Resolving indeterminacy of map on projective spaces induced by a linear map

Question

Let $V$ and $W$ be two $k$-vector spaces, where $k$ can be assumed as $\mathbb{C}$, and $f:V\longrightarrow W$ a non null $k$-linear map with kernel $K$. The map $f$ naturally induces a rational morphism $F:\mathbb{P}(V)\dashrightarrow\mathbb{P}(W)$, where $F$ is defined in the non-empty open set $\mathbb{P}(V)\backslash\mathbb{P}(K)$. Then via blowing up we resolve the map $F$, that is. Let $\pi: Bl_{\mathbb{P}(K)}(\mathbb{P}(V))\longrightarrow\mathbb{P}(V)$ be the blow up of $\mathbb{P}(V)$ along $\mathbb{P}(K)$, and $F':Bl_{\mathbb{P}(K)}(\mathbb{P}(V))\longrightarrow\mathbb{P}(W)$ be the map which resolves $F$ or more precisely $F = F'\circ\pi$.

I would like to know two things and any help by indicating books, solution any hint would let me happy.

1 - Is it simple to know what is the blow up in this specific case?

2- What is the pull back $(F')^*\mathcal{O}_{\mathbb{P}(W)}(1)$?

Thank you so much for any help!!

Answer 1

Let me assume $f$ is surjective (otherwise, replace $W$ by $Im(f)$). Then $$ Bl_{\mathbb{P}(K)}(\mathbb{P}(V)) = \mathbb{P}_{\mathbb{P}(W)}(K \otimes \mathcal{O} \oplus \mathcal{O}(-1)). $$ Next, if $H_V$ and $H_W$ are the pullbacks of the hyperplane classes of $\mathbb{P}(V)$ and $\mathbb{P}(W)$, respectively, and $E$ is the exceptional divisor of the blowup, then $$ H_W = H_V - E. $$

Answer 2

Here I use the tip ginven by @KReiser as an attempt of answer.

Let $p:\mathbb{P}^r\dashrightarrow\mathbb{P}^{r-k-1}$ with $0\le k<r$ be the projection $(a_0:\cdots:a_r)\longmapsto (a_{r-k}:\cdots:a_r)$. Such a rational map has base $K:=V(X_0,\cdots,X_k)\subsetneq\mathbb{P}^r$, that is, the domain of $p$ is $\mathbb{P}^r\backslash K$. The blow-up, $X:=Bl_{K}(\mathbb{P}^r)$, of $\mathbb{P}^r$ along of $K$ is \begin{eqnarray} X &\overset{\pi}{\longrightarrow}& \mathbb{P}^r\\ ((a_0:\cdots:a_r),(b_{k+1}:\cdots:b_r)) &\longmapsto& (a_0:\cdots:a_r), \end{eqnarray} where $X=V(X_iY_j-X_jY_i\mid k+1\leq i,j,\le r)\subset \mathbb{P}^r\times\mathbb{P}^{r-k-1}$, here we have given homogeneous coordinates $X_i$ to $\mathbb{P}^r$ and $Y_i$ to $\mathbb{P}^{r-k-1}$. The morphim \begin{eqnarray} X &\overset{P}{\longrightarrow}& \mathbb{P}^{r-k-1}\\ ((a_0:\cdots:a_r),(b_{k+1}:\cdots:b_r)) &\longmapsto& (b_{k+1}:\cdots:b_r), \end{eqnarray} resolves $p$, i.e, $p\circ\pi=P$. To see this, consider the open sets $$ U_i=\{Y_i\neq0,X_i\neq0\}\subsetneq X \mbox{ for } i=k+1,\cdots,r $$ In $U_i$ we have $ X_j=X_iY_j$ with $j=k+1,\cdots,r$. Then $p\circ\pi|_{U_i}$ is given by \begin{eqnarray} p\circ\pi((x_0:\cdots:x_k:x_iy_{k+1}:\cdots:x_iy_r),(y_{k+1}:\cdots:y_r)) &=& p((x_0:\cdots:x_k:x_iy_{k+1}:\cdots:x_iy_r)) \\ &=&(x_iy_{k+1}:\cdots:x_iy_r)\\ &=&(y_{k+1}:\cdots:y_r). \end{eqnarray} Which shows that $P|_{U_i}=p\circ\pi|_{U_i}$ for all $i=k+1,\cdots,r$. From this last expression we see that actually $P|_{U'_i}=p\circ\pi|_{U'_i}$ where $U_i\subsetneq U'_i:=\{Y_i\neq0\}$. But the $U'_i$ is a cover of $X$, which shows that $P=p\circ\pi$.

The exceptional divisor is given by $E=\pi^{-1}(K)=V(X_{k+1},\cdots,X_r)\subsetneq X$. We have a canonical immersion $(\pi,P): X \hookrightarrow\mathbb{P}^r\times\mathbb{P}^{r-k-1}$ which give us $P^*\mathcal{O}_{\mathbb{P}^{r-k-1}}(1)=\pi^*\mathcal{O}_{\mathbb{P}^r}(1)\otimes\mathcal{O}_{X}(-E)$.