Misunderstanding of the magnetic field

There are two main points of confusion in your reasoning:

1. Any field can be decomposed as a sum of two fields: one with zero divergence and one with zero rotor.
   You say "any field can be constructed using its divergence and rotational" which is kind of obscure to me.

2. If a field has zero divergence and zero rotor, it can still be finite! It does not have to be zero, if its divergence and rotor are zero.

So to sum up, the magnetic field outside of the wire has zero divergence (as everywhere) and zero rotor, but it is finite.

The statement that "We know that any field can be constructed using its divergence and rotational." is wrong (I assume that you mean "curl" when you say "rotational"). An easy counterexample are the following two fields

begin{alignat}{3} mathbf E_1 &= 2mathbf {hat i}quad &&;quad&&&mathbf E_2=3mathbf{hat j} nabla cdot mathbf E_1&=0 quad &&;quadnabla cdot &&&mathbf E_2 =0 nabla times mathbf E_1&=0quad &&;quad nabla times &&&mathbf E_2 =0 end{alignat}

As you can see that the curl and divergence of both $mathbf E_1$ and $mathbf E_2$ are the same, yet $mathbf E_1 notequivmathbf E_2$. So, there isn't any way of determining a vector field just from its curl and divergence.

You can reconstruct a vector field $mathbf F$ from its divergence $nabla cdot mathbf F$ and curl $nabla timesmathbf F$ and boundary conditions. More precisely, if $R$ is some bounded region on which $mathbf F$ is twice continuously differentiable, then

$$mathbf F = nabla Phi + nabla times mathbf G$$

where $$Phi(mathbf r) = frac{1}{4pi} int_R frac{nabla' cdot mathbf F(mathbf r')}{|mathbf r - mathbf r'|}d^3 r' -frac{1}{4pi}oint_{partial R} frac{hat n' cdot mathbf F(mathbf r')}{|mathbf r - mathbf r'|}dS$$ and $$mathbf G(mathbf r) = frac{1}{4pi}int_R frac{nabla'timesmathbf F(mathbf r')}{|mathbf r - mathbf r'|}d^3 r' - frac{1}{4pi}oint_{partial R} frac{hat n' times mathbf F(mathbf r')}{|mathbf r - mathbf r'|}dS$$

where $partial R$ is the boundary of $R$, and $hat n$ is the normal vector to the surface at the point $mathbf r'$. If the fields fall off faster that $1/|mathbf r-mathbf r'|$, then this can be extended to an unbounded region.

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In the static case, Ampere's Law says that $$nabla times mathbf B = mu_0 mathbf J$$

but this is not continuous at the boundary of the wire. As a result, the Helmholtz theorem applies separately to the interior region within the wire and the region outside it. Reconstructing the magnetic field outside the wire requires you to know the value of the magnetic field at the surface of the wire.