C1 to ground in series with the crystal and then another C2 to ground.
It's not obvious exactly what your circuit looks like, but to clarify my answer, I'll show you what imagine it looks like based on your description:
simulate this circuit – Schematic created using CircuitLab
Incidentally, this circuit, with C1 in series with the crystal, is not what I think of as an especially common oscillator circuit.
If the RF signal is AC why doesnt the AC signal just go straight to ground?
The very simple description of a capacitor is "at low frequencies it's an open circuit and at high frequencies it's a short circuit." And this is a reasonable model for a lot of cases. But saying you have an AC signal doesn't mean you have a low or a high frequency, it just means the frequency is not exactly 0 Hz. And what's happening here is that the frequency you're operating at is neither a "low" frequency or a "high" frequency, it's somewhere in between.
Between the two extremes you need to look at the impedance model of the capacitor:
\$Z = \dfrac{1}{j2\pi{}fC}\$
This also tells you what is meant by "low" and "high" frequencies. When the frequency is low enough that Z is so large it doesn't affect your circuit differently than an open circuit would, that's a "low" frequency. When the frequency is high enough that it doesn't affect your circuit differently than a short circuit would, that's a "high" frequency.
