To get started, you must define the polynomial ring you will be using. Here are some examples:

Assuming you've already defined your ring with variables x and y, here's how to define an ideal named I:

Here's how to find the square of I:

Note that the output for I^2 is not a minimal set of generators. Here's how to get a minimal set:

Given an ideal, here's how to find the ideals of the irreducible components of V(I), with output given as a list of ideals:

Here's how to find the radical of I:

Here's a useful operation: given ideals I and J, I:J is the ideal of all f in your ring R such that fg is in I for all g in J. So, to find the least power of x-y that is in ideal(x*y-x^2, y^2-x^3), keep trying ideal(x*y-x^2, y^2-x^3) : ideal((x-y)^n) until you find an n such that the output is the ideal (1) (you may want to use mingens to simplify the output):

Here's how to intersect a list of ideals:

When you're done, just quit:

So here's the homework:

[1] In Problem [1](b)(i) from Homework 2, use

Solution: We see from below, over the rationals, that m = 3 and n = 2, since m = 3 is the least m such that ideal(y+x*y-x^2,y^2+x*y^2-x^3) : ideal(x^m) = (1) and n = 2 is the least n such that ideal(y+x*y-x^2,y^2+x*y^2-x^3) : ideal(y^2) = (1):

[bharbour@math ~]$ M2 Macaulay 2, version 0.9 --Copyright 1993-2001, D. R. Grayson and M. E. Stillman --Singular-Factory 1.3b, copyright 1993-2001, G.-M. Greuel, et al. --Singular-Libfac 0.3.2, copyright 1996-2001, M. Messollen i1 : R = QQ[x,y] o1 = R o1 : PolynomialRing i2 : I = ideal(y+x*y-x^2,y^2+x*y^2-x^3) 2 3 2 2 o2 = ideal (- x + x*y + y, - x + x*y + y ) o2 : Ideal of R i3 : mingens (I : ideal(x)) o3 = | y x2 | 1 2 o3 : Matrix R <--- R i4 : mingens (I : ideal(x^2)) o4 = | y x | 1 2 o4 : Matrix R <--- R i5 : mingens (I : ideal(x^3)) o5 = | 1 | 1 1 o5 : Matrix R <--- R i6 : mingens (I : ideal(y)) o6 = | y x | 1 2 o6 : Matrix R <--- R i7 : mingens (I : ideal(y^2)) o7 = | 1 | 1 1 o7 : Matrix R <--- R[2] Given an algebraic set X, a theorem in commutative algebra says that Radical(I(X)) is the intersection of the ideals of the irreducible components of X. Demonstrate this in

Solution: Here we see that Radical(I(X)) and the intersection of the ideals of the irreducible components of X (working over the rationals) both give the same ideal, namely ideal(x-y, y2-y):

i9 : J = ideal(x*y-x^2, y^2-x^3) 2 3 2 o9 = ideal (- x + x*y, - x + y ) o9 : Ideal of R i10 : mingens radical J o10 = | x-y y2-y | 1 2 o10 : Matrix R <--- R i11 : mingens intersect decompose J o11 = | x-y y2-y | 1 2 o11 : Matrix R <--- R[3] Do problems #25(b) on p. 17 and #31 on p. 20. Use

Solution: First, consider #25(b) on p. 17. Since (y^2+x)(y^2-x) = y^4-x^2 and (y^2+x)(y - x)(y + x) = y^4-x^2*y^2+x*y^2-x^3, we know V(y^4-x^2, y^4-x^2*y^2+x*y^2-x^3) is the union of V(y^2+x) and V(y^2-x, (y - x)(y + x)). Now, V(y^2+x) is irreducible: it is infinite (in fact, a sideways parabola) and y^2+x is irreducible (since it has degree 1 regarded as a polynomial in x with coefficients that may involve y). Thus, by Corollary 2 on p. 19 of Fulton's book, V(y^2+x) is irreducible. Next, V(y^2-x, (y - x)(y + x)) is the intersection of the degenerate hyperbola (y - x)(y + x) = 0 with y^2-x = 0, which just gives three points, (0,0), (1,1) and (1, -1). Since (0,0) is already in V(y^2+x), the irreducible components of V(y^4-x^2, y^4-x^2*y^2+x*y^2-x^3) are: V(y^2+x), {(1,1)} and {(1, -1)}.

Here is what

i12 : decompose ideal(y^4-x^2, y^4-x^2*y^2+x*y^2-x^3) 2 o12 = {ideal (x - 1, y - 1), ideal (x - 1, y + 1), ideal(y + x)} o12 : ListAnd here is a graph:

Now consider #31(a) on p. 20. First we have V(y^2-x*y-x^2*y+x^3). But y^2-x*y-x^2y+x^3 = (y - x)(y - x^2), so by Cor. 3 on p. 20 (over the complexes) or by Cor. 2 on p. 19 (over the reals) the irreducible components of V(y^2-x*y-x^2y+x^3) are the line V(y - x) and the parabola V(y - x^2).

Here is what

i13 : decompose ideal(y^2-x*y-x^2*y+x^3) 2 o13 = {ideal(x - y), ideal(x - y)} o13 : ListAnd here is a graph:

Next there is #31(b): consider first V(x^3+x-x^2*y-y). Since x^3+x-x^2*y-y = (x^2 + 1)(x - y), the irreducible components of V(x^3+x-x^2*y-y) are the three lines V(x - y) and V(x - i) and V(x + i) over the complexes, and just the one line V(x - y) over the reals.

Here is what

i14 : decompose ideal(x^3+x-x^2*y-y) 2 o14 = {ideal(x + 1), ideal(x - y)} o14 : ListFinally, consider V(y^2-x^3+x). Clearly, y^2-x^3+x = 0 has infinitely many solutions over the reals (or the complexes): for any value of y, x^3 - x - y^2 has some root in x since it has odd degree in x. Thus by Corollary 2 on p. 19 V(y^2-x^3+x) is irreducible as long as y^2-x^3+x is. But y^2 + (x - x^3) is irreducible by "Eisenstein's criterion" (since k[x] is a domain, and x - x^3 is in the prime ideal (x), but not in (x^2)).

Here is what

i15 : decompose ideal(y^2-x^3+x) 3 2 o15 = {ideal(- x + y + x)} o15 : ListAnd here is a graph: