Subspace of $\Bbb C^2 $ as Riemann Surface I am reading on the construction of Riemann Surface. From the myriad of theorems and lemmas, I am unable to understand a clear 'method' of identifying Riemann surfaces. 
For example, consider $$X  = \{ (z, w) \in \mathbb{C}^2 \mid zw = 0 \}.$$ I have a hunch that this is not a Riemann Surface. One reason could be this: 
Set $P(z, w) = zw$.
Then $ \frac {\partial P}{ \partial w} = z $. This will be $0$ when $z = 0$. But given $z = 0$ , $P(z,w) = zw = 0$ for infinitely many values of $w$. 
However, for some reason which is not clear to me, this creates a problem. 
Can someone explain to me clearly 


*

*how to identify Riemann surfaces from equations such as above?

*is my reasoning (mentioned above) is in the right track

*how to find the atlas once I have identified a subspace (of $ \mathbb{C}^2 $ ) as a Riemann surface? For example $ X  = \{ (z, w) \in \mathbb{C}^2 \mid 3z - 14w^2 = 0 \} $ has what atlas? Is it $ \phi _1 = \sqrt {\frac{3}{14} z}, \phi _2 = -\sqrt {\frac{3}{14} z} $ ?

 A: For the particular surface $$X := \{zw = 0\} ,$$ we can see that the Jacobian of $P(z, w) = zw$, of which $X$ is a level set, is
$$\pmatrix{\frac{\partial P}{\partial z} & \frac{\partial P}{\partial w}} = \pmatrix{w&z}$$
and so vanishes only at $(0, 0)$, which is contained in $X$. Thus, if there is a problem, it will occur at that point. Put more precisely, $X - \{(0, 0)\}$ is a Riemann surface, as it is the level set at a regular value of the holomorphic function $P\vert_{\Bbb C^2 - \{(0, 0)\}}$.
Now, can you show that $X$ isn't a Riemann surface by considering the topology of $X$ near $(0, 0)$? (Note that on the other hand, $X$ is the union of the two Riemann surfaces $\{z = 0\}$ and $\{w = 0\}$.)
In general, finding an explicit atlas for a Riemann surface can be difficult, but note that the particular surface $$Y := \{3 z - 14 w^2 = 0\}$$ in (3) is the graph of a function $f(w)$, so $Y$ admits a preferred global chart, $$\Bbb C \to Y, \qquad w \mapsto (w, f(w)) .$$
One can use graph charts like the $\phi_1, \phi_2$ in your solution, but one must be a good deal more careful: The two functions $\phi_i$ are not smooth at $0$, and simply by writing down $\sqrt{\cdot}$ we must (at least implicitly) make a choice of branch cut. So, $\phi_1$ and $\phi_2$ do not together cover $Y$.
A: There's a Riemann surface for $z = w^2$ (or $w = \sqrt{z}$); for this case $$
P(z, w) = z - w^2
$$
and $\frac{\partial P} {\partial w} = 0$ for $w = 0$, so having a partial derivative that's zero isn't a surface-killer. So I guess that means that the answer to your question 2 is "no". 
As a kind of general rule, think about a curve in the plane, like, say, the unit circle. At almost every point (except $(\pm 1, 0)$) the map 
$$
(x, y) \mapsto x
$$
is a good coordinate function. That's because the tangent space to the curve projects to the $x$-axis nicely (i.e., projects to the whole axis rather than to a single point, as it does where there are vertical tangents). In those two "bad" places, we can tilt our heads and use the $y$-axis and the map
$$
(x, y) \mapsto y
$$
to map a neighborhood of each point in a nice way to the $y$-axis (which we regard as $\Bbb R$). 
This generally works: take a nice curve, and for most point $P$ on the curve, a neighborhood of $P$ projects to the $x$-axis nicely, and when it doesn't, some neighborhood projects to the $y$-axis nicely. 
There are exceptions, however: The curve defined by 
$$
t \mapsto \begin{cases} (-t^2, 0) & x \le 0 \\ (0, t^2) & x >  0\end{cases},
$$
for instance, projects badly (near $t  =0$, i.e., $x = y = 0$, onto both axes. 
As Travis notes, the critical thing is for your defining polynomial to have a rank-1 jacobian at every point (the same thing's true for curves in the plane!). 
And essentially the same argument applies: at almost every point, projection $(z, w) \mapsto z$ will be a chart; at points where it's not, projection $(z, w) \mapsto w$ will be chart. The transition function between these charts may, however, be complex. 
