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I want to show that $Z(y^2-x^3-ax-b)\subset \mathbb A^2$ is smooth if and only if $4a^3+27b^2\neq 0$. Let $f=y^2-x^3-ax-b$.

I know the following theorem: enter image description here

If we show that $f$ is irreducible then $I(X)=(f)$.Then the Jacobi matrix is $(-3x^2-a,2y)$. If we know that $X$ is irreducible, then $X$ is smooth exactly when this matrix has rank $1$. The matrix has rank $1$ whenever both terms are not zero. I don't see how this is true only when $4a^3+27b^2= 0$.

Things I am confused about:

  1. Why is $f$ irreducible?
  2. Why is $X$ irreducible?
  3. Why does the Jacobi have rank zero exactly when $4a^3+27b^2= 0$?
Mark Murray
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    Your characterization of the matrix having rank $1$ is wrong: you only need one of the terms to be nonzero. So a singular point happens if and only if $f$, $-3x^2-a$, and $2y$ are all zero at some common value of $x$ and $y$. Can you handle it from here? – KReiser Jan 24 '21 at 02:09
  • Why does a singular point also require $f$ to be zero? – Mark Murray Jan 24 '21 at 17:15
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    Because its needs to be on the variety. – Con Jan 24 '21 at 17:42
  • Ah yes! How can I see that $X$ is irreducible. I don't think it follows directly from $f$ being irreducible. – Mark Murray Jan 24 '21 at 17:49
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    If $f$ is irreducible, then it is prime as well. Hence the ideal generated by it in $k[x,y]$ is prime, which is equivalent to the algebraic set being irreducible. Are you aware of this fact? Are you even consideing it as algebraic set or even as a scheme? – Con Jan 24 '21 at 23:10

1 Answers1

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Let $f = x^3 + ax + b$, let $F = y^2 - f(x) = y^2-x^3-ax-b$, and let $X = Z(F)$. (Sorry for changing your notation, but I'll just confuse myself if I don't have $X$ given by $y^2 = f(x)$.)

  1. Irreducibility of $F$ basically follows from the fact $F$ is degree $2$ in $y$ and $x^3 + ax + b$ is not a square. If $F$ factors, then we can write it as a product of two monic, degree $1$ polynomials in $y$. Writing $$ (y + g(x))(y + h(x)) = y^2 - x^3 - ax - b $$ we find $h(x) = -g(x)$, so $y^2 - g(x)^2 = y^2 - x^3 - ax - b$ and hence $x^3 + ax + b = g(x)^2$. But $x^3 + ax + b$ is not a square in $k[x]$ (it has odd degree, for one thing), so this is impossible. (See here for more details.)

  2. As mentioned in the comments, since $F$ is irreducible and $k[x,y]$ is a UFD, then the ideal $(F)$ is prime. Since prime ideals correspond to irreducible varieties (see here, for example), then $X = Z(F)$ is irreducible.

  3. Note that $F_x = -f'(x)$ and $F_y = 2y$. Suppose that $(u,v)$ is a singular point of $X$, so $F(u,v) = F_x(u,v) = F_y(u,v) = 0$ by the Jacobian criterion. Then $0 = F_y(u,v) = 2v$, so $v = 0$. Thus $(u,v) = (u,0)$, and $u$ is a root of both $f'(x)$ and $F(x,0) = f(x)$. This implies that $u$ is a multiple root of $f$, hence the discriminant of $f$ is zero. The discriminant of the cubic $f(x) = x^3 + ax + b$ is $-(4a^3 + 27b^2)$, so the claim follows. Moreover, all the implications in this argument are reversible, so we get the iff statement.

Viktor Vaughn
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