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The riddle does not exist. If a question can be put at all, then it can also be answered.

Ludwig Wittgenstein


When you have eliminated the impossible, whatever remains, however improbable, must be the truth.

Sir Arthur Conan Doyle (Sherlock Holmes)


Dec
16
awarded  Famous Question
Dec
4
revised The affine line with two points removed
deleted 44 characters in body; edited tags
Dec
4
revised A birational map from $\mathbb{P}^1$ to an irreducible plane projective curve
deleted 12 characters in body; edited tags
Dec
4
accepted A birational map from $\mathbb{P}^1$ to an irreducible plane projective curve
Dec
3
answered How does this derivative notation work?
Dec
3
comment A birational map from $\mathbb{P}^1$ to an irreducible plane projective curve
@MarianoSuárez-Alvarez Can we just define it by $F([x : y]) : [x : y] \mapsto [x : y : -g/f]$?
Dec
3
comment A birational map from $\mathbb{P}^1$ to an irreducible plane projective curve
@MarianoSuárez-Alvarez If $z = -g(x, y)/f(x, y)$, the point would be $[x : y : -g(x, y)/f(x, y)]$.
Dec
3
comment A birational map from $\mathbb{P}^1$ to an irreducible plane projective curve
@MarianoSuárez-Alvarez Well, $z = -g(x, y)/f(x, y)$.
Dec
3
comment A birational map from $\mathbb{P}^1$ to an irreducible plane projective curve
@MarianoSuárez-Alvarez Algebra is not my strong suit. I cannot tell whether the answer is trivial or not.
Dec
3
comment A birational map from $\mathbb{P}^1$ to an irreducible plane projective curve
@MarianoSuárez-Alvarez This is my question.
Dec
2
comment A birational map from $\mathbb{P}^1$ to an irreducible plane projective curve
@MarianoSuárez-Alvarez Yes, but how do I come up with an explicit example of a birational curve?
Dec
2
asked A birational map from $\mathbb{P}^1$ to an irreducible plane projective curve
Dec
2
comment The affine line with two points removed
@Hoot Yes, I do.
Dec
2
asked The affine line with two points removed
Nov
26
comment Very ample divisors and the Riemann-Roch theorem
@rfauffar Thanks.
Nov
26
asked Linearly equivalent divisors and linear transformations
Nov
25
revised Very ample divisors and the Riemann-Roch theorem
added 15 characters in body
Nov
25
comment An algebraic curve $C$ as a cover of $\mathbb{P}^1$ over $\bar{\mathbb{Q}}$
Ok. I appreciate your help. Thank you.
Nov
25
comment An algebraic curve $C$ as a cover of $\mathbb{P}^1$ over $\bar{\mathbb{Q}}$
Honestly? It seems rather superfluous and it bewilders the reader. Why not simply say that there is a surjection $C \to \mathbb{P}^1$? It baffles me.
Nov
25
comment An algebraic curve $C$ as a cover of $\mathbb{P}^1$ over $\bar{\mathbb{Q}}$
Not really. Thank you. Could you explain what it means for an algebraic curve to cover $\mathbb{P}^1$? An intuitive explanation would suffice.