For $N_f$ numebr of massless quarks, we know that there are global symmetries $$ \frac{SU(N_f)_L \times SU(N_f)_R \times U(1)_V}{Z_{N_f}} $$ here $U(1)_V$ is the same as $U(1)$-Baryon number conservation. The axial $U(1)_A$ is anomalous.
Question: What would be the remained global flavor symmetries of massless quarks after gauging electromagnetic $U(1)_e$? We can take $Nf=3$ where we have 3 quarks like $u,d,s$.
[WARNING]: Notice that $U(1)_e$ is part of the vector global flavor symmetry $SU(N_f)_V$, and gauging result in the remained $SU(N_f-1)_V$ global flavor symmetry. Namely, $U(1)_e$ is part of $SU(N_f)_L \times SU(N_f)_R$. However, we note that $$SU(N_f)_L \times SU(N_f)_R \neq SU(N_f)_V \times SU(N_f)_A$$ because the left-right chiral flavor generators DO commute, but the vector and axial generators of $SU(N_f)_V$ and $SU(N_f)_A$ do NOT commute. So the answer won't be as simple as $$ \frac{SU(N_f-1)_V \times SU(N_f)_A \times U(1)_V}{Z_{N_f}} [\text{This is WRONG!}] $$
What is your answer?