ECDSA Compressed public key point back to uncompressed public key point From the ECDH demo here, if I generate a private key for Alice I can get _

P = 1175846487558108474218546536054752289210804601041


Which gives the following public key point.

X = 583857549063195252150226340830731484791130788759
Y = 1195037839477751118658084226553581900533276838164


For ECDSA public key compression the above coordinates would be prepended by 02 and the X coordinate would be used. So the compressed key for the above (based on X9.62) is:

02583857549063195252150226340830731484791130788759


If I only have the compressed key given above, how would I derive Y i.e. the uncompressed point?
All points on an elliptic curve verify, by definition, the curve equation, usually written as $Y^2 = X^3 + aX + b$, with two given $a$ and $b$ parameters (these two parameters actually define the curve). So, if you know $X$, you can use the curve equation to recompute $Y^2$. A square root extraction will yield $Y$ or $-Y$. The compressed point format includes the least significant bit of $Y$ in the first byte (the first byte is 0x02 or 0x03, depending on that bit): this bit is enough to know whether you got $Y$ or $-Y$ with the square root, and adjust in consequence.

All computations above are done in the base field, i.e. (usually) the integers modulo a prime integer. See this for square root extraction modulo a prime.



Edit: for binary curves, things are somewhat similar. We work in a field $GF(2^m)$, so addition is a XOR; the curve equation is $Y^2+XY = X^3+aX^2+b$ (generic equation for non-supersingular curves). When point $P = (X,Y)$, the opposite point is $-P = (X,Y+X)$. We assume that the point we are encoding and decoding has coordinate $X \neq 0$, because a point with such a coordinate $X = 0$ is equal to its opposite, i.e. has order $2$. In practice, when we work on such a curve, we really use a sub-group of order $r$ prime (generally, we arrange for the complete curve order to be equal to $2r$ or $4r$).

Since $X\neq 0$, we can define $Z=Y/X$, and divide the curve equation by $X^2$. This yields:
$$ Z^2+Z = X+a+\frac{b}{X^2} $$
Thus, given $X$, we can compute $Z^2+Z$. From that value, we can compute the two solutions, which are $Z$ and $Z+1$ (see below). Since $Y = XZ$, this yields $Y$ and $Y+X$, thus the points $P$ and $-P$.

This gives us the compression method for curves in binary fields: from $(X,Y)$, compute $Z=Y/X$; the compressed point then consists in $(X,z_0)$, where $z_0$ is the least significant bit of $Z$. When decompressing, you use the equation above to recompute $Z^2+Z$, then $Z$ (the knowledge of $z_0$ tells you which of the two solutions is the right $Z$), then $Y = XZ$.

To know how to solve for $Z$, we need to define the trace and the half-trace. We are in $GF(2^m)$; the trace of a value $x$ is:
$$ \mathrm{Tr}(x) = \sum_{i=0}^{m-1} x^{2^i} $$
We are in a field of characteristic $2$, which implies that squaring is an automorphism: $(uv)^2 = u^2v^2$ and $(u+v)^2 = u^2+v^2$ for all $u$ and $v$ (this is called the Frobenius automorphism). Moreover, for all $u$, $u^{2^m} = u$ (this is analogous to Fermat's little theorem, because the non-zero elements of $GF(2^m)$ are a multiplicative group of order $2^m-1$). This implies two important characteristics of the trace:


The trace is linear: for all $u$ and $v$, $\mathrm{Tr}(u+v) = \mathrm{Tr}(u) + \mathrm{Tr}(v)$.
The trace of any $u$ is either $0$ or $1$. Indeed, for all $u$, $\mathrm{Tr}(u^2) = \mathrm{Tr}(u)^2$ (by the Frobenius automorphism), and also $\mathrm{Tr}(u^2) = \mathrm{Tr}(u)$ (try it in the equation that defines the trace). Thus, for all $u$, $\mathrm{Tr}(u)^2 = \mathrm{Tr}(u)$. The equation $x^2=x$ is a polynomial equation of degree $2$, so it can have only (at most) $2$ solutions in a field, and $0$ and $1$ are solutions. Therefore, $\mathrm{Tr}(u)$ can only be $0$ or $1$.


Let's now define the half-trace. For now, I suppose that $m$ is odd (this is the normal case for binary elliptic curves, e.g. NIST curve B-233 is defined in $GF(2^{233})$). For any value $x$, the half-trace of $x$ is:
$$ H(x) = \sum_{i=0}^{(m-1)/2} x^{2^{2i}} $$
That is, we take the equation for the trace, but we keep only half of the operands in the sum.

Like the trace, the half-trace is linear. Also, for any $x$, we can see that:
$$ H(x)^2 + H(x) = \mathrm{Tr}(x) + x^{2^m} = \mathrm{Tr}(x) + x $$

Let's get back to our equation $Z^2+Z = \beta$ where $\beta = X+a+(b/X^2)$. Since the trace is linear, and since for $Z$ and $Z^2$ have the same trace, it follows that $\mathrm{Tr}(\beta) = \mathrm{Tr}(Z^2+Z) = \mathrm{Tr}(Z^2)+\mathrm{Tr}(Z) = 0$ (remember that addition is XOR in $GF(2^m)$). Thus, the equation can have solutions only if $\mathrm{Tr}(\beta) = 0$. Then, we have $H(\beta)^2+H(\beta) = \beta$, so the solutions for $Z$ are $H(\beta)$ and $H(\beta)+1$.

If $m$ is even (this will often be the case if using pairings over binary curves, although this is probably not a good idea since discrete logarithm in binary fields is a lot easier than previously expected), then most of the above hold, although we need another definition for the half-trace. Basically, we need a constant element $w$ such that $\mathrm{Tr}(w) = 1$, and we can define the half-trace as:
$$ H(x) = \sum_{i=0}^{m-1}\left(\sum_{j=0}^{i} x^{2^j}\right)w^{2^i} $$

In both cases ($m$ odd and $m$ even), the half-trace is linear, so it can be computed as a XOR of some precomputed values for the bits equal to $1$ of the source. This is rather inexpensive. The trace is even easier: it is simply a XOR of some of the input bits, and, in practice, often a XOR of very few of these bits.