Suppose I have an AC supply of 24 V and I convert it to DC by both a half wave rectifier and a full wave rectifier; then the half wave rectifier (1 diode) should give (under ideal conditions) peak voltage of 24 V, right?
No! When someone says "this AC supply is \$X\$ V", they mean that the effective voltage is \$X\$ V. Effective voltage is the RMS voltage, not the peak voltage of the AC.
Sine voltages. (1) peak voltage (amplitude), (2) peak-to-peak voltage, (3) effective voltage (RMS voltage) (4) period.
Image from wikipedia article on Alternating Current
Luckily, we know that when someone says "AC", they mean "sine wave" and for sines of any frequency, we know that the effective voltages is \$1/\sqrt2\$ of the peak voltage. So, when we multiply 24 V by \$\sqrt 2\$, we get the peak voltage – ca. 34 V.
Look at the figure above: what the half-wave rectifier does is simply "cut off" the half of the wave where the voltage is negative. The positive half is left unchanged.
Since the ideal diode (not the one you mention – the one that you can buy from Diodes Inc. is not an ideal diode, which only exists in theory, but it's close enough here) has no voltage drop over it, the peak of the half-wave rectified voltage also happens to be exactly the same peak voltage – so, \$\sqrt 2\$ times the effective voltage as you find on the output of your transformer (24 V), so in our case, ca 34 V.
Now I centre tap the transformer (which how I understand it should make the centre tap have 0 V and the ends have +12 V and -12 V (relatively of course)).
Now the full wave rectifier (with 2 diodes) would give (under all ideal conditions) a peak voltage of 12 V, right?
The thing about center tapping your transformer is this: Center-tapping means now you have three wires coming out of the transformer. But you haven't said how you want to connect the full-bridge rectifier. I'll guess you mean this (because if you ignored the center tap, then you could have just not center-tapped it):
and you measure your peak voltage over a load between +VDC and -VDC.
Then, again, the effective voltage of the AC is still 24 V. The peaks are still \$\sqrt 2\cdot 24\$ V high. Having both half waves did not change anything regarding the peaks, and all the center tap does is provide you a reference