Regularity of $H \cap X$ in Bertini's Theorem (Hartshorne CH. 2, Theorem 8.18)

Question

In Hartshorne's proof of Bertini's theorem (CH. 2, Theorem 8.18), I am having difficulty in the first part of the proof which tells us that the scheme $H \cap X$ is regular at every point.

I could follow till the statement that $x \in H \cap X \iff \varphi_x(f) \in \mathfrak{m}_x$.

(Q-1) I just want to confirm here that there is an abuse of notation here and the statement means $x \in H \cap X \iff \varphi_x(f) \in \mathfrak{m}_x/ \mathfrak{m}_x^2 = \overline{\mathfrak{m}_x} $.

I am not able to make any progress on the next statement that says "$x$ is non-regular on $H \cap X \iff \varphi_x(f) \in \mathfrak{m}_x^2$ because in that case the local ring $\mathcal{O}_x/ (\varphi(f))$ will not be regular."

(Q-2) I hope that RHS is 0 i.e. $\varphi_x(f) \in \mathfrak{m}_x^2$ is same as saying that $\varphi_x(f)=0$.

(Q-3) What will be the local ring of $H \cap X$ at $x$? Why does Hartshorne says that it will be $\mathcal{O}_x/(\varphi_x(f))$? (hoping that there is a typo: Intended $\varphi_x(f)$ but printed $\varphi(f)$)

My answer: It should be described as $\mathcal{O}_{X,x} / (\overline{f})$, where $\overline{f}$ is the image of $f$ in $\mathcal{O}_{X,x}$.

(Q-4) Prove that: $x$ is non-regular on $H \cap X \Rightarrow \varphi_x(f) \in \mathfrak{m}_x^2$ (or 0 based on my Q-2)

My attempt: Let us denote the local ring $A := \mathcal{O}_{X,x} / (\varphi_x(f))$ along with its maximal ideal $\mathfrak{m}$, I am not able to related how the fact that $\dim \mathfrak{m}/\mathfrak{m}^2 > \dim A $ leads to RHS?

(Q-5) Converse of Q-4?

Answer 1

It seems like you're getting really hung up on the notation while Hartshorne is being pretty free about identifying things with their image under quotient maps and the like. I think you're sophisticated enough to figure out what it should be if you insist on being really exacting like this - you're completely correct in your post when you're making statements about what the notation should be. In detail:

1. Correct, $x\in H\cap X$ iff $\varphi_x(f)\in\mathfrak{m}_x/\mathfrak{m}_x^2 \subset \mathcal{O}_{X,x}/\mathfrak{m}_x^2$.
2. Yes, that's correct, we are taking the quotient by $\mathfrak{m}_x^2$.
3. The local ring of $H\cap X$ at $x$ is the quotient of $\mathcal{O}_{X,x}$ by any local equation for $H$. Importantly, $f$ does not immediately restrict to an element of $\mathcal{O}_{X,x}$ without doing something to it, because it's homogeneous and therefore not a global regular function on $\Bbb P^n$ - instead, it's a section of $\mathcal{O}_{\Bbb P^n}(1)$. In order to get a local equation for $H$ in $\mathcal{O}_{X,x}$, you have to dehomogenize, which can be done by taking $f/f_0$ for any $f_0$ with $x\notin V(f_0)$, and it's immediate that this is independent of the choice of $f_0$: if $f_0,f_1$ are two possible choices, then $\frac{f_0}{f_1}$ is a unit in $\mathcal{O}_{X,x}$, so the ideals $(\frac{f}{f_0})$ and $(\frac{f}{f_1})$ are the same. (Yes, there's a typo, this should be $\mathcal{O}_{X,x}/(\varphi_x(f))$ - also, if you're being strict about notation, you'd need to include some sort of different notation here, since $\varphi_x$ is a map $V\to\mathcal{O}_{X,x}/\mathfrak{m}_x^2$ and therefore doesn't land in $\mathcal{O}_{X,x}$.)

Parts 4 & 5 are basically separate concerns.

4. (And 5) Use the Jacobian criteria: the Jacobian of $x\in X\cap H$ is the Jacobian of $x$ in $X$ with an extra row consisting of the partial derivatives of $\varphi_x(f)$. In order for $X\cap H$ to be 1 dimension smaller and $x\in X\cap H$ to be a regular point, we need the row we added to increase the rank by one. This happens iff $\varphi_x(f)$ is in $\mathfrak{m}_x$ but not $\mathfrak{m}_x^2$.