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I’ve been reviewing algebraic geometry by reading through A. Gathmann’s notes. Somehow I don’t understand few steps included in the following proof. I appreciate any help :) \

  1. How does $V(f_a)=V(h_a)$ imply that we can assume these two functions are the same?\
  2. Why are these two functions $$g_af_b=g_bf_a$$ both zero as he claims in the paragraph below the equation (*)?
    As for the first question, I understand that if their difference is in the ideal $I(X)$, then they can be used interchangeably. However, is it the case here? We do know that their vanishing sets are the same, but we are not so sure about how they behave on the distinguished set.
  • $\begingroup$ In fact, we can have $V (f) = V (h)$ while $f \ne h$. The easiest example is $h = -f$. But what is true is that inverting $f$ is the same as inverting $h$ in this case. $\endgroup$
    – Zhen Lin
    Mar 13, 2023 at 6:26
  • $\begingroup$ @ZhenLin Thank you for your answer! If you don’t mind, could you please elaborate your answer a bit? Here the implication is that, the same zero locus of two functions implies that they are the same. This confuses me the most. $\endgroup$
    – Mizutsuki
    Mar 14, 2023 at 7:20
  • $\begingroup$ It doesn't. The phrasing is bad. What the author means is that you can change your choice of $h_a$. $\endgroup$
    – Zhen Lin
    Mar 14, 2023 at 10:08
  • $\begingroup$ @ZhenLin I still do not see why this is the case here.. so we can invert two functions if their zero locuses are the same? $\endgroup$
    – Mizutsuki
    Mar 17, 2023 at 4:16
  • $\begingroup$ It isn’t obvious but it is true. It is a consequence of the Nullstellensatz. $\endgroup$
    – Zhen Lin
    Mar 17, 2023 at 5:55


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