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First Notation:

$f$ is a polynomial in $\mathbb{C}[x,y]$ such that $f=f_1\cdot...\cdot f_s$ is the decomposition of $f$ into relatively prime irreducible polynomials.

$f_x:=\frac{\partial f}{\partial x}$ and $f_y:=\frac{\partial f}{\partial y}$.

Let $\langle f_x, f_y\rangle =\bigcap\limits_ {i=1}^sq_i$ be the primary decomposition of $\langle f_x, f_y\rangle$.

We now define the ideals $a:=\bigcap\limits_ {i=1}^mq_i$ and $b:=\bigcap\limits_ {i=m+1}^sq_i$ so that each $q_i$, for $1\leq i\leq m$, corresponds to a singular point of $f$.

Furthermore, the primary ideals $q_i$, for $\:m+1\leq i\leq s$ correspond to the irreducible components of the variety $\mathfrak{V}(\langle f\rangle)$, which is a plane curve by definition.

I have already shown that the vector spaces $(q_i:\langle f\rangle)/q_i$, for $\:(m+1)\leq i\leq s$, have the dimension zero. I have also shown that for $1\leq i\leq m$, $dim((q_i:\langle f\rangle)/q_i)>0.$ (*)

So the statement I would like to prove says that if all the singular points of the curve $\mathfrak{V}(\langle f\rangle)$ have the multiplicity one, then the dimension of the vector space $\bigoplus\limits_{1\leq i\leq m}(q_i:\langle f\rangle)/q_i$ is equal to the number of the singular points of $f$. If the multiplicity of a singular point is more than one, then the dimension of $\bigoplus\limits_{1\leq i\leq m}(q_i:\langle f\rangle)/q_i$ is greater than the number of singular points. .

We know that if the primary ideal $q_i$, $1\leq i\leq m$, is of the form $\langle x-a, y-b \rangle$ then $\mathbb{C}[x,y]/q_i$ has the dimension one. Then by the (*) above I can say $(q_i:\langle f\rangle)/q_i$ has dimension one. But can I say if the multiplicity of a singular point of f is one, then the corresponding primary ideal is always of the form $\langle x-a, y-b \rangle$? When yes, why?

Thanks, Alex

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The proof of the statement about the dimension of $\mathbb{C}[x,y]/q_i$ being one is here:…;. – Alex Jun 29 '14 at 19:53

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