Why Read This Book? - Index of

342 Chapter 9 Rings
In the same way that we view elements of Z6 simply as {0, 1, 2, 3, 4, 5}, we can
view elements of Q[t]/(f ) simply as polynomials of the form at 2 + bt + c. But we
must consider how they add and multiply. Instead of using coset notation and
writing [(f ) + a1t 2 + b1t + c1]+[(f ) + a2t 2 + b2t + c2], we can just write
[a1t 2 + b1t + c1]+[a2t 2 + b2t + c2] =(f ) (a1 + a2)t 2 + (b1 + b2)t + (c1 + c2)
(9.76)
Adding two such polynomials cannot produce a sum of any larger degree, so
Eq. (9.76) is all that needs to be said about addition in the quotient ring. However,
for multiplication, let’s illustrate with a concrete example, where the details of
polynomial division have been omitted.
(4t 2 + 2)(3t 2 − 2t + 8) =(f ) 12t 4 − 8t 3 + 38t 2 − 4t + 16
=(f ) (6t − 4)f + 44t 2 − 38t + 36
=(f ) 44t 2 − 38t + 36
(9.77)
If we multiply two elements of the quotient ring as if they were polynomials
in Q[t], and we produce a product of degree at least three, we can apply the
division algorithm to subtract an appropriate multiple of f from the product to
produce an equivalent polynomial of the form at 2 + bt + c. Now it’s time for you
to practice this procedure.
EXERCISE 9.15.7 In Q[t], let f = t 4 + 2t + 1. Construct the form of elements
of Q[t]/(f ), and illustrate addition and multiplication.
Now for another very interesting example. Since Z3 is a field, Z3[t] is a
Euclidean domain, and we can construct the quotient ring Z3[t]/(f ) for f ∈ Z3[t]
in a similar way. Let’s use f = t 3 + t + 2, construct the quotient ring, and
look at addition and multiplication. By exactly the same reasoning as before,
Z3[t]/(f ) ={at 2 + bt + c : a, b, c ∈ Z3}. Notice this is a finite set. Each of a, b,
and c can take on values from {0, 1, 2}, soZ3[t]/(f ) has 27 elements. Adding
elements of Z3[t]/(f ) is easy:
(2t 2 + t + 2) + (t 2 + 2t + 2) =(f ) 3t 2 + 3t + 4 =(f ) 1 (9.78)
Doing multiplication would look like the following if we simplify the product by
way of the division algorithm.
(2t 2 + t + 2)(t 2 + 2t + 2) =(f ) 2t 4 + 5t 3 + 8t 2 + 6t + 4
=(f ) 2t 4 + 2t 3 + 2t 2 + 1
=(f ) (2t + 2)f
=(f ) 0
(9.79)