# Concrete syzygy generators

Consider $g_1=x^2, g_2=y^2, g_3=xy+yz\in k[x,y,z]$ with a field $k$. We consider the reverse lexicographic order, and put $x>y>z$. I want to find the generators of the syzygies.

Eisenbud CA book, p739, exercise 15.27, says that it is $$(y^2,-x^2,0),(0,x+z,-y),((x+z)y,0,-x^2).$$

However my computation yields $$(y^2,-x^2,0),(0,x+z,-y),(y,0,-x+z).$$

Which is the correct generators?

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Tom, you need to go back to some of your previous questions and accept the best answers by clicking the check mark between the voting arrows. That's how people can tell a question doesn't need any more attention. – Kevin Carlson Sep 27 '12 at 11:58
Thank you Kevin. – Tom Sep 27 '12 at 12:08
But in my understanding,I have voted for best answers of almost all of my questions. It is enough to click the upward arrow beside the answer, isn't it? Or don't I understand the system of stackexchange? – Tom Sep 27 '12 at 12:11
Oh, not the upward arrow but the check mark! I did it just right now! Thank you. – Tom Sep 27 '12 at 12:16
Thanks for doing that! – Kevin Carlson Sep 27 '12 at 12:33

Eisenbud has got it right here. You can check directly that your third proposed generator is not an element of the kernel of the map from the free $k[x,y,z]$-module on basis $g_1,g_2,g_3$ to $(g_1,g_2,g_3)\subseteq k[x,y,z]$: $(y,0,-x+z)$ maps to $x^2y-x^2y-xyz+xyz+yz^2=yz^2$. On the other hand $((x+z)y,0,-x^2)$ maps to $x^3y+x^2yz-x^3y-x^2yz=0$.
Thank you. I understood why mine is wrong and why his is correct. But how can we find his? I know its Grobner basis is $g_1,g_2,g_3,yz^2$. – Tom Sep 27 '12 at 13:17
I don't know anything about Grobner bases, but all it takes to find these is to put $g_i$ and $g_j$ in each other's coordinate for each pair $i,j$ and divide out the greatest common factor. I suppose that sort of method isn't robust enough to use in general. – Kevin Carlson Sep 27 '12 at 21:44