A *complex-oriented cohomology theory* $E^\*$ is a multiplicative cohomology theory with a choice of Thom class $x\in\tilde{E}^2(\mathbb{C}P^\infty)$ for the universal complex line bundle (which can be used to define generalised Chern classes for all complex vector bundles).


A *real-oriented cohomology theory* $F^\*$ a multiplicative cohomology theory with a choice of Thom class $x\in \tilde{F}^1(\mathbb{R}P^\infty)$ for the universal real line bundle (which can be used to define generalised Stiefel-Whitney classes for all real vector bundles).


**Question 0:** Is this correct?


**Question 1:** Are there any examples of cohomology theories which are both real and complex orientable?


**Question 2:** (Assuming a yes to Question 1) Are there any results/papers where the interaction between a real and complex orientation is used in an essential way? (I'm thinking perhaps about non-immersion results for real projective spaces.)


Thanks. 


**Update:** Neil's answer and Johannes' comments have answered my original questions: every real oriented cohomology theory is complex oriented, and in fact is a wedge of $H\mathbb{Z}/2$'s. Then let me ask a follow up question. Are there any complex-oriented theories which are not real-orientable but have $E^1(P^\infty)\neq 0$? 



Which cohomology theories are real- and complex-orientable?
A real-orientable ring spectrum $F$ admits a ring map from $MO$, and there is a straightforward ring map $MU\to MO$, so $F$ is also complex orientable. Moreover, $MO$ is a wedge of $H/2$'s, so $MO\wedge F$ is also a wedge of $H/2$'s, and $F$ is a retract of $MO\wedge F$ (by using the ring structure etc) so it is again a wedge of $H/2$'s. Thus, you don't expect to learn anything about (non)immersion from $F$ that you could not already learn from $H/2$ (although some kinds of bookkeeping may be simplified). 


**For the follow-up question:**


The mapping spectrum $E=F(S^1\_+,MU)$ is complex-orientable but not real-orientable and has 
$$ \tilde{E}^1(\mathbb{R}P^\infty) = 
 \tilde{MU}^1(\mathbb{R}P^\infty) \oplus \tilde{MU}^0(\mathbb{R}P^\infty) 
$$
Here $MU^{\ast}(\mathbb{R}P^\infty)=MU^\*[[x]]/{[2]}(x)$ with $|x|=2$ and $MU^{-2}=MU\_2\simeq\mathbb{Z}$ so the first summand is zero but the second is not.


