How to price a contract that is denominated in another currency using the Martingale Approach to the Black and Scholes theory? I am taking a course in asset pricing and I have the following problem at hand:

Suppose that the level of the UK FTSE100 index (in British pounds) evolves according to
$$\frac{\mathrm{d}S_t}{S_t}=\mu_S\mathrm{d}t+\sigma_S\mathrm{d}Z_{S,t}.$$  Assume that the dividend yield on the index is constant and equal to $\delta$. Suppose that the dollar value of a single British pound equals $X_t$, where
$$\frac{\mathrm{d}X_t}{X_t}=\mu_X\mathrm{d}t+\sigma_X\mathrm{d}Z_{X,t},\quad\mathrm{d}Z_{X,t}\mathrm{d}Z_{S,t}=\rho\mathrm{d}t.$$
Suppose that the interest rate on dollar-denominate bonds is constant at $r^{US}$, while the interest rate on the pound-denominated bonds is constant at $r^{UK}$.
Consider a derivative security that pays $S_T$ dollars at time $T$. Derive the market value of this security at time $0$.

In this course, we are emphasizing the ''Martingale Approach''. I usually look for a change of measure that will make the risky asset have the drift equal to the risk free rate and then I try to derive the dynamics of $S_T$ under this new measure. After I have done that, it's all a matter of taking expectations of known functions of something we the distribution of.
But here there are two differences that made me get lost. First, there are dividends. My intuition is that the dynamics for $S_t$ should take into account that it pays dividends. Why isn't the drift $\mu_S - \delta$?
Moreover, I understand that I should be studying the dynamics of $M_t = X_t S_t$, which is the value of the index in dollars. I derived the following dynamics:
$$\frac{dM_t}{M_t} = (\mu_S + \mu_X + \rho \sigma_X \sigma_Z)dt + \sigma_S dZ_{S, t} + \sigma_X dZ_{X, t}$$
But I am stuck here. Should I look for the measure $Q$ that will make $M_t/B_t$ a martingale? And how do I deal with dividends? Any solutions/hints are deeply appreciated! Thanks in advance.
 A: I believe I have figured out a solution myself. Here it goes:
Start noticing that we are in a complete markets world. We have two risky assets and two Brownian motions. Let $Z_t = (Z^S_{t}, Z^X_{t})'$ and $dB = d(B^S_t, B^X_t)' = dZ_t + (\eta_t, \gamma_t)'dt$ be the stacked Brownian motions under the physical measure and under the risk-neutral measure. We will characterize the processes $(\eta_t, \gamma_t)$.
What is really happening is that from the point of view of the US investor he has two risky at assets at his disposal: the British bond $U_t$ converted to American dollars $X_tU_t$ and the British index also converted in American dollars $M_t = X_tS_t$. First, I study the gain process of the British bond and make sure that it will be a martingale. Since it pays no coupons, the discounted gain process has evolution given by:
$$d\left(\frac{X_tU_t}{B_t}\right) = \frac{X_tU_t}{B_t}\left[(\mu_X + r^{UK} - r^{US} - \sigma_X \gamma_t)dt + \sigma_X dB^X_t \right]$$
Under the risk-neutral measure, if we let $\gamma_t = \frac{\mu_X + r^{UK} - r^{US} - \sigma_X}{\sigma_X}$, this process will be a martingale.
Now, we study the discounted gain process for the stock after making the exchange rate conversion. Taking into account the dividend yield, we can write that
$$d\left(\frac{M_t}{B_t}\right) = M_t\left[(\mu_X + \mu_S + \sigma_X\sigma_S \rho + \delta - r^{US} - \sigma_X\gamma_t - \sigma_S \eta_t)dt + (\sigma_S, \sigma_X)dB \right]$$
where the last term entails an inner product. Using our result for $\gamma_t$, if we let
$$\eta_t = \frac{\mu_S + \sigma_X\sigma_S \rho + \delta - r^{UK}}{\sigma_S}$$
we have that $M_t/B_t$ is also a martingale under the risk-neutral measure. Further, we notice that we can drop the subscript $t$ since these are constants and, as such, will trivially satisfy Novikov's condition.
Having fully characterized the risk-neutral measure, it's easy to see that
$$d\left(\frac{S_t}{B_t}\right) = \frac{S_t}{B_t}(r^{UK} - r^{US} - \delta - \sigma_X \sigma_S\rho)dt + \frac{S_t}{B_t}\sigma_s dB^S_t$$
Then, using the fact that Ito's integral is a martingale, the value $V_t$ of this derivative is given by:
$$V_t = S_t e^{(r^{UK} - r^{US} - \delta - \sigma_X \sigma_S\rho)(T-t)}$$
