. As I said in class, the authors neglect to state how they are defining operations on F[x]/

; but in complete analogy with the way operations were defined in Z_n and in factor-groups G/N, the definitions are [a(x)] [b(x)] = [a(x) b(x)] [a(x)] + [b(x)] = [a(x) + b(x)] Proposition 4.4.4 shows that these operations are well-defined. Once one has the operations defined in this way, the desired properties of these operations follow immediately from the corresponding operations of F[x]. For instance, in the display in the proof of Theorem 4.4.6, they verify commutativity of addition in F[x]/

using
commutativity of addition in F[x]. The first and last steps of that
display are the applications of the above definition of multiplication
of congruence classes; the middle step uses commutativity of
multiplication in F[x]. All the other laws are deduced in the same
way. You should take some law, such as distributivity, and go through
the calculation yourself, to check that it works.
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You ask whether the construction of the field of complex numbers
you learned in High School is rendered obsolete by the construction
of Example 4.4.2, p.191.
No. One of the points of the concept of isomorphism is that if two
objects are isomorphic, then mathematicians don't have to decide which
one to attach a given name to; as long as they agree that "A" means
a system that has the relevant properties, one of them can be thinking
of A as meaning a system constructed one way and the other can be
thinking of A as meaning a system constructed another way, and yet
they can talk about these systems without any disagreement. So if
one mathematician thinks of the complex numbers as the set of symbols
a + bi, and the other thinks of them as R[x]/ .
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You ask whether the u of Theorem 4.4.8 (p.191) might belong to F.
If the polynomial p has degree 1, then u will belong to F,
since a polynomial of degree 1 has a root in the given field.
(In that case, E = F[x]/ will come out isomorphic to F; so,
after we make our identification of F with its image in E, we
will have E = F.) If p has degree > 1, then, being irreducible,
it doesn't have a root in F, and u will lie outside of F.
You are right that if p already has a root in F, there is no
need to construct the extension F[x]/ . But in giving the
proof of the theorem, it's shorter to do things in one way for all
cases rather than saying "If p has degree 1, take E = F; if p
has degree > 1, let us construct E as follows."
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You ask how one proves that Q[x]/