Why does the Lorenz gauge $\partial_\mu A^\mu=0$ eliminate the spin-0 part of $A^\mu$? Schwartz' QFT textbook states on page 116 that:

Since $\partial_\mu A^\mu=0$ is a Lorentz-invariant condition, it has to remove a complete representation, which with one degree of freedom can only be the spin-0 component.

I don't understand this. He claims that in $4=3\oplus 1$, we removed the 1, instead of changing the 3 to a 2. Why is that?
More context:
This chapter talks about how to construct a Lagrangian for a massive spin-1 particle. We use the four-vector $A^\mu$, but a naive guess for the Lagrangian like
$$ \mathcal L = -\frac{1}{2} \partial^\mu A^\nu \partial_\mu A_\nu + \frac{m^2}{2}A_\mu A^\mu\tag{8.17}$$
leads to a set of four uncoupled equations of motion, i.e. we decomposed the four degrees of freedom in $A^\mu$ into $4=1\oplus 1\oplus 1\oplus 1$. With a term like $\partial_\mu A^\mu$ in the Lagrangian, we force $A^\mu$ to transform as a vector, i.e. the decomposition $4=3\oplus 1$ is guaranteed. He then claims that (see the quotation) that the gauge condition removes the "1" part, instead of reducing the "3" part to "2".
 A: Let us decompose $A$ in Fourier modes using the Lorentz-invariant measure (due to the equation of motion $A^\mu$ depends only on the spatial components of $p$, whereas $p^0= \sqrt{\vec{p}^2 + m^2}$),
$$A^\mu(x) = \int_{\mathbb{R}^3} e^{ipx} \hat{A}^\mu(p) \frac{d\vec{p}}{p^0}$$
where $p = (p^0,\vec{p})$. Now we can parametrize $\hat{A}^\mu(p)$ using the components it takes in the rest frame with the particle:
$$A^\mu(x) = \int_{\mathbb{R}^3} e^{ipx} \Lambda^\mu_\nu(k)\hat{A}_r^\nu(p) \frac{d\vec{p}}{p^0}$$
where $\Lambda(p)$ is the pure Lorentz transform from the rest frame to the used reference frame. In other words $p_\mu {\Lambda(p)^\mu}_\nu = m \delta^\nu_0$.
The condition $\partial_\mu A^\mu(x)=0$ reads
$$0= \partial_\mu A^\mu(x) = i\int_{\mathbb{R}^3} e^{ipx} p_\mu 
\Lambda^\mu_\nu(k)\hat{A}_r^\nu(p) \frac{d\vec{p}}{p^0} = i\int_{\mathbb{R}^3} e^{ipx} m\hat{A}_r^0(p) \frac{d\vec{p}}{p^0}$$
Namely $\hat{A}_r^0(p)=0$. As a consequence:
$$A^\mu(x) = \int_{\mathbb{R}^3} e^{ipx}  \sum_{a=1}^3\Lambda^\mu_a(k)\hat{A}_r^a(p) \frac{d\vec{p}}{p^0}\:.$$
Notice that if $R\in SO(1,3)$ is a spatial rotation,  $R\Lambda(p) =  \Lambda(Rp)R$. You see that the component $\hat{A}_r^0$ remains unchanged as $R$ acts only on the spatial components. In this sense we have a scalar.
