I have a nitpicky question about notation in Hartshorne's proof of his Theorem 3.4 in his AG book; in particular, it is in his proof of part (b).
The point in contention is the line
One checks easily that $\phi_i^*(\mathfrak m'_P)=\mathfrak m_P\cdot S(Y)_{(x_i)}$.
I already made my way through some confusion regarding $S(Y)_{(x_i)}$ v.s. $S(Y)_{((x_i))}$ thanks to this post. My question here is different.
In short, I am getting stuck at the notation $\mathfrak m_P\cdot S(Y)_{(x_i)}$. At first, I thought the multiplication was taking place in the larger ring $S(Y)_{x_i}$, since, by abuse of parenthesis, I thought we had something like
$$(\mathfrak m_P)\cdot (S(Y)_{(x_i)}).$$ That is to say, since the homogeneous generators of $\mathfrak m_P$ have strictly positive degree, the product $\mathfrak m_P\cdot S(Y)_{(x_i)}$ does not live in $S(Y)_{(x_i)}$. I now believe this interpretation of the notation is wrong, even though it is the first interpretation that suggested itself to me.
I now believe the correct interpretation of the notation $\mathfrak m_P\cdot S(Y)_{(x_i)}$ is `the degree zero part of the localization of $\mathfrak m_P$ with respect to the homogenous element $x_i$, sitting in the ring $S(Y)_{(x_i)}$.' That is to say, $$\mathfrak m_P\cdot S(Y)_{(x_i)} \subset S(Y)_{(x_i)}.$$ This second interpretation agrees with my understanding of the image of $\mathfrak m'_P$ under the map $\phi_i^*$, which seems to be composed exclusively of (degree zero) elements in the ring $S(Y)_{(x_i)}$, not the larger ring $S(Y)_{x_i}$.
If the second interpretation is correct, it would be clearer to me if
$$\mathfrak m_P\cdot S(Y)_{(x_i)}$$
were instead written as
$$(\mathfrak m_P)_{(x_i)}$$
or something.
For a ring homomorphism $\varphi : R \to S$ and an $R$-ideal $I$, it is common to write $I \cdot S$ or $IS$ or $I^e$ for the extension of $I$ to $S$, i.e. $\varphi(I)S$. This is the ideal in $S$ generated by the elements which are the images of elements of $I$.
However, Hartshorne is slightly abusing notation in this case. There is no natural ring map from $S(Y)$ to $S(Y)_{(x_i)}$; instead $\mathfrak{m}_P \cdot S(Y)_{(x_i)} = (\mathfrak{m}_P \cdot S(Y)_{x_i}) \cap S(Y)_{(x_i)}$ means the extension of $\mathfrak{m}_P$ to the localization $S(Y)_{x_i}$, then intersected with the subring of degree $0$ elements in $S(Y)_{x_i}$. So your second interpretation is correct.