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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?

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2 Answers

up vote 9 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.

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Beautiful! This is precisely the sort of thing I had in mind. –  Justin Campbell May 11 '12 at 23:55
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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: $$x=u \\ y=v/u \;\text {if } u\neq 0 \quad \text {or} \quad y=z/w \; \text {if } w\neq 0$$
Edit
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.

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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
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@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
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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
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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
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