A has_many :bs; what is the Rails idiom to save @a and @.bs in one transaction ?

Suppose I have two models:

class A < ActiveRecord::Base
  has_many :bs

class B < ActiveRecord::Base
  belongs_to :a

Now, if I say in a controller
  b1 = @a.build(params[:b])
I can build a new B associated to an A.

If the @ is also new, I need to also do this
  b1.a = @a
but if the @a is not new, then its id is filled in for me.

(1) Why? Even though there is no @a.id to set the b1.a to, there is
an @a object that I can set b1.a to, and I do. Couldn't build do this
for me so @a.build(...) would work in both cases?

Either way, if I now say
Then not only are the values of @a that may have been updated saved,
the new object b1 is also saved to the database. Not only that, this
all happens in the same transaction. Yea!

However, instead if I were to update an existing b as follows:
  b1 = @a.bs[1].update(params[:b])
and then
then this time, the updated values of b1 are not saved to the

(2) Why? Wouldn't it be cleaner to update the associated b values
into the database as well during the save of the @a ?

Further, if I want the @a and the associated @a.bs to all be sent to
the database in the same transaction, as far as I can tell, I have the
following choices.

In my controller save the @a and @a.bs in one transaction:
ActiveRecord::Base.transaction do
  @a.bs.each do |b|

Or perhaps to be more elegant I could push all that into a method
save_it! on class A.

Or I can try to figure out how to get the @a.bs to save in the same
transaction as the @a. I suppose I can do that as follows:

class A < ActiveRecord::Base
  before_save do |a|
    bs.each do |b|

This is a bit better, but -- amazingly -- I have somehow at times
gotten into an infinite loop doing this, though it was a bit more
complex than that. Even though the documentation says that if I save
on the belongs_to side, the object belonged-to will not be saved, I
had a case where saving the associated objects like this ended up
recursing on itself. Is this idiom supposed to work and was I
therefore getting an infinite loop because (a) I was not doing it
quite right, or (b) is this idiom not good.

(3) What is the right idiom for this situation, where I want to save
the @a and the associated bs in one transaction?