Download MA554 Workshop 3

MA554
Workshop 3
The ring Z/n of integers modulo n
So far we have considered Z/n as an abelian group under addition. In that case, we
defined [a]n + [b]n = [a + b]n . In a similar way, we define multiplication by [a]n [b]n = [ab]n .
1. In Z/7, show that [2]7 [4]7 = [1]7 . (Interpret this as saying that both 2 and 4 have
multiplicative inverses in this ring.) Find a multiplicative inverse for [3]7 .
(In fact, Z/7 is a field—this is because 7 is a prime number—so that every non-zero
element has a multiplicative inverse; but you don’t need to check that.)
2. Give the definition of zero divisor in a ring. Which elements of Z/10 are zero divisors?
Show that any element of Z/10 that is not a zero divisors has a multiplicative inverse.
3. Let n be a fixed positive integer. Prove that Z/n is a ring under addition and
multiplication. (You don’t need to show that it is an abelian group under + because that
was on sheet 3.)
Polynomial rings
4. Let f = t3 + 2t2 + 3t + 4 and g = t2 + 1. (These are elements of the ring R[t], for
example.) By using long division of polynomials, find polynomials q, r so that
f = qg + r
and r = 0 or deg r < deg g.
Repeat this exercise with f = t4 − 2t3 + 3t − 2 and g = t − 1.
Repeat this exercise with f = t4 − 2t3 + 3t − 2 and g = 2t2 − 1.
5. Show that R[x] is a ring (under addition and multiplication).
What does it mean to say that a ring R is an integral domain?
Show that R[x] is an integral domain. [Hint: if f, g are non-zero polynomials, consider
their leading terms and that of their product f g.]
Rings of matrices
6. The set Mat(2, R) of 2 × 2 matrices with real entries is a ring under usual matrix
addition and multiplication. Say which matrix is the additive identity (the ‘zero’ of the
ring) and which is the multiplicative identity (the ‘one’ of the ring).
Find two matrices A, B ∈ Mat(2, R) for which AB 6= BA. Deduce that Mat(2, R) is
not a commutative ring.
Find a zero divisor in Mat(2, R). [Hint: you’re not allowed to use the zero matrix by
definition, but you can try using matrices that have many zero entries.]
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