Grassmann paradox weirdness I'm running into an annoying problem I am unable to resolve, although a friend has given me some guidance as to how the resolution might come about.  Hopefully someone on here knows the answer.
It is known that a superfunction (as a function of space-time and Grassmann coordinates) is to be viewed as an analytic series in the Grassmann variables which terminates.  e.g. with two Grassmann coordinates $\theta$ and $\theta^*$, the expansion for the superfunction $F(x,\theta,\theta^*)$ is
$$F(x,\theta)=f(x)+g(x)\theta+h(x)\theta^*+q(x)\theta^*\theta.$$
The product of two Grassmann-valued quatities is a commuting number e.g. $\theta^*\theta$ is a commuting object.  One confusion my friend cleared up for me is that this product need not be real or complex-valued, but rather, some element of a 'ring' (I don't know what that really means, but whatever).  Otherwise, from $(\theta^*\theta)(\theta^*\theta)=0$, I would conclude necessarily $\theta^*\theta=0$ unless that product is in that ring.
But now I'm superconfused (excuse the pun).  If Dirac fields $\psi$ and $\bar\psi$ appearing the QED Lagrangian 
$$\mathcal{L}=\bar\psi(i\gamma^\mu D_\mu-m)\psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$
are anticommuting (Grassmann-valued) objects, whose product need not be real/complex-valued, then is the Lagrangian no longer a real-valued quantity, but rather takes a value which belongs in my friend's ring???  I refuse to believe that!!
 A: The Lagrangian may be proved to be real but the individual factors in its terms, such as $\psi$, are neither real nor complex. They're anticommuting. There are no "particular" elements of this set of anticommuting numbers that one could "enumerate" (except for zero) and they can't appear as final predictions for observable quantities but it still makes a perfect sense to do algebra with them. A product of an even number of anticommuting variables is commuting which means that it may take particular values that may be measured and compared with theoretical predictions.
I think that I am not the only one who doesn't really understand what you're asking about but there is a chance that the answer is either in the previous paragraph or the text below:

http://motls.blogspot.com/2011/11/celebrating-grassmann-numbers.html?m=1

A: Let us sort out some terminology issues first. If the fermionic fields in your Lagrangian are Grassmanian, that means that the Lagrangian is classical, i.e. second quantization has not been performed yet. You can write a classical Lagrangian using c-number fermionic fields, but, as far as I understand, it is generally recognized now that one should use the classical Lagrangian with Grassmanian fermionic fields.
I also ran into the issue that you describe some time ago. I may be mistaken, but my conclusion was that indeed, the Lagrangian is not real, for the reasons that you give in your question. On the other hand, it is not obvious why this is necessarily bad.
EDIT: Maybe, to avoid ambiguity, I should have written that the Lagrangian is not real-valued
A: A supernumber $z=z_B+z_S$ consists of a body $z_B$ (which always belongs to $\mathbb{C}$) and a soul $z_S$ (which only belongs to $\mathbb{C}$ if it is zero), cf. Refs. 1 and 2.  
A supernumber can carry definite Grassmann parity. In that case, it is either 
$$\text{Grassmann-even/bosonic/a $c$-number},$$ 
or 
$$\text{Grassmann-odd/fermionic/an $a$-number},$$ 
cf. Refs. 1 and 2.$^{\dagger}$ The letters $c$ and $a$ stand for commutative and anticommutative, respectively.  
One can define complex conjugation of supernumbers, and one can impose a reality condition on a supernumber, cf. Refs. 1-4. Hence one can talk about complex, real and imaginary supernumbers. Note that that does not mean that supernumbers belong to the set of ordinary complex numbers $\mathbb{C}$. E.g. a real Grassmann-even supernumber can still contain a non-zero soul.
An observable/measurable quantity can only consist of ordinary numbers (belonging to $\mathbb{C}$). It does not make sense to measure a soul-valued output in an actual physical experiment. A soul is an indeterminate/variable, i.e. a placeholder, except it cannot be replaced by a number to give it a value. A value can only be achieved by integrating it out!
In detail, a supernumber (that appears in a physics theory) is eventually (Berezin) integrated over the Grassmann-odd (fermionic) variables, say $\theta_1$, $\theta_2$, $\ldots$, $\theta_N$, and the coefficient of the fermionic top monomial $\theta_1\theta_2\cdots\theta_N$ is extracted to produce an ordinary number (in $\mathbb{C}$), which in principle can be measured.
E.g. the Grassmann-odd (fermionic) variables $\psi(x,t)$ in the QED Lagrangian should eventually be integrated over in the path integral.
References:


*

*planetmath.org/supernumber.

*Bryce DeWitt, Supermanifolds, Cambridge Univ. Press, 1992.

*Pierre Deligne and John W. Morgan, Notes on Supersymmetry (following Joseph Bernstein). In Quantum Fields and Strings: A Course for Mathematicians, Vol. 1, American Mathematical Society (1999) 41–97.

*V.S. Varadarajan, Supersymmetry for Mathematicians: An Introduction, Courant Lecture Notes 11, 2004.
--
$^{\dagger}$ In this answer, the words bosonic (fermionic) will mean Grassmann-even (Grassmann-odd), respectively. 
A: One thing to add which is often confusing. Physicists use the symbol $\psi$ for both the real-valued wave-function of single-particle quantum mechanics and the grassman valued dirac field in the action. 
One can see a connection like this. The wavefunction $\psi$ of a spin 1/2 Grassman valued field $\Psi$:
$$\frac{\psi[\Psi]}{\psi_0[\Psi]} = \alpha + \int\Psi(x)\psi(x)dx+\int\Psi(x)\Psi(y)\psi(x,y)dxdy+... $$
Where $\psi_0[\Psi]$ is the vacuum wave function. Then $\psi(x)$ can in some approximate sense be thought of as the single-particle wave function (real valued).
So we see that only the wavefunctions $\psi(x), \psi(x,y),...$ are important and real valued. The fields $\Psi$ are simply placeholders.
