How Do You Decompose the Jacobian of a Polynomial in Algebraic Geometry?

[Fermat1]

Let $m$ be an integer ${\geq}3$ and $f(x,y,z)=y^m(x+y^3)-z^3$ be in $k[x,y,z]$. Find the singular points of $f$ and find a minimal primary decomposition of the jacobian of $f$.

I find the set of singular points of $f$ to be {$(x,0,0): {x\in k} $} and the jacobian to be $\langle y^m,mxy^{m-1},-3z^2\rangle$. How do I find a minimal primary decomposition?

Thanks

 

[jvicens]

for your post! You are correct in identifying the singular points of $f$ as the set $(x,0,0)$ for any $x\in k$. This is because the partial derivative with respect to $y$ is always $mx^{m-1}$, which is only zero when $x=0$. And since $x$ appears in every term of $f$, setting it to $0$ will result in a singular point.

To find a minimal primary decomposition of the jacobian, we can use the following steps:

1. Calculate the radical of the jacobian ideal: In this case, the radical is simply $\langle y,z\rangle$, since these are the only prime ideals that contain all the generators of the jacobian.

2. Determine the minimal primes of the radical: The minimal primes of $\langle y,z\rangle$ are $\langle y\rangle$ and $\langle z\rangle$.

3. Decompose the radical into primary ideals: Since $\langle y\rangle$ and $\langle z\rangle$ are already prime, they are their own primary decompositions.

4. Check for redundancy: In this case, there is no redundancy since the minimal primes are already in their prime forms.

Therefore, the minimal primary decomposition of the jacobian of $f$ is $\langle y\rangle \cap \langle z\rangle = \langle y,z\rangle$.