Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am looking for an isolated non-normal singularity on an algebraic surface. One obvious example occurs to me: the union of two $2$-dimensional affine subspaces of $\mathbb{A}^4$ which meet in a point. But this seems like "cheating." Can someone provide an irreducible example?

share|cite|improve this question
up vote 13 down vote accepted

Identify two points in the affine plane over $\mathbb C$: it is the affine variety given by the algebra $$\{ f\in\mathbb C[x,y] \mid f(0,0)=f(0,1)\}.$$ Its normalization is the affine plane.

share|cite|improve this answer
Beautiful! This is precisely the sort of thing I had in mind. – Justin Campbell May 11 '12 at 23:55

If $k$ is a field and $A=k[x,xy,y^2,y^3]\subset k[x,y]$, then $X=Spec (A)$ is a surface with $P= (x,xy,y^2,y^3)$ as only (and thus isolated) singularity, non normal because $A$ is non normal.
Here are some details:

I) $A$ is non normal because $y\in Frac(A)\setminus A$ is integral over $A$ (it satisfies $T^2-y^2=0$)
II) The morphism $$f:\mathbb A^2_k \to \mathbb A^4_k:(x,y)\mapsto (u=x,v=xy,w=y^2,z=y^3) $$ has as image $X$, the surface $X\subset \mathbb A^4_k$ defined by the three equations $$ u^2w=v^2,\: u^3z=v^3, \:w^3=z^2 $$ III) If we put $O=(0,0)\in \mathbb A^2_k$ and $P=(0,0,0,0) \in X\subset \mathbb A^4_k$, the morphism $f$ restricts to an isomorphism $f_0:\mathbb A^2_k\setminus \lbrace O \rbrace \stackrel {\cong}{\to} X\setminus \lbrace P\rbrace$.
Its inverse $$f_0^{-1}:X\setminus \lbrace P\rbrace \stackrel {\cong}{\to} \mathbb A^2_k\setminus \lbrace O \rbrace: (u,v,w,z)\mapsto (x,y)$$ is given by: $$x=u \\ y=v/u \;\text {if } u\neq 0 \quad \text {or} \quad y=z/w \; \text {if } w\neq 0$$
The surface $X$ is not isomorphic to QiL's beautifully simple example since the normalization map for $X$ is bijective, and it isn't for QiL's surface.
However Justin remarks that my $A$ can be described as the ring of polynomials $f(x,y)\in k[x,y]$ such that $f_y(0,0)=0$, whereas QiL requires $f(0,0)=f(0,1)$
I find this analogy very interesting and I am grateful to Justin for having pointed it out, since I had not noticed it at all.

share|cite|improve this answer
Always, I enjoy reading your fruitful answers/comments. – Ehsan M. Kermani May 12 '12 at 18:50
Thanks, @ehsanmo – Georges Elencwajg May 12 '12 at 19:03
@GeorgesElencwajg: Thanks! In the spirit of QiL's answer, I think it's interesting to notice that $A = \{ f \in k[x,y] \ | \ f_y(0,0) = 0 \}$. – Justin Campbell May 13 '12 at 18:22
By analogy with curves: the nodal cubic, like QiL's surface, can be thought of as the spectrum of $\{ f \in k[x] \ | \ f(0) = f(1) \}$, and the cuspidal cubic, like your surface, is none other than the spectrum of $\{ f \in k[x] \ | \ f'(0) = 0 \}$. – Justin Campbell May 13 '12 at 18:37
Dear Justin: ah yes it seems obvious once you have explained it :-) (By the way I have incorporated your remark in an Edit to my answer) – Georges Elencwajg May 13 '12 at 18:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.