I think atyy is right (also with his opinion on Greiter's article, which I think is a very good one). I forgot, what the debate about it was about, but I also read it pretty carefully, and couldn't find obvious flaws.
On the other hand, since a theory that is invariant under local gauge symmetries, it is also invariant under the global one. This implies the applicability of the Noether theorem and thus the conservation of the corresponding Noether charges, and this in turn implies that the conservation of these charges is a necessary but not sufficient condition for local gauge invariance.
There's a very important difference in spontaneous symmetry breaking of global and local gauge symmetries. If you break the global gauge symmetry that is not a special case of a more comprehensive local gauge symmetry, the QFT necessarily contains massless (zero) modes in the physical spectrum, the famous Nambu-Goldstone bosons related to this spontaneously broken symmetry. How many of such Goldstone bosons occur depends on the group structure of the original group and the unbroken subgroup that acts among the eigenvectors of the (degenerate!) groundstate of the system. One important example related to elementary-particle physics is the light-quark sector of QCD, which has an approximate chiral symmetry. Taking only up and down quarks as "light", we have an ##\mathrm{SU}(2)_L \times \mathrm{SU}(2)_R## global (chiral) gauge symmetry (in the limit of massless quarks). The strong interaction leads, however, to the formation of a quark condensate, i.e., the vacuum expectation value ##\langle \overline{\psi} \psi \rangle \neq 0##. This VEV is invariant under the vector transformations, ##\psi \rightarrow \exp(-\mathrm{i} \vec{\alpha} \cdot \vec{\tau}) \psi## but not under the axial-vector transformations ##\psi \rightarrow \exp(-\mathrm{i} \vec{\beta} \cdot \vec{\tau} \gamma^5) \psi##. Thus the original symmetry is spontaneously broken to the usual strong-isospin subgroup ##\mathrm{SU}(2)_V##. The corresponding Goldstone modes are the pions. Their non-zero mass is only due to the (small) current quark masses of the light quarks u and d.
On the other hand, a chiral symmetry ##\mathrm{SU}(2)_{\text{wiso}} \times \mathrm{U}(1)_{\text{Y}}## is also at work in the electroweak sector of the Standard Model, but this chiral symmetry is also local. In a loose sense you can say, the Higgs sector breaks this symmetry spontaneously, but the implications are totally different from a case of a spontaneously broken symmetry that's "only" local! There are no Nambu-Goldstone modes, but they are in some sense "eaten up" by the gauge fields. In the case of the weak-isospin symmetry the symmetry-breaking pattern is from the original symmetry to a remaining ##\mathrm{U}(1)_{\text{em}}##. Thus there would be 3 Goldstone modes, corresponding to the continuous degeneracy of the ground state, if the symmetry was only global. But now in the electroweak theory this symmetry is local, and thus you can parametrize the Goldstone modes (here more accurately dubbed "would-be-Goldstone modes") as a gauge-transformation unitary matrix, and this lumps these degrees of freedom into the gauge fields, when choosing a special gauge fixing, known as "unitary gauge" (for reasons that become obvious) soon. So first of all in this gauge the would-be-Goldstones do not occur as massless particles in the physical spectrum described by the theory but they become the (spatially) longitudinal part of the gauge bosons, and in turn these become massive. The gauge symmetry is still not really broken, and thus the gauge invariance holds in this clever way although the gauge bosons become massive! That came to the rescue of a big puzzle in electroweak theory and earned Higgs and Englert the Nobel Prize after the discovery of a Higgs boson, which is indeed necessarily occurring in the physical particle spectrum of this kind of theory.
This procedure also shows that there is in fact no degeneracy of the ground state, because we have a local gauge transformation, the acting of the gauge group doesn't lead to a different ground state but to another representant of the one ground state that is realized as an equivalence class of vectors modulo local gauge transformations (such as the electromagnetic field in classical electrodynamics is represented by an equivalence class of four-vector potentials with two four-vector potentials representing the same physical em. field if they are connected by a gauge transformation). That's another reason for the fact that there are no massless Goldstone modes in this case. It's better to say the gauge theory is "Higgsed" than to say one spontaneously breaks the local gauge symmetry, which in fact you don't as just explained.
Now the unitary gauge has a drawback. It is called unitary, because it makes the physical particle spectrum manifest on the tree level of the theory (or in the Lagrangian, describing the QFT). That's good, but only on the first glance! Now the propagators of the gauge fields in this gauge have a propagator with a piece ##\propto p^{\mu} p^{\nu} /[M^2(p^2-M^2+\mathrm{i} 0^+]##, which has not the nice power counting as a usual bosonic propagator. It now seems as if the Higgsed gauge theory is no longer renormalizable, but that's not right! What should be renormalizable are the physcially relevant S-matrix elements, not necessarily the propagators and vertex functions used to calculate them.
Now fortunately, one can fix the gauge differently from the unitary gauge, and 't Hooft came up with a very clever choice during his PhD work, the socalled renormalizable ##\xi## gauges. They depend on a parameter ##\xi##, and for finite values of ##\xi## it leads to nicely renormalizable propagators and vertex functions in the usual sense of perturbative Dyson renormalizability. In this gauge, however you have both, massive gauge fields, would-be Goldstone modes and, in addition, Faddeev-Popov ghosts. This sounds very complicated, and in fact it is! But the would-be Goldstone modes are ghosts similar to the Faddeev-Popov ghosts, but they are represented as usual and not Grassmann fields in the path-integral formalism. Thus, roughly speaking, what happens in this class of gauges is the following: You have one unphysical degree of freedom in the (massive) gauge-boson fields and for each "broken one-parameter symmetry trafo" a would-be Goldstone mode, both of which are compensated in loops of Feynman diagrams by the Faddeev-Popov ghosts. So you are left with the 3 physical gauge-boson degrees of freedom. Now, since one deals with so many "ghost particle degrees of freedom" one might worry about the unitarity of the S matrix, but there's no danger because of local gauge symmetry! The S-matrix elements are independent of the ##\xi## parameter, and the 't Hooft gauges are such that the limit ##\xi \rightarrow \infty## makes the ##R_{\xi}## gauges into the unitary gauge, i.e., the theory becomes manifestly renormalizable in the usual Dyson sense (for the Green's and proper vertex functions) and the S matrix stays unitary! This proof for the renormalizability of local Higgsed gauge theories made up not only a great doctoral thesis for 't Hooft but was the breakthrough for the electroweak standard model (discovered by Glashow, Salam, and Weinberg before) and earned 't Hooft and Veltman their Nobel prize.
