Normalization of Algebraic and Arithmetic Curves

Normal schemes are nice, and happily there is a process for taking a scheme and producing “the same scheme but normal”, namely normalization. This can be thought of as a mild analogue of resolution of singularities (whose goal is to produce “the same scheme but nonsingular”), and indeed in the case of curves normalization succeeds in resolving singularities.

The normalization of an integral scheme $X$ is constructed as follows. If $X={\rm Spec} A$ , the normalization of ${\rm Spec} A$ is ${\rm Spec} \widetilde{A}$ , where $\widetilde{A}$ is the integral closure of $A$ in its field of fractions $K(A)$ . In general $X=\bigcup{\rm Spec} A_i$ with ${\rm Spec} A_i$ integral; we can normalize each affine piece and then glue these together to obtain a normalization of $X$ . The normalization some universal property of being terminal among dominant morphisms from integral normal schemes to $X$ , so is unique up to unique isomorphism.

The examples here are exercises 9.7.E, 9.7.F, 9.7.H of Vakil’s notes (and are maybe also just standard examples, but that’s where the inspiration to do them came from).

First let’s normalize the nodal cubic ${\rm Spec} k[x,y]/(y^2-x^3-x^2)$ . By intersecting with lines through the double point at the origin we can find a map from $\mathbb{A}^1$ : setting $y=tx$ ,

$\displaystyle y^2=x^3+x^2\text{ becomes }(t^2-1)x^2=x^3,$

so that $x=t^2-1$ and $y=t^3-t$ . In fact the map

$\displaystyle \mathbb{A}^1_k \to{\rm Spec} k[x,y]/(y^2-x^3-x^2)$

$\displaystyle t^2-1 \mapsfrom x$

$\displaystyle t^3-t \mapsfrom y$

is the normalization.

Let $K$ be the field of fractions of $k[x,y]/(y^2-x^3-x^2)$ . First note that $t=\frac yx$ is integral over $k[x,y]/(y^2-x^3-x^2)$ , because e.g. $t^2-x-1=0$ . Furthermore $k[x,y]/(y^2-x^3-x^2)\subset k[t]\subset K$ (in particular they have the same field of fractions) because of the relations $x=t^2-1$ and $y=t^3-t$ and $t=\frac yx$ . And finally $k[t]$ is integrally closed, because it is a unique factorization domain. Thus $k[t]$ is the integral closure of $k[x,y]/(y^2-x^3-x^2)$ in its field of fractions.

Let’s examine the normalization map more closely; in particular, the fiber over the node. The node is

$\displaystyle {\rm Spec} \left(k[x,y]/(y^2-x^3-x^2)\right)_{(x,y)}\Big/(x,y)={\rm Spec} k \to{\rm Spec} k[x,y]/(y^2-x^3-x^2)$

$\displaystyle 0 \mapsfrom x$

$\displaystyle 0 \mapsfrom y,$

and the fiber over this point is

$\displaystyle {\rm Spec} k\times_{{\rm Spec} k[x,y]/(y^2-x^3-x^2)}{\rm Spec} k[t]\cong{\rm Spec} k\otimes_{k[x,y]/(y^2-x^3-x^2)}k[t].$

Since $t^2=x+1$ , every element of $k[t]$ can be written $f+tg$ for some $f=\sum a_ix^i$ and $g=\sum b_ix^i$ in $k[x]$ . We can transfer $x$ through the tensor, and on the left hand side it becomes zero, so

$\displaystyle 1\otimes (f+tg)=1\otimes (a_0+tb_0)=a_0(1\otimes 1)+b_0(1\otimes t).$

An element of the form on the right is zero precisely when $a_0=b_0=0$ (since $t$ is not in $k[x,y]/(y^2-x^3-x^2)$ ), so the representation is unique. We can also compute that the multiplication is

$\displaystyle \Big(a(1\otimes 1)+b(1\otimes t)\Big)\Big(a'(1\otimes 1)+b'(1\otimes t)\Big) =(aa'+bb')(1\otimes 1)+(ab'+a'b)(1\otimes t).$

This realizes $k\otimes_{k[x,y]/(y^2-x^3-x^2)}k[t]$ as $k\times k$ , via the isomorphism

$\displaystyle k\otimes_{k[x,y]/(y^2-x^3-x^2)}k[t] \to k\times k$

$\displaystyle a(1\otimes 1) +b(1\otimes t) \mapsto (a+b,a-b).$

Thus the fiber over the node is ${\rm Spec} k\coprod{\rm Spec} k$ , two points with residue field $k$ , i.e. normalization splits the node into two points.

The cuspidal cubic ${\rm Spec} k[x,y]/(y^2-x^3)$ is similar: essentialy the same argument shows that its normalization is also $\mathbb{A}^1$ , with the following map.

$\displaystyle \mathbb{A}^1_k \to{\rm Spec} k[x,y]/(y^2-x^3)$

$\displaystyle t^2 \mapsfrom x$

$\displaystyle t^3 \mapsfrom y$

Here the cusp is

$\displaystyle {\rm Spec} \left(k[x,y]/(y^2-x^3)\right)_{(x,y)}\Big/(x,y)={\rm Spec} k \to{\rm Spec} k[x,y]/(y^2-x^3)$

$\displaystyle 0 \mapsfrom x$

$\displaystyle 0 \mapsfrom y,$

and the fiber over this point is

$\displaystyle {\rm Spec} k\times_{{\rm Spec} k[x,y]/(y^2-x^3)}{\rm Spec} k[t]\cong{\rm Spec} k\otimes_{k[x,y]/(y^2-x^3)}k[t].$

The same analysis as above (using $t^2=x$ ) shows every element of this tensor product can be uniquely written $a(1\otimes 1)+b(1\otimes t)$ . We can also compute that the multiplication is

$\displaystyle \Big(a(1\otimes 1)+b(1\otimes t)\Big)\Big(a'(1\otimes 1)+b'(1\otimes t)\Big) =aa'(1\otimes 1)+(ab'+a'b)(1\otimes t).$

This realizes $k\otimes_{k[x,y]/(y^2-x^3-x^2)}k[t]$ as $k[T]/T^2$ , via the isomorphism

$\displaystyle k\otimes_{k[x,y]/(y^2-x^3-x^2)}k[t] \to k[T]/T^2$

$\displaystyle a(1\otimes 1) +b(1\otimes t) \mapsto a+bT.$

Thus the fiber over the node is ${\rm Spec} k[T]/T^2$ , one point with nilpotents, i.e. normalization straightens out the cusp into a single point by somehow adding first-order behavior.

For one last geometric example, consider the cubic ${\rm Spec} k[x,y]/(y^2-x^3+x^2)$ . This has a node at the origin which appears as an isolated point e.g. if we are working over $\mathbb{R}$ , but looks like a self-intersection over $\mathbb{C}$ . That is, it has two distinct tangents, but they may live over a quadratic extension of the base field.

The normalization is again ${\rm Spec} k[t]$ with $t=\frac yx$ , and the fiber over the origin is ${\rm Spec} k[T]/(T^2+1)$ . So in this case the behavior of the normalization at the singular point depends on the ground field; if $\sqrt{-1}\in k$ , then the node is split into two copies, but if $\sqrt{-1}\notin k$ then the fiber is a single point with residue field a quaratic extension of $k$ .

Now consider the arithmetic curve ${\rm Spec} \mathbb{Z}[15i]$ . The field of fractions of $\mathbb{Z}[15i]$ is $\mathbb{Q}(i)$ , whose ring of integers is $\mathbb{Z}[i]$ , so this is the integral closure of $\mathbb{Z}[15i]$ (since $\mathbb{Z}\subset\mathbb{Z}[15i]\subset\mathbb{Z}[i]$ ). Thus the normalization is given by

$\displaystyle {\rm Spec}\mathbb{Z}[i]\to{\rm Spec}\mathbb{Z}[15i]$

corresponding to the inclusion $\mathbb{Z}[15i]\to\mathbb{Z}[i]$ .

If $\mathfrak{p}$ is a prime of $\mathbb{Z}[15i]$ lying over a prime $p\neq3,5$ of $\mathbb{Z}$ , then $\mathfrak{p}\mathbb{Z}[i]+\mathbb{Z}[15i]=\mathbb{Z}[i]$ . In this case

$\displaystyle \mathbb{Z}[i]/\mathfrak{p}^e\cong\mathbb{Z}[15i]/\mathfrak{p},$

which shows that the fiber of ${\rm Spec}\mathbb{Z}[i]\to{\rm Spec}\mathbb{Z}[15i]$ over $\mathfrak{p}$ is

$\displaystyle {\rm Spec}\mathbb{Z}[i]\otimes_{\mathbb{Z}[15i]}(\mathbb{Z}[15i]/\mathfrak{p}) \cong{\rm Spec}\mathbb{Z}[i]/\mathfrak{p}^e \cong{\rm Spec}\mathbb{Z}[15i]/\mathfrak{p},$

just a copy of the point $\mathfrak{p}$ .

The primes $3$ and $5$ remain prime in $\mathbb{Z}[15i]$ , for

$\displaystyle \mathbb{Z}[15i]/3\cong\mathbb{Z}/3\hspace{2em}\text{ and }\hspace{2em} \mathbb{Z}[15i]/5\cong\mathbb{Z}/5.$

Recall from our knowledge of prime splitting in quadratic fields that

$\displaystyle \mathbb{Z}[i]/3\cong\mathbb{F}_9\hspace{2em}\text{ and }\hspace{2em}\mathbb{Z}[i]/5\cong\mathbb{Z}/5\times\mathbb{Z}/5$

Thus the fiber of the normalization ${\rm Spec}\mathbb{Z}[i]\to{\rm Spec}\mathbb{Z}[15i]$ over $3$ is

$\displaystyle {\rm Spec}\mathbb{Z}[i]\otimes_{\mathbb{Z}[15i]}\mathbb{Z}[15i]/3 \cong{\rm Spec}\mathbb{Z}[i]/3 \cong{\rm Spec} \mathbb{F}_9,$

and the fiber over $5$ is

$\displaystyle {\rm Spec}\mathbb{Z}[i]\otimes_{\mathbb{Z}[15i]}\mathbb{Z}[15i]/5 \cong{\rm Spec}\mathbb{Z}[i]/5 \cong{\rm Spec} \mathbb{Z}/5\times\mathbb{Z}/5 \cong{\rm Spec}\mathbb{Z}/5\coprod{\rm Spec}\mathbb{Z}/5.$

So we can imagine $5$ to be a node, because normalization splits it into two copies of itself. We can also imagine $3$ to be a node, but of the “isolated point” type, in analogy with the isolated real point of $y^2=x^3-x^2$ .

Algebr. Teahouse J. Math.

Normalization of Algebraic and Arithmetic Curves

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