I have started reading basic texts of algebraic geometry and have come across a problem that I would like some help with.
A plane algebraic curve (PAC) of degree $d$ is the set of points over the complex numbers, whose coordinates satisfy a polynomial equation of degree $d$. For example, a general PAC of degree $2$ will be of the form $ax^2+bxy+cy^2+dx+ey+f=0$.
An equation for a PAC of degree $d$ is determined by $D(d)$ coefficients where $D(d)=\binom{d+2}{2}$. Here, $D$ is just the number of coefficients in a general polynomial equation of degree $d$ in $2$ variables. If I want a curve to contain a point $P$, that gives me a linear equation/constraint on the coefficients. So, if I specify $D(d)-1$ (which is $\approx d^2/2$) points, this determines a curve uniquely ($-1$ because scalar multiples of an equation specifies the same curve) containing these points. For example, for $d=2$, we have $\binom{2+2}{2}-1=5$ points determine a conic.
Bezout's theorem states that two distinct PACs $C_1$ and $C_2$ of degree $d$ meet at $d^2$ points (counting multiplicities).
However, from the third para, these $d^2$ points can at most specify one PAC uniquely. From Bezout, we have at least $C_1$ and $C_2$ which contain these points. Contradiction!
I had some ideas why my argument may not work but I don't know how to develop them. Firstly, perhaps the $d^2$ points from Bezout are such that they do not yield independent equations, so there is no uniqueness because the linear algebra argument in the third para fails.
Secondly, the $d^2$ points that I get from Bezout need to be distinct in order for my argument to work, but I thought one could assume that is what happens in a 'general' situation.
The failure of your argument as you guess is exactly because
I think the simplest illustration is the case of plane cubics. Here there are ${3 + 2 \choose 2}=10$ coefficients, so 9 general points determine a unique cubic.
On the other hand, it is also true that 2 general cubics intersect in $3^2=9$ points. However, these points are definitely not general in the sense of the previous paragraph --- that is, they do not give linearly independent equations on coefficients of cubics. Indeed, any cubic that passes through 8 of the intersection points must also pass through the 9th (exercise!) so these 9 points really give 8 linear conditions.
For a starker illustration of the same point, you can look more closely at your dimension estimates: as you say, the family of plane curves of degree $d$ has dimension roughly $d^2/2$. So if you pick $d^2$ general points they should not lie on any curve of degree $d$. But clearly some sets of $d^2$ points do lie on a curve of degree $d$! The moral is just that when you pick a set of points with some special property ("intersection of 2 curves" or "lie on a curve of degree $d$") then you shouldn't really expect a statement that is stated for general sets of points to apply to that set.