Is electromagnetic vector field a sum of E and B? I have a hard time to fully understand (classical) electromagnetic field theory with respect to Helmholtz's decomposition. Let me start from Helmholtz's theorem:

Any vector field of class $C^{\infty}$ in $R^3$ can be docomposed into sum of >two other fields: one curl-free and one divergence free.
$\bf{F}=\bf{F_1}+\bf{F_2}$
but (due to some vector operator identities) we can rewrite $F_1$ and $F_2$ to
$\bf{F_1}=-\nabla F_3$
$\bf{F_2}=\nabla\times\bf{F_4}$
where
$F_3$,$\bf{F_4}$ are scalar and vector fields respectively

Now going to electrodynamic we know that in stationary case
$\bf{E}=-\nabla\phi$
and
$\bf{B}=\nabla\times\bf{A}$
It fits very well so we can write that electromagnetic field is equal
$\bf{F_{EM}}=\bf{E+B}=-\nabla\phi+\nabla\times\bf{A}$
or can we? Why in none of my books nor in the net there is written that EM field is just $\bf{E+B}$? For example wikipedia states that EM is combination of $\bf{E}$ and $\bf{B}$. Yes, of course it is combination (from Maxwell equations) but that is not precise statement. Obviously nowhere I could find any equation for EM field (treated as one single vector field).
So, can someone please elaborate what this EM field is with respect to $\bf{E}$ and $\bf{B}$ in the context of Helmholtz decomposition?
 A: 
If fits very well so we can write that electromagnetic field is equal
$\bf{F_{EM}}=\bf{E+B}=-\nabla\phi+\nabla\times\bf{A}$
or can we?

No! For the love of god, no! 
Do not just add those fields together... it's not a useful quantity.
In the SI system $E$ and $B$ have different units. Another good indicator that you don't want to just add them together....
Additionally, the electric field is not always longitudinal (i.e., not always equal to the gradient of a scalar $-\nabla \phi$). In general it can have a transverse component:
$$
\vec E = -\nabla \phi - \frac{\partial \vec A}{\partial t}
$$
A: If the field is not stationary, curl of $\vec{E}$ does not vanish. So generally you cannot identify electromagnetic field with the curl-free part of the decomposition.
However, you can indeed introduce a complex vector combination of electric and magnetic field, in a certain system of units it is $\vec{E}+i\vec{H}$. This is the so-called Riemann-Silberstein vector (http://en.wikipedia.org/wiki/Riemann%E2%80%93Silberstein_vector ). It is sometimes very useful (for example, I used it in my recent articles (http://arxiv.org/abs/1502.02351 and http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf (published in J. Math. Phys.)). However, it is a vector only under the transformations of the rotation group, not of the entire Lorentz group. 
A: I will find the Maxwell word in the Oersted Medal Lecture 2002: Reforming the Mathematical Language of Physics by Hestenes, on pages 25/26 
That formidable text presents a better math formalism for physics, imo.  
Starting with $ F(x,t) = E(x,t) + i B(x,t) $  ...
The 4 equations of Maxwell (64..67) that describe two viewpoints ( E and B ) of a single entity 'EM' field can be expressed with only one equation (63):  
$$(\frac{1}{c}\partial_t+\nabla) F= \rho - \frac{1}{c}J$$
The word Helmholtz , etc, etc, is not present in that formalism. Complex numbers, vectors, matrices, tensors, etc, are particular viewpoints that Geometric Algebra integrates.  
A: Let me try this more clearly than the other answers, which aren't wrong. You ask:

So, can someone please elaborate what this EM field is with respect to $\vec E$ and $\vec B$ in the context of Helmholtz decomposition?

There is no "EM field in the context of Helmholtz decomposition". Helmholtz just says that every vector field $\vec V$ is decomposable as curl and gradient of two other fields, i.e. 
$$\vec V = \vec \nabla \phi + \vec \nabla \times \vec A $$
You can do this for the electric or the magnetic field, of course, but this isn't particularly enlightening as to the nature of "the EM field". A field should behave nicely under transformations, and special relativity with its action on the electric and magnetic fields shows us that we should not add them together, but seek a quantity that transforms nicely under Lorentz transformations instead:
"The electromagnetic field" is equivalently the gauge four-potential $A$ (consisting of the scalar electrostatic potential in the temporal and the magnetic vector potential in the spatial entries) or its derivative, the field strength tensor $F = \mathrm{d}A$. Electric and magnetic fields become part of the tensor as
\begin{align}
F^{0i} & = E^i \\
F^{ij} & = \sum_k\epsilon^{ijk}B^k
\end{align}
This is "the EM field", but it has nothing to do with Helmholtz decomposition, since electromagnetism is properly looked at in the four-dimensional setting of special relativity, for which only the general Hodge decomposition may be applied, of which Helmholtz is a special case, but even this has nothing to do with it.
This EM field acts on the four-velocity, reproducing the Lorentz force by
$$ \frac{\mathrm{d}p}{\mathrm{d}t} = q F(u)$$
where $u$ is the four-velocity, and $(F(u))_\mu = F_{\mu\nu}u^\nu$.
A: Actually the electromagnetic field can be seen as a tensor. The combination Wikipedia talks about is this, $E$ and $B$ are organised in an antisymmetric matrix $F_{\mu\nu}$ with $\mu,\nu = 0,\ldots, 4$ so the number of independent components is $6$.
A: You are free to define the vector $\bf{F_{EM}}$, but I don't believe this vector would have any value.  It wouldn't obey any simple laws, and it would not be found to have any practical use in the lab.
