I've got a couple rather geometric questions about the following setup.

Let $X$ be a scheme over an algebraically closed field ($\mathbb{C}$, say) with the action of a torus $T$, such that the action has finitely many fixed points. $X$ may have some mild singularities but is normal, Cohen-Macaulay, etc. Let $\mathscr{L}$ be a $T$-equivariant line bundle on $X$.

We can form the T-fixed subscheme $X^T$. In general, the $\mathbb{C}$ points of $X^T$ will be some discrete set, but $X^T$ need not be reduced, so we should think of $X^T$ as some nonreduced scheme supported at some finite set of points.

Furthermore, let me consider a rational curve $Y \hookrightarrow X$, stable under the $T$ action, joining two $T$-fixed points $p_1$ and $p_2$ in $X$. If I assume that $Z$, the component of $X^T$ supported at $p_2$, is nonreduced, I could consider the scheme $Y'=Y \cup Z$ (if you press me for more precision I think I should make a fiber product $Y'=Y \times_{p_2} Z$).


Question 1: In the above setup, am I justified in viewing this scheme $Y'$ as $Y$ along with some embedded points (corresponding to the non-reduced points over $p_2$)?

Question 2: Let $I$ be the ideal sheaf of $X^T$ in $X$. Assume that I know how $\mathscr{L}$ restricts to $Y$; $\mathscr{L}|_Y=\mathscr{O}_{\mathbb{P}^1}(n)$, say.

Now consider the restriction $(I \otimes \mathscr{L})|_{Y}$. The sections of this sheaf over $Y$ will be some subset of $\Gamma(Y,\mathscr{O}_Y(n))$, vanishing at the $T$-fixed points $p_1$ and $p_2$. Note moreover that $I \otimes \mathscr{L}|_Z=0$, since $I$ vanishes identically on the $T$-fixed locus. My question is, do I gain information about orders of vanishing of sections $s \in \Gamma(Y, (I \otimes \mathscr{L})|_Y)$ at $p_2$?

In other words, if a section vanishes at an embedded point of the curve, can I see that reflected in higher orders of vanishing of that section restricted to the closed points of the curve?

I'm happy to expand or clarify any bit of this (it's rather sketchy) and would be nearly as happy with a good reference as a direct answer.