Example I.4.9.1 in Hartshorne (blowing-up)

Question

Let $Y$ be the irreducible curve of $\mathbb{A}^2$ given by $y^2 = x^2(x+1)$. Let $t,u$ be homogeneous coordinates of $\mathbb{P}^1$. Then the total inverse image of $Y$ under the blowing-up $\phi: X \rightarrow \mathbb{A}^n$ of $\mathbb{A}^n$ at the origin $O=(0,0)$ is a subset of $\mathbb{A}^2 \times \mathbb{P}^1$ and given by the equations $y^2 = x^2(x+1), x u = y t$. Note that by definition the blowing-up of $Y$ at $O$ is the closure of $\phi^{-1}(Y-O)$ and is denoted by $\tilde{Y}$.

Hartshorne, first considers the open set $t \neq 0$, sets $t=1$, treats $u$ as an affine parameter and arrives at the exceptional curve and the equations $y=ux, u^2=x+1$, and he states that the latter give precisely $\tilde{Y}$.

Question 1: Why do the equations $y=ux, u^2=x+1$ define $\tilde{Y}$?

My effort: Since $Y$ is irreducible, the set $Y-O$ is not closed and so $\phi^{-1}(Y-O)$ will not be closed. Now $\tilde{Y}$ is inside the total inverse image of $Y$, i.e. $\phi^{-1}(Y)$, and so any point in $\tilde{Y} - \phi^{-1}(Y-O)$ must be a point of $\phi^{-1}(O)$. Finally, note that $V(y-ux,u^2-x-1) = \phi^{-1}(Y-O) \cup \left\{(0,0,1),(0,0,-1)\right\}$. How can i complete the argument now?

Question 2: Accepting that $\tilde{Y}=\phi^{-1}(Y-O) \cup \left\{(0,0,1,1),(0,0,-1,1)\right\}$, Hartshorne states that the points of $\tilde{Y}$ that intersect $\phi^{-1}(O)$, i.e. points $(0,0,1),(0,0,-1)$, (assuming that $t=1$), give the slopes of the branches of $Y$ through the origin $O$. How can we see that? What is a general proof of this statement?

Question 3: By considering now the open set $u \neq 0$, we can set $u=1$ and the equations of $\phi^{-1}(Y)$ become $x = t y, y^2 = x^2(x+1)$ which give $y = 0$ and $1 = t^2(ty+1)$. What does the information given my these equations signify and why is Harsthorne now analyzing them?

Answer 1

I'd like to make some complement to Mohamed Hashi's answer.

Let $X$ be the blow-up of $\mathbb{A}^2$ at $O=(0,0)$,that is $$X=\{((x,y),(u:t))\in \mathbb{A}^2 \times \mathbb{P}^1\mid xu=yt\}.$$ Let $\mathbb{A}^1_u,\mathbb{A}^1_t$ be the affine cover of $\mathbb{P}^1$,then $$\mathbb{A}^2 \times \mathbb{P}^1=(\mathbb{A}^2 \times\mathbb{A}^1_u) \cup(\mathbb{A}^2 \times\mathbb{A}^1_t).$$ We have the following isomorphism: $$X_u=X\cap(\mathbb{A}^2 \times\mathbb{A}^1_u)\longrightarrow \mathbb{A}^2,$$ where $((x,y),(1:t))\longmapsto (y,t)\longmapsto ((yt,y),(1:t))$.Under this isomorphism,we may write $$\varphi^{-1}(Y)\cap X_u=\{(0,t))\in \mathbb{A}^2\mid t \in k\}\cup \{(y,t)\in \mathbb{A}^2\mid 1=t^2(yt+1)\},$$ $$\varphi^{-1}(Y-O)\cap X_u=\{(y,t)\in \mathbb{A}^2\mid 1=t^2(yt+1)\}- \{(0,\pm 1))\},$$ since $\overline{\varphi^{-1}(Y-O)}^X\cap X_u=\overline{\varphi^{-1}(Y-O)\cap X_u}^{X_u}$ is closed in $ X_u\cong\mathbb{A}^2$,we have $$\overline{\varphi^{-1}(Y-O)}^X\cap X_u=\{(y,t)\in \mathbb{A}^2\mid 1=t^2(yt+1)\},$$ hence $\tilde{Y}_u=\tilde{Y}\cap(\mathbb{A}^2 \times\mathbb{A}^1_u)=\{(y,t)\in \mathbb{A}^2\mid 1=t^2(yt+1)\}.$

Similarly,under the isomorphism $X_t=X\cap(\mathbb{A}^2 \times\mathbb{A}^1_t)\cong\mathbb{A}^2$,we get $$\tilde{Y}_t=\tilde{Y}\cap(\mathbb{A}^2 \times\mathbb{A}^1_t)=\{(x,u)\in \mathbb{A}^2\mid u^2=x+1)\}.$$ Now come back to our question.

Question 1:why $\tilde{Y}=\tilde{Y}\cap(\mathbb{A}^2 \times\mathbb{A}^1_t)$?

Note that $\mathbb{A}^2 \times \mathbb{P}^1=(\mathbb{A}^2 \times\mathbb{A}^1_t) \cup\{((x,y),(1:0))\in \mathbb{A}^2 \times \mathbb{P}^1\},$ so \begin{equation} \begin{split} \tilde{Y}&=\tilde{Y}_t\cup (\tilde{Y}\cap\{((x,y),(1:0))\in \mathbb{A}^2 \times \mathbb{P}^1\})\\ &=\tilde{Y}_t\cup (\tilde{Y}\cap\{((x,y),(1:0))\in \mathbb{A}^2 \times \mathbb{P}^1\cap X\})\\ &=\tilde{Y}_t\cup (\tilde{Y}_u\cap\{((0,y),(1:0))\in X_u\}). \end{split} \end{equation} But under our isomorphism $X\cap(\mathbb{A}^2 \times\mathbb{A}^1_u)\cong\mathbb{A}^2$,we know that $$\{((0,y),(1:0))\in X_u\}=\{(y,0)\in \mathbb{A}^2\mid y\in k\},$$and $$\tilde{Y}_u=\{(y,t)\in \mathbb{A}^2\mid 1=t^2(yt+1)\}.$$ Therefore $\tilde{Y}_u\cap\{((0,y),(1:0))\in X_u\}=\emptyset$,we conclude that $\tilde{Y}=\tilde{Y}_t$.

Question 3:what about $\tilde{Y}_u$?

In fact,we have $\tilde{Y}=\tilde{Y}_u \cup \{((-1,0),(0:1))\in \tilde{Y}_t\}$.

Question 2: Pionts of $\varphi^{-1}(O)$ are in 1-1 correspondence with the set of lines through $O$ in $\mathbb{A}^2$,points $((0,0),(0,\pm 1))$ correspond to lines $y=\pm x$ whose slope are $\pm 1$.Note that lines $y=\pm x$ are the tangent of the two branches of $Y$ at $O$.

Added:there is a more direct approach concerning question 1.In the same way,we may look at the following isomorphism:$$X_u\longrightarrow Z(x-yt)\subset\mathbb{A}^3,$$ where $((x,y),(1:t))\longmapsto (x,y,t)\longmapsto ((yt,y),(1:t))$.Under this isomorphism,we may write \begin{equation} \begin{split} \varphi^{-1}(Y)\cap X_u&=Z(y^2-x^3-x^2,x-yt)\subset \mathbb{A}^3,\\ \varphi^{-1}(Y-O)\cap X_u&=Z(1-t^2-yt^3,x-yt)-(0,0,\pm 1)\subset\mathbb{A}^3,\\ \overline{\varphi^{-1}(Y-O)}^X\cap X_u &=\overline{\varphi^{-1}(Y-O)\cap X_u}^{X_u}=Z(1-t^2-yt^3,x-yt), \end{split} \end{equation} hence we have $\tilde{Y}_u=\tilde{Y}\cap(\mathbb{A}^2 \times\mathbb{A}^1_u)=Z(1-t^2-yt^3,x-yt)$.

Similarly,$\tilde{Y}_t=\tilde{Y}\cap(\mathbb{A}^2 \times\mathbb{A}^1_t)=Z(1+x-u^2,xu-y)$. And since \begin{equation} \begin{split} \tilde{Y}\cap\{((x,y),(1:0))\in \mathbb{A}^2 \times \mathbb{P}^1\} &=\tilde{Y}\cap\{((x,y),(1:0))\in \mathbb{A}^2 \times \mathbb{P}^1\}\cap X_u\\ &=\tilde{Y}_u\cap\{((0,y),(1:0))\in X_u\}\\ &=\tilde{Y}_u\cap\{((0,y,0)\in \mathbb{A}^3\}\\ &=\emptyset. \end{split} \end{equation} Illustration: 1.The first equality holds because$\tilde{Y}\cap\{((x,y),(1:0))\in \mathbb{A}^2 \times \mathbb{P}^1\}$ is in fact a subset of $X_u$.

2.The second equality holds because in $X_u$,we have $u=1$ and $x=yt$.

3.In the third equality,we used the isomorphism $X_u\cong Z(x-yt)$.

3.The last equality holds because $\tilde{Y}_u=Z(1-t^2-yt^3,x-yt)$,and $$Z(1-t^2-yt^3,x-yt)\cap\{((0,y,0)\in \mathbb{A}^3\}=\emptyset.$$

Finally,we see that $\tilde{Y}=\tilde{Y}_t\cup(\tilde{Y}\cap\{((x,y),(1:0))\in \mathbb{A}^2 \times \mathbb{P}^1\})=\tilde{Y}_t\cup \emptyset=\tilde{Y}_t$.

Edit3:In this example,points in $\tilde{Y} \cap \varphi^{-1}(O)$ correspond to tangent of the two branches of $Y$ at $O$,we will see that this is true for any curves in $\mathbb{A}^2$.

Let $Y=Z(f)$ be a curve in $\mathbb{A}^2$,where $f(x,y)=f_r+\cdots+f_d$ and $f_i$ are homogeneous polynomials in $x,y, ~\forall r\leq i\leq d$.We will assume that $r\geq 2$ so that $O=(0,0)$ is a singular point of $Y$.As usual,since $X_u\cong\mathbb{A}^2$,we may write \begin{equation} \begin{split} \varphi^{-1}(Y)\cap X_u&=\{(0,t))\in \mathbb{A}^2\mid t \in k\}\cup \{(y,t)\in \mathbb{A}^2\mid \frac{f(yt,y)}{y^r}=0\},\\ \varphi^{-1}(Y-O)\cap X_u&= \{(y,t)\in \mathbb{A}^2\mid \frac{f(yt,y)}{y^r}=0\}-\{(0,t)\in \mathbb{A}^2\mid f_r(t,1)=0\},\\ \tilde{Y}_u=\overline{\varphi^{-1}(Y-O)}^X\cap X_u &=\{(y,t)\in \mathbb{A}^2\mid \frac{f(yt,y)}{y^r}=0\}\\ \tilde{Y}_u \cap \varphi^{-1}(O) &=\{(0,t)\in \mathbb{A}^2\mid f_r(t,1)=0\} \end{split} \end{equation} Similarly,we have $$\tilde{Y}_t \cap \varphi^{-1}(O) =\{(0,u)\in \mathbb{A}^2\mid f_r(1,u)=0\}.$$ Next we glue $\tilde{Y}_u \cap \varphi^{-1}(O)$ and $\tilde{Y}_t \cap \varphi^{-1}(O)$ to see that$$\tilde{Y} \cap \varphi^{-1}(O)\cong D=\{(x:y)\in \mathbb{P}^1\mid f_r(x,y)=0\}.$$In waht follows,we let $D_1=\{(1:y)\in \mathbb{P}^1\mid f_r(1,y)=0\}$, $D_2=\{(x:1)\in \mathbb{P}^1\mid f_r(x,1)=0\}$.The glueing is carried out by the following isomorphism: $$\tilde{Y}_u \cap \varphi^{-1}(O)\cong D_2:(0,t)\mapsto (t:1),$$ $$\tilde{Y}_t \cap \varphi^{-1}(O)\cong D_1:(0,u)\mapsto (1:u).$$ Now we see that $\tilde{Y} \cap \varphi^{-1}(O)\cong D=\{(x:y)\in \mathbb{P}^1\mid f_r(x,y)=0\},$which is a finite set.Further more,we note that $f_r(x,y)$ is a homogeneous polynomial of degree $r$ in $k[x,y]$ and $k$ is algebraically closed,so $f_r(x,y)$ can be factored into linear terms:$$f_r(x,y)=(a_1x-b_1y)\cdots(a_rx-b_ry).$$ It follows that $$\tilde{Y} \cap \varphi^{-1}(O)=\{(b_1:a_1),...,(b_r:a_r)\}. $$ So we can say that points $(b_i:a_i)$ in $\tilde{Y} \cap \varphi^{-1}(O)$ corresponde to the so called tangent directions $a_ix-b_iy$.But we know that tangent directions are exactly the tangents of branchs of $Y$ through $O$.

We conclude that points in $\tilde{Y} \cap \varphi^{-1}(O)$ corresponde to the tangents of branchs of Y through $O$.

Answer 2

$u$ and $t$ represent the slope of the tangent. In the first case $u=\frac{y}{x}$ assuming $x\neq 0$. so dividing the original equation by $x^2$ we get $u^2=x+1$. This should answer question $1$. At the origin there are two tangents, given by $x+y=0$ and $x-y=0$. The intersection of $\tilde{Y}$ at $(0,0,\pm 1)$ gives $x=y=0$ so $u^2=1$ and $u=\pm 1$ the slopes of the two tangents. The point is that the origin with two tangents, now corrisponds to two point each with a single tangent. And the two tangents project onto the tangents and the origin. For question 3 we now preform the same argument for $t=\frac{x}{y}$ the inverse slope.

Answer 3

What is a strict transform? Let $X\subset \mathbb{A}^2$ be any affine variety containing the origin. Then $\varphi^{-1}(X)$ consists of two irreducible components. Namely $\tilde{X}$, the closure of $\varphi^{-1}(X-O)$, and $\varphi^{-1}(O)\cong \mathbb{P}^{1}$. The first is called the strict transform and the second the exceptional curve. So the exceptional curve in the blow-up is the irreducible component of the blow-up that lies completely above the origin.

Question 1:

For reference:

$$f(x,y)=y^2-x^2(x + 1), \quad g(x,y,u,t)=xu-ty. \tag 1$$

Your method, Reference of proof strategy:

This method consists of two parts. Let $f,g$ be the equations for which you want to show $V(f,g)=\varphi^{-1}(Y-O)$.

First, show that $\varphi^{-1}(Y-O)$ is not closed. You say: since $Y$ is irreducible, the set $Y−O$ is not closed and so $\varphi{−1}(Y−O)$ will not be closed. Indeed, by irreducibility of $Y$, $Y-O$ cannot be closed in $\mathbb{A}^2$ since points are closed. Where do you want $\varphi^{−1}(Y−O)$ to not be closed in? I image $\varphi^{−1}(Y)$. Since $\varphi^{−1}(Y−O)$ is always closed in itself. To infer this, you need $\varphi$ to be a closed map restricted to $\varphi^{−1}(Y)$. But we only have that it is a homeomorphism on $\varphi^{−1}(Y−O)$, so this path is useless. Here completeness of $\mathbb{P}^1$ comes into play. By completeness, all projection maps are closed maps! So this holds.

Second, notice that if you add $d$ points you get $V(f,g)$, with $0<d\in \mathbb{N}$. If $d=1$, you are done since you found the smallest closed subset containing your subset. It being the smallest is guaranteed as $d=1$. If $d>1$, you can show that $V(f,g)$ is irreducible and you are done too. The reasoning being, the closure of $\varphi^{-1}(Y-O)$ cannot be strictly contained in $V(f,g)$, that would imply that $V(f,g)$ is not irreducible.

This method only works if you already know what the equations are. My body starts to ache when I need to prove irreducibility, so I only use this tactic for $d=1$.

What I would do:

First, note that the total inverse image $\varphi^{-1}(Y)$ is defined by $(1)$. Then look one affine chart at a time. We will look in in the affine chart where $t\neq 0$. Then we can assume that $t=1$. The equations in this chart become $$y^2=x^2(x + 1), \quad xu=y.$$ This defines a closed set in $\mathbb{A}^3$ consisting of two irreducible parts: $x=0$, $y=0$, $u$ arbitrary and $$(\frac{y}{x})^2=u^2=x+1, \quad xu = y.$$ The first part lies completely above the origin, hence is part of the exceptional curve. Then the other has to be part of the strict transform. We will look in in the affine chart where $u\neq 0$. Then we can assume that $u=1$. The equations in this chart become $$y^2=x^2(x + 1), \quad x=ty.$$ This defines a closed set in $\mathbb{A}^3$ consisting of two irreducible parts: $x=0$, $y=0$, $t$ arbitrary and $$1=(\frac{x}{y})^2(x+1)=t^2(x+1), \quad x = ty.$$ The first part lies completely above the origin, hence is part of the exceptional curve. Then the other has to be part of the strict transform. Note that in the second part, the strict transform part, there is no point with $t=0$. Hence all the points already where there in the affine chart $t\neq 0$. Now we can conclude that the affine curve $$\frac{y}{x}^2=u^2=x+1, \quad xu = y$$ in $\mathbb{A}^3$ is isomorphic to the strict transform $\tilde{Y}$. You can also see that the strict transform intersects the pre-image of the origin in two points: $x=y=0$ and $[t:u]=[1:1]$ or $[1:-1]$.

I take it for granted, but it still is a good thing to play with I think!

a) Why do you get the total inverse image $\varphi^{-1}(Y)$ of $Y$ in $\mathbb{A}^2\times \mathbb{P}^1$ by the equations (as stated in the example) $$y^2=x^2(x + 1), \quad xu=ty.$$

Question 2:

A blow-up provides a projective space above the origin that keeps track of the slopes of the branches in the origin of the variety. Outside the pre-image of the origin, it keeps track which line goes through a point and the origin [note that projective points are lines through the origin]. This is done through the blow-up equations $x_jy_i=x_iy_j$.

You need some more background for a rigorous proof of this. Same goes for me, but on the level of HAG I, this was enough for me. Just compute a lot of blow-ups and it will become more natural.

Question 3:

When inspecting a projective variety, one often looks standard open affine at a time. In this case the strict transform of Y is contained in one open affine chart, so Harsthorne does not inspect the other for $Y$ anymore(in my copy at least). Usually you need to inspect both charts and glue them to get your total blow-up variety. So these equations describe the affine curve $\tilde{Y}\cap U$ where $U$ is the affine open corresponding to $t \neq 0$.

Look at Example 6.22 on page 62 of the notes by B.Moonen for a more worked out example of the blow-up for this curve.

Glueings

You can glue varieties along opens by giving an isomorphism from the open of variety A to an open of variety B. Projective varieties come with standard glueings between their standard affine opens. For example, let $V_t$ be the affine chart where $t\neq 0$ and $V_u$ where $u\neq 0$ in $\mathbb{P}^1$ with coordinates $[t:u]$:

$$\mathbb{A}^1\setminus O \cong _u \cap V_t \rightarrow V_u \cap V_t \cong \mathbb{A}^1 \setminus O,$$ $$\frac{t}{u} \cong [\frac{t}{u},1] \mapsto [1,\frac{u}{t}] = \frac{u}{t}.$$

This glueing is therefore sometimes denoted by $x \mapsto x^{-1}$. For the blow-up, you always have the same glueings since this is only based on the containing space $\mathbb{A}^2 \times \mathbb{P}^1$. Let $[t:u]$ be the projective coordinates again and $V_u, V_t$ the corresponding standard affine opens. Then we get:

$$\mathbb{A}^3\setminus V(\mbox{projection on third coord}) \cong V_u \cap V_t \rightarrow V_t \cap V_u \cong \mathbb{A}^3\setminus V(\mbox{projection on third coord}),$$ $$(x,y,\frac{t}{u}) \cong ((x,y),[\frac{t}{u},1]) \mapsto ((x,y),[1,\frac{u}{t}]) \cong (x,y,\frac{u}{t}).$$

If we actually look in the blow-up, then we have the equation $xu=ty$. So we have $x\frac{u}{t}=y$ and vice versa, so we can drop one coordinate. In other words, in $\mathbb{A}^3$ the variety cut-out by $xy=z$ is isomorphic to $\mathbb{A}^2$. Using this and removing the redundant information, let $X$ be the blow-up:

$$V_u \cap X \cap V_t \rightarrow V_t \cap X \cap V_u,$$ $$(y\frac{t}{u},y,\frac{t}{u}) \mapsto (x,x\frac{u}{t},\frac{u}{t}).$$

I hope this helps.