= Solution
For $m_u=m_d=0$ and electromagnetism neglected, the classical flavour symmetry is
$$
U(2)_L\times U(2)_R
\simeq SU(2)_L\times SU(2)_R\times U(1)_V\times U(1)_A
$$
up to finite quotients. The <chiral anomaly> breaks $U(1)_A$ quantum mechanically, while $U(1)_V=U(1)_B$ remains. The quark condensate
$$
\langle\overline q_Rq_L\rangle\ne0
$$
spontaneously breaks
$$
SU(2)_L\times SU(2)_R\longrightarrow SU(2)_V.
$$
Small quark masses and electromagnetism explicitly break parts of this approximate symmetry.
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