Compute a Gröbner basis for $I=\langle f_1,f_2,f_3\rangle$.

Question

Using lexicographic order compute a Gröbner basis for $$I=\langle f_1=xy^2-xy+y,f_2=xy-z^2,f_3=x-yz^4\rangle\subset \Bbb R[x,y,z]$$

I was strictly using these notes to compute a Gröbner basis.

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When the computation gets messy I used a computer software at least to know the answer, but the software gave me $$\{z^8-z^6-2 z^4-z^2, y-z^6+2 z^4, x-z^6+z^4+z^2\}$$
as a Gröbner basis, which I wouldn't even imagine, because the elements in $I$ are not included. I was expecting the answer to be like $\langle f_1,f_2,f_3,\ldots\rangle$.

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Edited: How can I compute a Gröbner basis by hand?

Answer 1

The Gröbner basis depends on the monomial order you choose. If you use lexicographical order you do get (in M2; commented out -- is the answer):

R=QQ[x,y,z,MonomialOrder=>Lex];
I=ideal(x*y^2-x*y+y,x*y-z^2,x-y*z^4);
gens gb I -- | z8-z6-2z4-z2 y-z6+2z4 x-z6+z4+z2 |
J=ideal oo
I==J -- true

But if you use graded reverse lexicographical order (default for M2):

R=QQ[x,y,z];
I=ideal(x*y^2-x*y+y,x*y-z^2,x-y*z^4);
gens gb I -- | y2-2z2+x+y xy-z2 x2-2z2-2x+y yz2-z2+y xz2-2z2-x+y z4-z2-x+y |
J=ideal oo
I==J -- true

In either case the ideals are the same, what needs to be true for a Gröbner basis (that it is a generating set and) that the initial terms generates the initial ideal for that monomial order.

This means that your original generators could be part of the list but that the initial terms aren't needed for generating the initial ideal.