Eisenbud & Harris's The Geometry of Schemes proof of Prop I-18 There seems to be an error in the proof of Proposition I-18.  The
first inset equation (line 7) on page 20 only holds in $X_{f_a f_b}$. But it is
used on line 15, where it needs to hold in $X_{f_b}$.  This seems to
invalidate the proof for rings with zero divisors.
As an example, use the scheme in Exercise I-20(b).  Take $f_a$ to be
the image (coset) of $x$ in $\mathbb{C}[x]/(x^2-x)$ and $f_b$ the
image of $x-1$.  Then $f_a f_b = 0$, $X = X_{f_a} \cup X_{f_b}$ and
$X_{f_a} \cap X_{f_b} = X_{f_a f_b} = \varnothing$.  So for any $g_a \in
R_{f_a}$ and $g_b \in R_{f_b}$ there must be a $g \in R$ that becomes
$g_a$ in $X_{f_a}$ and $g_b$ in $X_{f_b}$. 
To be specific take $g_a = 1/f_a$ and  $g_b = 1/f_b$.
Then $g$ is the image of $2x-1$ in $\mathbb{C}[x]/(x^2-x)$ 
but the proof of Proposition I-18 doesn't produce it.  Am I confused, or is the proof wrong?
 A: There is no error: the ends of the equation in line 7 (i.e. $f_b^Nh_a = f_a^Nh_b$) are meant to hold in $R$. The point is that $h_a, h_b$, and $N$ should be chosen for all the intermediate equalities to hold as well. So in your example $R = \mathbb{C}[x]/(x^2-x)$ with $f_a = x$, $f_b = x - 1$, $g_a = 1/f_a$, $g_b = 1/f_b$, taking $h_a = 1 = h_b$ will not work, as this does not satisfy $f_b^N h_a = (f_af_b)^N g_a$ in $R$. 
In fact, we see that the simplest way to satisfy the equation above is by taking $h_a = f_a$, $h_b = f_b$. Thus $N = 2$ (note that $f_a$ is idempotent, but $f_b$ is not, as $f_b^2 = -f_b$), which yields $e_a = 1$, $e_b = 1$, so $g = e_ah_a + e_bh_b = f_a + f_b = 2x-1$, and indeed $f_b^2g = f_b^3 = f_b = f_b^2g_b$ in $R_{f_b}$ (and similarly in $R_{f_a}$).
A: The equality on line 7 holds in the ring $R$, that is, on the whole space $X$. The point is that by definition of a localization, if $x$ and $y$ are elements in $R$, then their images in $R_f$ are equal if and only if $f^Na = f^Nb$ in $R$. So that equality is true because the images of $g_a$ and $g_b$ are equal in $R_{f_af_b}$ but the equality itself is in $R$. So once the equality holds in $R$ it also holds in $R_{f_a}$ and $R_{f_b}$. 
There's no issue in your example. Let $\mathbb{C}[x]/(x^2 - x) = R$. Then by Chinese remainder theorem, $R = \mathbb{C}[x]/(x) \times \mathbb{C}[x]/(x - 1) = R_1 \times R_2$. We have $R_x = R_2$ and $R_{x-1} = R_1$ with the natural maps being the same as the projections. Then the statement of the proposition is just that any $g_1 \in R_1$ and $g_2 \in R_2$ are the images of $(g_1,g_2) \in R_1 \times R_2$ which by Chinese remainder theorem we can take to be $g \in R$. 
Geometrically, what is happening here is that $R$ corresponds to two distinct points in $\mathbb{A}^1$ and the statement is just that a function on $R$ is uniquely determined by it's value on each of the two points and that conversely if we pick values on each point we can find a function on $R$ that takes those values at the points because the points are two separate connected components. In this case the partition of unity in the proof of the proposition is $x - (x-1)$. 
