MATH 433 March 27, 2015
Problem 1.
Let M be the set of all 2 × 2 matrices of the form n 0 k n
, where n and k are integers. Under the operations of matrix addition and multiplication, does this set form a ring? Does
M form a field? Explain.
Solution: M is a commutative ring, but not a field.
We have n 0 k n
+
− n k n ′ 0 k ′ n ′
0 n
=
= n + n ′ k + k ′
− n 0
− k − n
0 n + n ′
,
, n 0 k n n ′ 0 k ′ n ′
= nn ′ 0 kn ′ + nk ′ nn ′
.
It follows that the set M is closed under matrix addition, taking the negative, and matrix multiplication.
Besides, the multiplication is commutative on M . Clearly, the set M contains the zero matrix. The associativity and commutativity of the addition, the associativity of the multiplication, and the distributive law hold on M since they hold on the set of all 2 × 2 matrices. Thus M is a commutative ring.
The ring M is not a field since it has zero-divisors (and zero-divisors do not admit multiplicative inverses).
For example, the matrix
0 0
1 0
∈ M is a zero-divisor as
0 0 0 0
1 0 1 0
=
0 0
0 0
.
Problem 2.
For any λ ∈ Q and any v ∈ Z let λ ⊙ v = λv if λ is an integer and λ ⊙ v = 0 otherwise. Does this “scalar multiplication” make the additive Abelian group Z into a vector space over the field Q ? Explain.
Solution: No, Z is not a vector space over Q .
One reason why Z is not a vector space over Q is that the scalar multiplication ⊙ does not distribute over addition. To be precise, the axiom ( λ + µ ) ⊙ v = λ ⊙ v + µ ⊙ v fails. Indeed, let v be an arbitrary odd integer. Then ( 1
2
+ 1
2
) ⊙ v = 1
2
⊙ v + 1
2
⊙ v as ( 1
2
+ 1
2
) ⊙ v = 1 ⊙ v = v is odd while 1
2
⊙ v + 1
2
⊙ v is even.
Remark.
The only information about the scalar multiplication ⊙ used in the above counterexample is that 1 ⊙ v = v and 1
2
⊙ v into a vector space over Q .
is an integer. It follows that the additive group Z , in principle, cannot be made