
    – No_way
    Sep 30 '15 at 14:13
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Take the map $\mathbb{C}[x,y,z]\to\mathbb{C}[t]$ given by $x\mapsto t^3,y\mapsto t^4, z\mapsto t^5$. It satisfies your hypothesis, but it is well known that the kernel is a 3-generated ideal and not a two generated ideal.

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