Skip to main content
2 votes
Accepted

Normalization of nodal singularity of a curve and canonical divisor.

I will assume your nodal curve is embedded in $\mathbb{P}^2$. Then blowing up $\mathbb{P}^2$ at the nodal point will exactly give the normalization of our curve (by strict transformation). Let me call ...
Elk's user avatar
  • 656
2 votes
Accepted

Smoothing projective nodal curve, is the general fiber smooth?

Let $f:X \to T$ be a flat and proper morphism, locally of finite presentation. By Tag 01V8, $f$ is smooth if and only if the fibers of $f$ are smooth. The set of points in $X$ where $f$ is not smooth ...
Dori Bejleri's user avatar
  • 5,235
2 votes
Accepted

What is a curve at $y=\infty$ mean?

It is explained to some extent in the link. Work in the projective plane, so introduce a variable $z$ and make every term in the polynomial homogeneous of degree 3: $$F(x,y,z) = xyz + ax^3 + bx^2z + ...
Ted Shifrin's user avatar
1 vote

Finiteness of the intersection multiplicity of plane algebraic curves

Because $k[x,y]\subset\mathcal{O}_{0}$ and $F,G$ have no common component there is a polynomial in $k[x]$ say $p(x)\in (F,G)$ and as $F(0)=G(0)=0$, we have $p(0)=0$. Thus $p(x)=x^{n}q(x)$ for some $q(...
uzumymw's user avatar
  • 21
1 vote

Theorem 1 in Fulton's algebraic curves section 3.2

This is from section $2.4$: The ideal $m_P(V ) = \{f ∈ \mathcal O_P(V ) | f (P) =0\}$ is called the maximal ideal of $V$ at $P$. It is the kernel of the evaluation homomorphism $f \mapsto f (P)$ of $\...
Ajin Shaji Jose's user avatar
1 vote

How to show there are only two discrete valuation rings with quotient field $k(x)$?

The case where $k[x] \subseteq S$ is easy. Let $m$ be maximal ideal of $S$. Thus $k[x] \cap m$ is a prime ideal in $k[x]$. Since $k[x]$ is a PID, $k[x] \cap m$ is either zero ideal or a maximal ideal. ...
Ajin Shaji Jose's user avatar

Only top scored, non community-wiki answers of a minimum length are eligible