I.4.4_ERRATA

6/1/2021

From part (c):

Hence, we've constructed an isomorphism from
an open set of Y to P^{1}, making them birationally equivalent. Done!

(BTW: I caught this error by glancing back at 3.1e. If it were that open set of Y were isomorphic to P

What you're actually supposed to do becomes clear upon graphing it:

There's a singularity in the middle, so of course we should blow it up, like in part (b). Using coordinates u,t for P

xu | = y | ||

y^{2} | = x^{2}(x + 1) | ||

Subbing the first equation into the second yields...

x^{2}u^{2} | = x^{2}(x + 1) | ||

⇐⇒x^{2}u^{2} - x^{2}(x + 1) | = 0 | ||

⇐⇒x^{2}(u^{2} - (x + 1)) | = 0 | ||

⇐⇒x^{2}(u^{2} - (x + 1)) | = 0 | ||

Using the open set x≠0, we can ignore x

xu | = y | ||

x | = u^{2} - 1 |

i.e. it's the curve given by (u,u

(u,u^{2} - 1,u^{3} - u) |
(u,u^{2},u^{3}) |

is bijective onto the TCC, and it and its inverse maps are morphisms thanks to this:

So Y is birational to the TCC, and the TCC itself is birational to P

Also, welcome to June! I think June is a pretty cute name, tbh.