Matching bosonic and fermionic degrees of freedom in Wess-Zumino Lagrangian

Question

In Wess-Zumino model, supersymmetric Lagrangian in addition to the ordinary complex scalar field $\phi$ contains auxiliary field $F$ (also complex scalar) to match the degrees of freedom of the Weyl (or Majorana) spinor, meaning 4. But from equations of motion for $F$'s we have $$F=-W^{\dagger},\;\;\;F^\dagger=-W,$$ where $$W=M\phi+\frac{1}{2}y\phi^2;$$ $M$ is mass and $y$ is Yukawa coupling.

Here goes my question: since $W$ depends on $\phi$ and that makes $F$ depend on $\phi$, doesn't that reduce the bosonic degrees of freedom back to 2?

P.S. From "Supersymmetry in particle physics" by Ian Aitchison, 5th chapter.

Answer 1

In case anyone has the same question, I found the answer.

On-shell and off-shell degrees of freedom have to be counted separately. So off-shell means we don't use EOM, thus F fields do not depend on $\phi$, and degrees of freedom are 4, both bosonic and fermionic. But if we are on-shell, Weyl equations reduce fermionic degrees of freedom to 2, and EOM for F's reduce the bosonic degrees of freedom to 2.