Calculating a free resolution of $\mathbb Q[x,y,z]/I$ where $I = (x,y,z)$

Question

Let $R=\mathbb Q[x,y,z]$ and $I = (x,y,z)$. I am trying to find the minimal free resolution of $R/I$.

This is what I have got:

$R \rightarrow R/I$ whose kernel is $I$, which is generated by 3 elements, so I would need $R^3$ to map to $R$ by sending each generator of $R^3$ to $x, y$ and $z$ respectively. I am stuck at finding the generator of the kernel of this map. Also how to perform this in Macaulay 2 ?

Answer 1

Although in the comments the essential hint has been given, to use the Koszul-complex, I give the calculation with Macaulay2 below. As you see, it is simply the command 'res (R^1/id1)' in line i4. Note that 'R/id1' instead of 'R^1/id1' would give an error, as the argument of res must be a module, not a ring. With the commands C.dd_(number) you get the maps between the modules in the complex. The generators that you are looking for are in C.dd_2

i1 : R=QQ[x,y,z]

o1 = R

o1 : PolynomialRing

i2 : id1 = ideal gens R

o2 = ideal (x, y, z)

o2 : Ideal of R

i4 : res (R^1/id1)

      1      3      3      1
o4 = R  <-- R  <-- R  <-- R  <-- 0

     0      1      2      3      4

o4 : ChainComplex

i5 : C= oo

      1      3      3      1
o5 = R  <-- R  <-- R  <-- R  <-- 0

     0      1      2      3      4

o5 : ChainComplex

i6 : C.dd_1

o6 = | x y z |

             1       3
o6 : Matrix R  <--- R

i7 : C.dd_0

o7 = 0

                    1
o7 : Matrix 0 <--- R

i8 : C.dd_2

o8 = {1} | -y -z 0  |
     {1} | x  0  -z |
     {1} | 0  x  y  |

             3       3
o8 : Matrix R  <--- R

i9 : C.dd_3

o9 = {2} | z  |
     {2} | -y |
     {2} | x  |

             3       1
o9 : Matrix R  <--- R