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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
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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
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1 Answer

up vote 2 down vote accepted

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 $x^3y+x^2yz-x^3y-x^2yz=0$.

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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
    
Thank you. I understood. I should have read CA book more carefully. But now I have learned meuch better. Thanks –  Tom Sep 28 '12 at 12:15
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