Representation of free sigma-algebras

Question

In his Lectures on Boolean Algebras, Halmos states the following theorem (p. 102):

Theorem 14 For every set $I$, there exists a free $\sigma$-algebra generated by $I$, and, in fact, that algebra is isomorphic to the $\sigma$-field of all Baire sets in the Cantor space $2^I$.

I don't understand this theorem.

By a theorem of Loomis, every (Boolean) $\sigma$-algebra is isomorphic to the quotient of a $\sigma$-field by a $\sigma$-ideal (a theorem that Halmos states himself on the same page). So, why is the free $\sigma$-algebra generated by $I$, as a $\sigma$-algebra, not isomorphic to the quotient of a $\sigma$-field by a $\sigma$-ideal?

PS: Halmos calls a Baire set an element of the $\sigma$-field generated by the clopen sets.

Answer 1

So, why is the free $\sigma$-algebra generated by $I$, as a $\sigma$-algebra, not isomorphic to the quotient of a $\sigma$-field by a $\sigma$-ideal?

Theorem 14 says no such thing at all. It says the free $\sigma$-algebra generated by $I$ is isomorphic to a certain $\sigma$-field. This in no way means that it can't also be isomorphic to the quotient of a $\sigma$-field by a $\sigma$-ideal.

(In fact, Theorem 14 immediately implies that the free $\sigma$-algebra generated by $I$ is isomorphic to the quotient of a $\sigma$-field by a $\sigma$-ideal, since it is isomorphic to the quotient of itself by the trivial $\sigma$-ideal $\{0\}$.)