Suppose we have the curve $Z\subset \mathbb{P}^2$ given by the equation $y^2z-x^3=0.$ I have to find a basis for the tangent space at $(0:0:1)$, but I find the definition hard:
Let $X$ be a variety and $a\in X$. Then $T_X(a)=(\mathfrak{m}/\mathfrak{m}^2)^*$ where $m\subset O_X(U)$ ($O_X$ is the sheaf of regular functions) is the maximal ideal of $x$ and $U$ an affine open containing $x$.
To use this in my situation, we have $X=Z$ and the open set containing $(0:0:1)$ should be $U=Z\cap \{(a:b:1)\colon a,b\in k\}$. So now I have to know how $O_Z(U)$ looks like. I believe it is just $k[x,y]/(y^2-x^3)$ but I am not sure if this is true and how to show it if it is true. So $\mathfrak{m}=(x,y)$ and so $\mathfrak{m}/\mathfrak{m}^2$ has basis $\{x,y\}$.
My question: is this actually correct and are there easier ways to compute tangent spaces?
Yes, $O_X(U) \cong k[x,y]/(y^2 - x^3)$. To give this full justification depends on your foundations, I guess, but you can think of this in the following way: The equation $y^2z - x^3$ cuts out some points in projective space, and convenient covers are given by the planes $z = 1$, $x = 1$, $y = 1$, which in projective coordinates means the set of points $[x:y:z]$, with $z \not = 0$, etc. (Mentally I am imagining the space of all lines through the origin in 3-space, and the lines that intersect these planes are those in the cover. This may remind you of an old trick for representing affine transformations of a plane as linear ones, by embedding your plane at $z = 1$ in 3-space...)
On the space of points with projective coordinates $[x:y:z]$, where $z \not = 0$, the ratios $x/z$ and $y/z$ become honest functions (you can plug in points and get specific numbers), and these are affine coordinates on this patch (the points on this patch is the data of a plane, $x$ and $y$ can be arbitrary numbers, and we have given two functions that spit out the coordinates). Or, again remembering that our space is that of lines in 3-space, for lines with nonzero z-coordinate, $x/z$ and $y/z$ are well defined "slopes" of those lines, and these two numbers uniquely and freely determine any one of the lines that intersects the plane $z = 1$.
Then the equation $y^2z - x^3 = 0$ is equivalent (dividng both sides by $z^3$, which we have assumed to be nonzero) to $(y/z)^2 - (x/z)^3 = 0$, and on this patch the variety we have defined is given by $k[(y/z),(x/z)]/((y/z)^2 - (x/z)^3 )= 0$. At this point is just convenient to set $z = 1$, but you could just as well call $y/z = Y$ and $x/z = X$.
Now let $m = (X,Y)$, so $m^2 = (X^2, XY, Y^2)$. The equation $Y^2 - X^3$ involves no linear pieces, so is zero after modding out by $m^2$. What is secretly going on here is that $Y^2 - X^3$ can be thought of as the Taylor expansion of $Y^2 - X^3$ around the origin, and the you can compute that its derivatives with respect to $X$ and $Y$ are both zero, hence the Taylor expansion has no linear part (and it also has no constant part since $(0,0)$ is on the curve defined by this equation), and hence this equation becomes $0$ $mod m^2$. Similarly, if $f$ is some polynomial defining a plane curve passing through the origin, then you can Taylor expand around $(x,y)$ to see what equations are imposed on $m /m^2$. If you think this through you will arrive at the Jacobian criterion for the smoothness, in particular you will recover the calculus fact that the tangent space of a hypersurface is always normal to the gradient vector - though you will have to interpret this correctly (in multivariable calculus various identifications are made between the tangent and cotangent space, once you undo them you will have a statement that makes algebraic-geometric sense).
A good reference for all of this stuff, done in a very geometric fashion, is book I of Shafarevich "Varieties in Projective Space."