Fibers

peccavi_2006 · **Posted:** Wed, 14 Dec 2011 14:04:03 UTC

I'm reading through the chapters in Algebraic Geometry, Harris, about Blow-ups and Resolving Singularities and he keeps mentioning fibers (for instance, in one example he says
the fiber over that point [the origin in $\mathbb{A}^{2}$ ] is a copy of $\mathbb{P}^{1}$ corresponding to the lines through that point), however, he doesn't seem to ever actually define what a fiber is, only something called the fiber product (which I assume is also something I will need).

So if someone could tell me what he means that would be much appreciated (I also noticed he never actually explains what a homogeneous equation is, which I know is a really basic concept (and we have subsequently covered it in lectures so it doesn't matter), but still!)

Thank you very much!

outermeasure · **Posted:** Wed, 14 Dec 2011 14:49:44 UTC

peccavi_2006 wrote:

I'm reading through the chapters in Algebraic Geometry, Harris, about Blow-ups and Resolving Singularities and he keeps mentioning fibers (for instance, in one example he says
the fiber over that point [the origin in $\mathbb{A}^{2}$ ] is a copy of $\mathbb{P}^{1}$ corresponding to the lines through that point), however, he doesn't seem to ever actually define what a fiber is, only something called the fiber product (which I assume is also something I will need).

So if someone could tell me what he means that would be much appreciated (I also noticed he never actually explains what a homogeneous equation is, which I know is a really basic concept (and we have subsequently covered it in lectures so it doesn't matter), but still!)

Thank you very much!

Suppose $p\colon E\to B$ . The fibre over $b\in B$ is $p^{-1}(b)$ in topology. However, in AG you need a bit more but that requires you to know about schemes (you have $\mathop{\mathrm{Spec}}k(b)\hookrightarrow B$ , and the fibre of

over

is $E\times_B\mathop{\mathrm{Spec}}k(b)$ ).

A homogeneous equation is exactly what you think it means --- the total degree of every term appearing in the equation is the same.

So you blow up $\mathbb{A}^2$ at the origin, getting $\mathop{\mathrm{Bl}}_{(0,0)}\mathbb{A}^2\subseteq\mathbb{A}^2\times\mathbb{P}^1\to\mathbb{A}^2$ , where $\mathop{\mathrm{Bl}}_{(0,0)}\mathbb{A}^2=V(\{x_iX_j-x_jX_i\mid 1\leq i,j\leq 2\})$ , where x_1,x_2

are the coordinates for $\mathbb{A}^2$ and X_1,X_2

are the coordinates of $\mathbb{P}^1$ . The fibre over any point of $\mathbb{A}^2-\{(0,0)\}$ is still a point, but the fibre over (0,0) is the entire $\mathbb{P}^1$ since there isn't any condition on the X_1,X_2

.

peccavi_2006 · **Posted:** Wed, 14 Dec 2011 15:30:04 UTC

So if I had a point $p \in \mathbb{A}^{n}$ for some arbitrary n > 0

, would the map $\mathop{\mathrm{Bl}}_{p} \mathbb{A}^{n} \rightarrow \mathbb{A}^{n}$ be an isomorphism away from p, and at p, would that generalise to the fiber over p being $\mathbb{P}^{n-1}$ ?

outermeasure · **Posted:** Wed, 14 Dec 2011 16:14:05 UTC

peccavi_2006 wrote:

So if I had a point $p \in \mathbb{A}^{n}$ for some arbitrary n > 0

, would the map $\mathop{\mathrm{Bl}}_{p} \mathbb{A}^{n} \rightarrow \mathbb{A}^{n}$ be an isomorphism away from p, and at p, would that generalise to the fiber over p being $\mathbb{P}^{n-1}$ ?

If you are blowing up a point, yes. If you blow up along a subvariety you replace the subvariety by its projectivised normal bundle.

peccavi_2006 · **Posted:** Wed, 14 Dec 2011 17:25:35 UTC

Excellent - thanks outermeasure

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Fibers

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