Question

Let $\mathbb G_a$ denote the additive group of complex numbers.

Definition: Let $V \subset Y$ be a dense open subset of the affine variety $Y$ and $\pi : P \longrightarrow V$ a $\mathbb G_a$-principal bundle. An affine extension is an affine $\mathbb G_a$-variety $\hat P$ together with a morphism $ \hat \pi: \hat P \longrightarrow Y$ and an equivariant open embedding $ \iota: P \hookrightarrow \hat P$, such that the diagram

$$ \begin{array}{ccc} P & \hookrightarrow & \hat P \\ \downarrow & & \downarrow \\ V & \hookrightarrow & Y \end{array} $$ is commutative and $\iota (P)= \hat \pi^{-1}(V).$

I am interested in affine extensions of the trivial $\mathbb G_a$-bundle over the affine plane punctured at the origin, i.e. $\mathbb A^2_*:=Sp(\mathbb C[x,y])\setminus{\mathbf o}$ with $\mathbf o:=(x,y)$, and I have the following description:

Proposition: If $P \longrightarrow \mathbb A_*^2$ is the trivial bundle, then any affine extension $\hat P \longrightarrow \mathbb A^2$ is of the form $$ \hat P = Sp (A), $$ where $$ A= \bigoplus_{\nu=0}^\infty \mathfrak m_\nu t^\nu \subset \mathbb C [x,y,t], $$ with a decreasing sequence $(\mathfrak m_\nu)$, $\nu \in \mathbb N$, of ideals $\mathfrak m_\nu \subset \mathbb C [x,y]$, such that

1. $ \mathfrak m_\nu \cdot \mathfrak m_\lambda \subset \mathfrak m_{\nu + \lambda} $ for all $\nu, \lambda\in \mathbb N$,

2. $\mathfrak m_0= \mathbb C [x,y]$, and

3. $V(\mathfrak m_\nu) \subset \mathbf o$ for $\nu > 0$.

On the other hand, every finitely generated $\mathbb C$-algebra of that form defines an affine extension of the trivial bundle.

Question: Does somebody know of a criterion on the sequence $(\mathfrak m_\nu)$$_{\nu \in \mathbb N}$ so that $A$ becomes normal?

Examples: a) If $\mathfrak m_\nu= \mathbb C [x,y]$ for all $\nu$ we have $\hat P \cong \mathbb A^2 \times \mathbb G_a$.

b) If $\mathfrak m_\nu =(x^m,y^n)^\nu$, we have $ A = \mathbb C [x,y,x^mt, y^nt]. $

In the second example, one can see that $A$ is normal for instance if $\mathfrak m_\nu=(x^2,y)^\nu$, but not if $\mathfrak m_\nu=(x^2,y^2)^\nu$ -- in the latter case I think the normalization would be defined by the sequence $\mathfrak m_\nu=(x^2,xy,y^2)^\nu$. Since $A = \mathbb C [x,y,x^mt, y^nt]\cong\mathbb C[x,y,u,v]/(x^mv-y^nu)$ in example b), $Sp(A)$ is a hypersurface in $\mathbb C^4$, so normality is equivalent to singularities being of codimension at least two.

Edit: If the question is difficult in general, I am also interested in the following special case: For which monomial ideals $\mathfrak m\subset\mathbb C[x,y]$, is the ring defined by the sequence $\mathfrak m^\nu$ normal? In this situation I would expect something like: $A$ is normal iff the support of $\mathfrak m^\nu$ consists of all lattice points in the convex hull of the support of $\mathfrak m^\nu$ in $\mathbb R^2$. Here the support of $\mathfrak m^\nu$ is the set of pairs $(k,l)\in\mathbb N^2$ such that $x^ky^l\in\mathfrak m^\nu$.

Answer 1

Hi Isac, One simple criterion can be given which is analogous to Rees' valuations corresponding to ideals. Set $R := \mathbb{C}[x,y]$, define $\nu: R \to \mathbb{N} \cup \infty$ as $\nu(f) := \max\lbrace k: f \in m_k\rbrace$. Then I believe $A$ is normal if $\nu(f^k) = k\nu(f)$ for all $f \in R$. For other criteria, I would look into Kei-ichi Watanabe's articles.

Edit: Here is a general approach.

Claim 1: $A$ is integrally closed iff every homogeneous (with respect to the grading of $A$) element in $\mathbb{C}[x,y,t]$ (i.e. an element of the form $f(x,y)t^k$) which is integral over $A$ is in $A$.

By Claim 1 the integral equation of $ft^k$ (for $f \in R$) over $A$ be of the form $z^d + \sum_{i=0}^d g_ez^{d-e} = 0$ for some $g_e \in m_{ek}$. This proves Claim 2 below.

Claim 2: $A$ is integrally closed iff $\bar m_k := m_k$ for all $k \geq 1$, where

$\bar m_k := \lbrace f \in m_k: f^d + \sum_{i=0}^d g_ez^{d-e} = 0$ for some $d \geq 0$ and $g_1, \ldots, g_d \in m_{ek}\rbrace$.

In the special case that $m_k$ is a monomial ideal for each $k$, it suffices to prove the integral condition only for monomials in $m_k$. And in the special special case that $m_k = m^k$ for some monomial ideal $m$, it follows that your claim is true, i.e. $A$ is integrally closed iff the support of $m^k$ contains all the monomials in the cone spanned by its monomials.

Answer 2

The $\nu$-formulation is a little off, you need to incorporate $t$ as well. Let $ord(\alpha) = \sup\lbrace k: \alpha \in m_k t^k\rbrace$, and let $\overline\nu(f) = \limsup\lbrace ord(f^k)/k : k \in N\rbrace$ (related to Section 6.9 in Huneke-Swanson book). At least if $m_k = I^k$ for some ideal $I$, then for all $f$, if $\overline \nu(f) = ord (f)$, then $A$ is normal.

In a two-dimensional regular local ring or two-dimensional polynomial ring such as here, $k[x,y][It]$ is integrally closed iff $I$ is integrally closed (Chapter 14 in Huneke-Swanson).