Explicit calculation of the twisted sheaf in a blow-up Let $X$ be the nodal rational curve, let $Y:= \mathbb{P}^1 \overset{\nu}{\to} X$ be the normaliztion map. If we let $\omega_X$ be the sheaf of regular differentials on $X$ then we have the obvious inclusion
$$ \nu_*\omega_{Y} \hookrightarrow \omega_X. $$
Viewing $\nu_*\omega_{Y} \otimes \omega_X^\vee$ as an ideal sheaf $I$ of $\mathcal{O}_X$ we can construct the blowup $\pi: \tilde X \to X$ corresponding to $I$.
The fiber over the node is a newly formed $\mathbb{P}^1$, call it $E$. The blowup construction gives us a canonical line bundle on $\tilde X$ denoted by $\mathcal{O}_{\tilde X}(1)$. My question is

What is the degree of $\mathcal{O}_{\tilde X}(1)|_E$?

 A: It is in Vakil's notes that you can not only calculate the fiber of a blowup by restricting the graded algebra to the point, but you can also calculate the pullback of $\mathcal{O}(1)$ to that fiber with the same construction.
Therefore, we need only focus our attention on the node $n \in X$. Pulling back the ideal sheaf $I$ to the node, we get a 2-dimensional vector space. Over a point, there are no mysteries: The Proj construction gives the usual $\mathbb{P}^1$ together with its degree 1 line bundle.
PS: I was expecting a more interesting answer because of some contextual reasons, which is why I asked. Sorry for this rather dull solution.
PPS: Something more interesting happens on $Y$ whose map to $X$ factors through $\tilde X$. There is a canonical map $\mathcal{O}(1) \to \mathcal{O}$ on $\tilde X$, which by the previous argument has to be zero on the exceptional component $E$. However, it is clearly an isomorphism away from $E$. It follows that $\mathcal{O}(1)|_{Y} \simeq \mathcal{O}_Y(-p-q)$ if $p,q$ are the preimages of the node.
