electromagnetism - Physical degrees of freedom of the Electromagnetic field

As I understand it, the classical source-free electric, $\mathbf{E}$ and magnetic, $\mathbf{B}$ wave equations are solved by solutions for the electric and magnetic fields of the following form: $$\mathbf{E}=\mathbf{E}_{0}e^{i (\mathbf{k}\cdot\mathbf{x}-\omega t)}$$ $$\mathbf{B}=\mathbf{B}_{0}e^{i (\mathbf{k}\cdot\mathbf{x}-\omega t)}$$ Naively counting the degrees of freedom (dof) at this point it would appear that the electromagnetic field has 6 dof.

However, is it correct that Maxwell's equations provide 4 constraints: $$\mathbf{k}\cdot\mathbf{E}_{0}=0 \\ \mathbf{k}\cdot\mathbf{B}_{0}=0$$ $$\mathbf{E}_{0}=-\frac{1}{\sqrt{\mu_{0}\varepsilon_{0}}}\mathbf{k}\times\mathbf{B}_{0}$$ and $$\mathbf{B}_{0}=\sqrt{\mu_{0}\varepsilon_{0}}\mathbf{k}\times\mathbf{E}_{0}$$

Thus reducing the number of physical dof to 2?!

If the above is correct what do these remaining dof correspond to? Are they simply the two possible polarisation (unit) vectors $\mathbf{\epsilon}_{1}$, $\mathbf{\epsilon}_{2}$ that one can construct such that $$\mathbf{k}\cdot\mathbf{\epsilon}_{1}=\mathbf{k}\cdot\mathbf{\epsilon}_{2}=0$$ and $$\mathbf{k}\times\mathbf{\epsilon}_{1}=\mathbf{\epsilon}_{2}\\ \mathbf{k}\times\mathbf{\epsilon}_{2}=-\mathbf{\epsilon}_{1}$$ and hence $\lbrace\mathbf{k},\;\mathbf{\epsilon}_{1},\;\mathbf{\epsilon}_{2}\rbrace$ form an orthornormal basis, such that the general solutions for $\mathbf{E}$ and $\mathbf{B}$ are linear combinations of $\mathbf{\epsilon}_{1}$ and $\mathbf{\epsilon}_{2}$?!