Concrete Bezout's Theorem Calculation

You're leaving out an important part of Bezout's theorem: you need to count the intersections with multiplicity. The intuition is then that if two curves intersect with multiplicity $m$ at a point $P$, we can slightly jiggle our curves to see $m$ actual intersections there.

To start, let's calculate some intersection multiplicities to see what should happen. Remember that the intersection multiplicity at a point $P$ of curves cut out by equations $f,g\in\mathcal{O}_{\Bbb P^2,P}$ is $\dim_k \mathcal{O}_{\Bbb P^2,P}/(f,g)$.

• $P=[0:0:1]$: the ring in question is $k[x,y]_{(x,y)}/(y-x^3,y-x^2)$. Making the change of coordinates $y\mapsto y+x^2$, this is isomorphic to $k[x,y]_{(x,y)}/(y+x^2-x^3,y)\cong k[x]_{(x)}/(x^3-x^2)$, and as $x^3-x^2=(x-1)x^2$ and $x-1$ is a unit, we have that the dimension of this ring as a $k$-vector space and thus the intersection multiplicity is two.
• $P=[1:1:1]$: the ring in question is $k[x,y]_{(x-1,y-1)}/(y-x^3,y-x^2)$. Making the same coordinate change as last time, we have that this ring is isomorphic to $k[x,y]_{(x-1,y)}/(y+x^2-x^3,y)\cong k[x]_{(x-1)}/(x^2(x-1))\cong k$, so the intersection multiplicity is one.
• $P=[0:1:0]$: the ring in question is $k[x,z]_{(x,z)}/(z-x^2,z^2-x^3)$. This is isomorphic to $k[x]_{(x)}/(x^4-x^3)$ by the map sending $z\mapsto x^2$, so we get that the intersection multiplicity is three.

At the first point, it's not so hard to see what happens when we wiggle: considering $y=x^3$ and $y=x^2-c$ for small positive $c$, we get two roots. The second point needs no perturbation, and the third is a little interesting. Considering $z=x^2$ and $z^2=x^3+c$ for small $c$, we see that our equation reduces to $x^4=x^3+c$, which for small $c$ is approximately $(x-1)(x^3+c)$, which can be seen to have exactly three roots near $x=0$. So everything is as it should be.

KReiser has given a nice explanation of what's going on with your original system of equations. Let me discuss how to correctly investigate the perturbation with $\alpha$ and $\beta$ that you tried.

First, you made a mistake when homogenizing the second equation: it should be $$yz^2 = x^3 + \beta z^3,$$ not $$yz = x^3 + \beta z^3,$$ since every term needs to have degree $3$.

Now let's look at what happens when $x=1$. The equations become $$yz=1+\alpha z^2$$ and $$yz^2=1+\beta z^3.$$ These can be combined to get $$z+\alpha z^3=1+\beta z^3$$ which for generic $\alpha$ and $\beta$ will have $3$ different solutions for $z$, all of which are nonzero. For each value of $z$ you can then solve for $y$ (since $z$ is nonzero), so this gives $3$ solutions to the system.

In your attempt, you proceeded to consider the case $y=1$, and discarded it as redundant because you can "eliminate" $y$. This doesn't make sense, though. Remember that a solution to the original homogeneous system of equations is a triple $[x:y:z]$ which is defined up to scaling, and you are putting them in one of your affine planes by scaling so that one variable is $1$. So, a solution $[x:y:z]$ can be scaled to be in the $y=1$ plane as long as its $y$-coordinate is nonzero. This solution will also show up in the $x=1$ plane as long as its $x$-coordinate is nonzero, since you could instead scale it to make the $x$-coordinate nonzero. This has nothing to do with whether $y$ can be "eliminated" from the system; rather, what is relevant is whether the resulting values of $x$ and $z$ you get are $0$ or not.

Here's how I would think about it instead. We've already found the solutions where $x$ is nonzero (so we can scale to make $x=1$), so let's consider solutions where $x=0$. As long as $y$ is nonzero, we can scale such a solution to make $y=1$, so let's set $x=0$ and $y=1$ in the equations to get $$z=\alpha z^2$$ and $$z^2=\beta z^3.$$ For generic $\alpha$ and $\beta$, this system has only $1$ solution, $z=0$.

The only remaining case is if $x$ and $y$ are both $0$, in which case $z$ must be nonzero (since we are not allowed to have all three variables $0$), and thus we can scale so $z=1$. For generic $\alpha$ and $\beta$ this does not solve the equations, so we get no solutions in this case.

So in total, we got only $4$ solutions, not $6$! What's wrong? Well, it turns out that even with your $\alpha$ and $\beta$ perturbation, these polynomials are not sufficiently "generic", i.e. they still intersect with multiplicity. Specifically, the intersection $[x:y:z]=[0:1:0]$ that we found in the second case actually has multiplicity $3$. To get a sense of why this is the case, let's look at it in the $y=1$ plane, in which our equations become $$z=x^2+\alpha z^2$$ and $$z^2=x^3+\beta z^3.$$ Isolating $x$ in the first equation and cubing, we write $x^6$ as some degree $6$ polynomial in $z$. Isolating $x$ in the second equation and squaring, we write $x^6$ as some other degree $6$ polynomial in $z$. Setting these equal, we get a degree $6$ equation in $z$ which ought to have $6$ solutions. Actually, though, it has only $4$, since $z=0$ is a triple root. We can see this quickly by observing that both of the expressions we get for $x^6$ will be divisible by $z^3$. On the other hand, if we perturbed the equations a little more (say by adding $\gamma y^2$ and $\delta y^3$ terms which would become constant terms when we set $y=1$), the degree $6$ polynomial in $z$ that we would get would generically have no repeated roots, so we would get $6$ different solutions for $z$. For each one, we could then solve uniquely for $x$ (though it takes a bit of work to see we really do get solutions that work in the original system), giving $6$ different intersection points of our curves.