Groebner Basis Question: Where are the other two zeros hiding?

Question

Consider the system of polynomial equations: $$f_1=x^2+y^2-1=0$$ $$f_2=(x+y)(x-y)(2x-y)=0$$ Obviously there are 6 real solutions, which are the intersecting points of the lines $y=x, y=-x,$ and $y=2x$ with the circle of radius one.
The reduced Groebner basis for the Ideal $$V(I)=\langle f_1,f_2 \rangle$$ with respect to lexicographic ordering is though: $$\{x^2+y^2-1,4xy^2-2x-2y^3+y,10y^4-13y^2+4\}$$ Now if if look at the leading terms of the three members of the basis, we see the "pure" monomials $x^2$ and $y^4$ and according to for example Bernd Sturmfels in "What is a groebner Basis?", there should be 2*4=8 complex zeros, if we count them with multiplicity. Where are the last 2 zeros? Are they complex? Are they multiple zeros?
I would also greatly appreciate a good reference on the topic multiplicity of zeros in the multivariate case.

Answer 1

You shouldn't look at the pure monomials, but at all the initial monomials. When you only look at the pure monomials and compute $2*4 = 8$ it's really like deciding that $4xy^2-2x-2y^3$ is unimportant and that you can remove it.

If I am reading the same thing (https://math.berkeley.edu/~bernd/what-is.pdf), they say that there is an isomorphism of $K$-vector spaces $K[x,y]/(I) \to K[x,y]/in(I)$.

Now since you have a Grobner basis $(g_1,g_2,g_3)$ for $I$, $in(I) = (in(g_i)) = (x^2,xy^2,y^4)$. $K[x,y]/in(I)$ turns out to have dimension $6$ (so the variety is $0$-dimensional, and the number of points is $6$) and a basis is what they call the standard monomials, $\{1,x,y,xy,y^2,y^3\}$. If you remove $xy^2$ you do get an $8$-dimensional space.

In fact the initial monomials of the elements in the Grobner basis tell you clearly how the points are sorted when you look at their $y$-coordinate then at their $x$-coordinate.

The polynomial in $y$ of degree $4$ tells you that there are $4$ different possible values for $y$.

The polynomial starting with $x(2y^2-1)$ telss you that unless $2y^2-1=0$ then $x$ can only have a single value. Of course $(2y^2-1)$ divides the polynomial of degree $4$ in $y$ (said another way, it says $x=h(y)$ for some rational fraction $h$ that has $2$ poles among the roots of the polynomial in $y$), or else this wouldn't be a Grobner basis, so for $4-2 = 2$ values of those values of $y$ there is a single possible value for $x$.

The polynomial starting with $x^2$ tells you that for any $y$ there is at most two possible values of $x$.

So you have in total $1+1+2+2 = 6$ points (counted with multiplicity).

Try doing a change of variables and express things in terms of $(x,z=y+2x)$, look at the Grobner basis in this new point of view and you should get an initial ideal $(x,z^6)$ saying there are $6$ possible $z$ values and only $1$ $x$-value for all of the $z$-values.