I am totally stack in how to do this exercise : How can I find the irreducible components of $ V(X^2 - XY - X^2 Y + X^3) $ in $A^2$(R)?

Given an algebraic set, what is the consideration I have to make to find them? I know the definitions of components and irreducible components but these don't help in to solve this kind of exercise in the end.

Can anyone suggest a good reference book also?



2 Answers 2


Write your polynomial as a product of irreducible polynomials. The vanishing loci of these irreducible polynomials are then the irreducible components of your variety.


$V(f)$, the zero set of a polynomial $f \in R[x,y]$, is given as set of points such that $f(x,y) = 0$. If this polynomial is reducible, i.e. $f=gh$, the points $(x_0, y_0)$ in $V(f)$ are such that $g(x_0, y_0) = 0$ or $h(x_0, y_0) = 0$. This means that $V(f)$ is a union of $V(g)$ and $V(h)$.

In your case, $f(x,y)=x^2-xy-x^2y+x^3=x(x+1)(x-y)$. Zeros of this polynomial are all the points for which any factor of $f$ is zero. Thus we see that $V(f) = V(x) \cup V(x+1) \cup V(x-y)$. Geometrically, we can imagine this in $\mathbb{R}^2$ as a union of three lines, the y-axis, diagonal line through the origin passing through the first quadrant and parallel to the y axis passing through the point $(-1,0)$.

As for the books, very nice introductory book about algebraic curves which is freely available is by Fulton: http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf. Quick google search also turned up these notes which seem to contain some nice examples and theorems ilustrating what is going on: http://csclub.uwaterloo.ca/~mlbaker/get.php?name=LW-1135-pmath764notes.pdf.

  • $\begingroup$ Would this change if we consider $A^2(C)$? I don't think we have used anywhere the fact that we are in the field of real numbers $\endgroup$
    – User
    Mar 23, 2014 at 9:34
  • $\begingroup$ You are right, this will work in C as well. Actually I think this will work in any field (not just characteristic 0). $\endgroup$
    – BoZenKhaa
    Mar 23, 2014 at 9:36

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