Thanks for updating the answer Ken! One more small nit: $\int |f(x)| dt \ge |\int f(x) dt |$ in general, so the $=$ on the 4th line should be a $\ge$.

Note that if $g$ is scaled by a function $h$, then $|\cdot|_g$ is scaled by $\sqrt{h}$. Therefore, $h$ should be defined as $df(\operatorname{grad} f) + 1$ rather than its square root. Also note that the use of Cauchy-Schwarz needs absolute values around $\langle \cdot, \cdot \rangle_g$ (so that $L_{\tilde g}$ is non-negative). (I tried to edit the answer to fix the mistakes above, but it got rejected.)

Thank you! I had assumed that "Intersection Theory" would be too modern to talk about conservation of number, but I should have looked it up.

Nice, that works! But a simpler choice might be $x(t) = 1$ and $y(t) = t^2$.

Thank you for taking a look at the problem, but I'm not sure if further discussion would be fruitful, especially since I already had my original question answered.

I've verified to my satisfaction that $(x - x_0, y)^2 = (x - x_0)$ in $A$, simply by expressing the generators in one side in terms of the other; the proof in the paper is along the same lines. The proof doesn't rely on $k$ being algebraically closed. So I disagree with your claim, and besides you haven't addressed the issue I pointed out in your earlier claimed disproof. You seem to be getting hung up on the geometric language used in the first paragraph. If you'd like, you can ignore it, and treat the question purely in ring-theoretic terms.

We may be talking past each other, but my issue is with the claim that "If $y_0=0,f(x_0)=0$ it follows $IJ≠N$". I agree with the general point that working with non-algebraically-closed needs special tools, or passing to the algebraic closure.

Also note that Lemma 10 of the linked paper provides a formally-checked proof using straightforward ideal properties of a generalization of the statement which doesn't rely on $y_0$ being non-zero, so unless I missed something in my simplification (which is not unlikely), or this paper is wrong, I'm a bit skeptical that the statement is false.

Hmm, I'm not sure I follow the $y_0 = 0$ case. I agree that $A/N$ has nilpotent elements, but since $y_0 = 0$, $I = J$, and so $I$ and $J$ are not comaximal anymore, and thus we can't conclude that $A/IJ = A/I^2$ is congruent to $k \oplus k$, and thus doesn't have nilpotent elements.

(Also, I guess that explains my earlier confusion with what Lucellia was saying.)