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Let $X$ be an integral scheme of finite type over a field. Then there is a surjective finite map $\tilde{X} \to X$ from the normalization $\tilde{X}$ of $X$.

Is this going to be bijective?

In the simplest non-normal case, namely the spectrum of $k[x^2, x^3] = k[t,u]/(t^3 -u^2)$, the map is bijective, because the curve is geometrically just a cubic curve (the set of all $(v^2, v^3)$ in affine 2-space, geometrically). I can't find it asserted anywhere that the map is necessarily bijective, though.

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    $\begingroup$ Akhil, another example to think about the normalization of an order in a quadratic field, which you can compute "by hand" (since we can write down the orders explicitly). Loosely speaking, consider the pair of primes $P, Q$ over a split prime $p$ in the quadratic field and the order of elements $a$ in the quadratic integer such such that $a(P) = a(Q)$ in the common residue field $\mathbf{F}_p$. It's like "gluing" two transverse branches. Then you can make analogues using quadratic (separable) extensions of $k(x)$ instead, and you'll see you get the nodal example suggested by Emerton. $\endgroup$
    – BCnrd
    Commented Oct 4, 2010 at 0:45
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    $\begingroup$ If you're willing to give up integral, k[t,u]/(tu) is an even simpler counterexample, and formally isomorphic to Emerton's. $\endgroup$
    – Allen Knutson
    Commented Oct 4, 2010 at 1:06
  • $\begingroup$ That's a nice example; thanks for pointing it out. In the second comment, what does the integral closure of $k[t, u]/(tu)$ mean though? $\endgroup$
    – Akhil Mathew
    Commented Oct 4, 2010 at 4:10
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    $\begingroup$ Dear Akhil, Any reduced ring has embeds into its total quotient ring (the localization at all non-zero divisors), which (if the original ring is Noetherian) is just a product of finitely many fields. One can then just define the normalization in this ring in the usual way (elements integrally dependent over the original ring). See Matsumura's (newer) book for a discussion of integral closure and normality in this level of generality. (Formally, one could do the same for non-reduced rings as well, but the practical outcome is a bit of a mess in that case, so it is probably better to ... $\endgroup$
    – Emerton
    Commented Oct 4, 2010 at 4:27
  • $\begingroup$ ... stick to reduced rings.) $\endgroup$
    – Emerton
    Commented Oct 4, 2010 at 4:27

3 Answers 3

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Your example is not really the simplest case, in that a cusp is not the simplest possible curve singularity. Rather, the simplest curve singularity is a node, e.g. as in $y^2 = x^3 - x^2$. There are then two branches passing through the singularity, and the normalization map separates them. (Exercise: The normalization is again given by a map $t \mapsto x(t),y(t)$; find this map!)

This is the typical phenomenon with normalization: it separates the different branches passing through a singularity. The situation of having just a single branch (and hence having a bijective normalization map) is somewhat unusual (and has a special adjective to describe it: unibranch).

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    $\begingroup$ Thank you! Some calculations suggest the normalization is $t \to (t^2 + 1, t(t^2 + t))$, which is as you say not bijective. $\endgroup$
    – Akhil Mathew
    Commented Oct 4, 2010 at 0:07
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    $\begingroup$ Dear Akhil, Yes. In fact my exercise was a little poorly chosen; $y^2 = x^3 + x^2$ has better looking real points, in that the normalization is given by $t \mapsto (t^2 - 1, t(t^2 - 1))$, and you can see the node forming when $t = \pm 1$. In the case of $y^2 = x^3 + x^2$, you don't see the two branches defined over $\mathbb R$, because the node corresponds to $t = \pm i$, and so while the node is defined over $\mathbb R$ (it corresponds to the conjugate pair $t = \pm i$), the branches are only defined over $\mathbb C$ (if we move away from $\pm i$ we get points that are only defined over ... $\endgroup$
    – Emerton
    Commented Oct 4, 2010 at 2:34
  • $\begingroup$ ... $\mathbb C$). In the real points, this corresponds to the singularity $(x,y) = (0,0)$ being isolated. (One says that $y^2 = x^3 - x^2$ is a non-split nodal cubic over $\mathbb R$, whereas $y^2 = x^3 + x^2$ is a split nodal cubic.) $\endgroup$
    – Emerton
    Commented Oct 4, 2010 at 2:36
  • $\begingroup$ Ah. Yes, I see what you mean; the real curve passes through the origin twice. $\endgroup$
    – Akhil Mathew
    Commented Oct 4, 2010 at 3:59
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If $X$ is reduced and finite type over an algebraically closed field of characteristic zero, then there is a largest finite map $Y \to X$ which is bijective (where $Y$ is also reduced). This map is the seminormalization, $Y = X^{SN}$. In fact, it always factors the normalization map $X^N \to X^{SN} \to X$ where $X^N$ is the normalization.

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No, the map isn't generally bijective. For curves, normalization is the same as resolution of singularities. Look at a nodal cubic curve $y^2=x^2(x-1)$ in the plane. Over the node, the normalization has two points.

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