Why is the change in internal energy in an isothermal system for an ideal gas zero? It seems to contradict the fundamental equation If I consider the fundamental equation, and take the derivative of it, I'll get a differential for internal energy:
dU = TdS - PdV

In an isothermal system, temperature is constant. However, if the pressure and volume is changing, as well as presumably the entropy, then it seems like dU should not be equal to zero, unless TdS = PdV. So my question comes in two parts: 1) Am I using the right equation, and 2) If so, is TdS = PdV?
I've read many explanations online as to why dU = 0 in an isothermal system, but none of the answers touch upon the equation written above.
 A: The starting point is indeed $dU=TdS-PdV$.  But, since the equilibrium state of a pure substance is determined by specifying any two intensive properties, $$dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV$$So combining these equations, we have:
$$dU=T\left(\frac{\partial S}{\partial T}\right)_VdT-\left[P-T\left(\frac{\partial S}{\partial V}\right)_T\right]dV$$But since, by definition, $$\left(\frac{\partial U}{\partial T}\right)_V=C_v$$we must have:$$dU=C_vdT-\left[P-T\left(\frac{\partial S}{\partial V}\right)_T\right]dV$$Now, from the equation for the variation in Helmholtz free energy $dA=-SdT-PdV$ (which represents an exact differential), it follows that $$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$$Therefore, we have in general that:$$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$
But, for an ideal gas, the term in brackets is exactly equal to zero.  Therefore, for an ideal gas $dU=C_vdT$, irrespective of volume changes.
A: (Assuming we're talking about an ideal gas) The internal energy of an ideal gas is only dependent on temperature.  This is because there is only one contribution to this energy, which is kinetic energy, which in turn is only temperature dependent.  If the internal energy depended on volume / pressure (e.g. increase in internal energy because the gas molecules get closer together) then the gas is no longer ideal.
So to answer your questions,

Am I using the right equation

Yes, the equation $\text{d}U=T\text{d}S-p\text{d}V$ is valid.

is TdS = PdV?

Yes (for an isothermal change of an ideal gas).
A: It is right to say that $u$ doesn't change for an ideal gas during any isothermal process, since $du=c_{v}dT$ and $dT=0 \rightarrow dU=0$. Now consider 
$Tds=du+pd\nu$. You get $Tds=pd\nu \rightarrow ds=\frac{pd\nu}{T}$. You can get this formula also in another way, if you consider a rev process you have: $ds=\frac{\delta q_{rev}}{T}$ and from the first principle $du=\delta q - \delta l \rightarrow \delta q = \delta l=pd\nu $ finally you obtain still $ds=\frac{pd\nu}{T}$ that confirms that all is right.
