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The book "Algebraic Curves" by Fulton is available free for download on his website.

On page 27, Fulton constructs an isomorphism which is used several times throughout the book. His construction is very concrete.

Does anyone know what is really going on? Where does this isomorphism really come from?

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up vote 4 down vote accepted

Are you talking about Proposition 6? This is a form of the Chinese remainder theorem. Geometrically it says that if $V(I)$ is finite we can identify $k[x_1, ... x_n]/I$ (the ring of functions on $V(I)$) with the direct product of the local rings of functions on each of the points in $V(I)$. In other words, a function on $V(I)$ (in the generalized sense) is determined by what it does in a "neighborhood" of each point.

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Proposition 6 (page 27) of Fulton's Algebraic Curves is the one you refer to, I'll include it here for convienience:

Let $I$ be an ideal of $k[X_1,\cdots, X_n]$ (k algebraically closed), and suppose $V(I)=\{P_1,\cdots, P_N\}$ is finite. Let $\mathcal{O}_i = \mathcal{O}_{P_i} (\mathbb{A}^n ).$ Then there is a natural isomorphism $$\frac{k[X_1,\cdots,X_n]}{I} \cong \prod_{i=1}^N \frac{\mathcal{O}_i}{I\mathcal{O}_i}.$$

Recall that $\mathcal{O}_P (V)$ denotes the ring of rational functions $V\to k$ defined at $P\in V.$ The isomorphism is given by $f(X_1,\cdots, X_N)+I \mapsto ( f(X_1,\cdots, X_N)+I\mathcal{O}_1, \cdots, f(X_1,\cdots, X_N)+I\mathcal{O}_N. $ It is easy to see this is a well defined ring homomorphism but harder to see why it is invertible.

The "moral" behind the isomorphism is that that if we know the behavior of a polynomial $f\in k[X_1,\cdots, X_n]$ near each point in $V(I)$ then we can actually recover $f+I.$ This is the best we could expect since elements of $I$ behave like $0$ near $V(I).$

To try to see the actual content behind this isomorphism, it is useful to look at some special cases, for example when $n=1.$ In this case, for any proper ideal $I$ (the theorem is boring if $I=(1)$) we can write $I=(f)$ where $f$ is a monic polynomial of degree greater than $0.$ It could be useful to go through the following in the special case $f(X)=X^3-1$ (no repeated roots), then $f(X)=X^2(X-1)$ and then for general $f.$ Let $n$ now denote $\deg f.$

Since $k$ is algebraically closed, we can write $f(X) =\displaystyle \prod_{i=1}^m (X-a_i)^{\lambda_i}$ where $a_i$ are the distinct roots of $f$ with multiplicity $\lambda_i$ (so $\sum \lambda_i = n$). We have $V(I) = \{ a_1,\cdots, a_m\}.$ Let $\mathcal{O}_i = \mathcal{O}_{a_i} (\mathbb{A}).$

We want to extract enough information from a given $$(g_1(X)+I\mathcal{O}_1, \cdots, \ g_m(X)+ I\mathcal{O}_m)$$ to determine a class $p(X)+I$ whose functions imitate that local information and it suffices to determine a polynomial of degree $n-1$. We do this by letting $p(X)$ be the unique polynomial of degree $n-1$ such that the $n$ conditions hold:

  1. $p(a_i)=g_i(a_i)$ for all $i\leq m$
  2. $p'(a_i)=g_i'(a_i)$ for every double root
  3. $p''(a_i)=g_i''(a_i)$ for every triple root etc.

With some (elementary but not instant) calculation one can verify that these values $g^{(l)}_i (a_i)$ are indeed well defined in that the values exist, do not depend on the choice of coset representative, and that this mapping is the inverse of the natural ring homomorphism, which is thus an isomorphism.

Let us try a case in two variables now: $I=(X^2,XY,Y^2)$ so $V(I)=\{ (0,0)\}.$ Let $\mathcal{O}=\mathcal{O}_{(0,0)}(\mathbb{A}^2).$ Using only local information near $(0,0)$ we want to invert the map $$\varphi : p(X,Y)+I \mapsto p(X,Y)+I\mathcal{O}.$$ We need only determine an appropriate linear polynomial, since $I$ kills higher terms. Suppose we are given the coset $g(X,Y)+I\mathcal{O}.$ To imitate functions in this coset near $(0,0)$ choose the polynomial $g(0,0)+g_X(0,0) X+g_Y(0,0)Y.$ Check this is a well defined choice so that we now have a map $\phi :\mathcal{O}/I\mathcal{O} \to k[X,Y]/I.$ Finally, check this map inverts the homomorphism. For example, verifying that $ \varphi \circ \phi$ is the identity amounts that showing that for all $g(X,Y)\in \mathcal{O},$ there exists $r_i\in \mathcal{O}$ such that $$ g(X,Y) - g(0,0) - g_X(0,0) X - g_Y(0,0) Y = X^2r_1 + XYr_2+Y^2r_3.$$ Once we've done that, we've confirmed that $\varphi$ is an isomorphism.

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