School of Mathematics and Statistics MT4517 Rings & Fields ...

Solution to Exercise 3.5. It is straightforward to verify that R = { r + s √ 2 : r, s ∈ Q } is a
commutative ring with respect to the operations + and ∗ in R. It remains to prove that every
non-zero element of R is a unit. Let r + s √ 2 ∈ R. Then
1
r + s √ 2 = r − s √ 2
(r + s √ 2)(r − s √ 2) = r − s√ 2
r 2 − 2s 2 ∈ R
is the multiplicative inverse for r + s √ 2 in R. Thus R is a field.
Let Σ be the smallest subfield of R containing √ 2. We will prove that R = Σ. First note that,
since 1 ∈ Σ, n ∈ Σ for all n ∈ Z. Since Σ is a field, it follows immediately that Q ⊆ Σ. Since Σ
contains √ 2 and is closed under multplication and summation, we obtain that R = { r + s √ 2 :
r, s ∈ Q } ⊆ Σ. So, R ⊆ Σ. On the other hand, R is a subfield of R that contains √ 2. Hence, by
the choice of Σ, we have Σ ⊆ R. Therefore R = Σ, as required.
Solution to Exercise 3.6. Let I be the ideal in R generated by √ 2. Then
1 = √ 2 ·
1
√
2
∈ I
and so I = R.
Solution to Exercise 3.7. Let R be a commutative ring with identity whose only ideals are {0}
and R. In order to prove that R is a ring it suffices to show that every non-zero element in R is
a unit. So let r ∈ R \ {0}. Then the ideal generated by (r) = R. Hence there exists t ∈ R such
that rt = 1 and r is a unit.
Now, let R = Z/(6). Then R is a commutative ring with identity. All the non-invertible
elements of R are A = {0, 2, 3, 4}. Since 3 − 2 = 1 /∈ A, it follows that A is not an ideal in R.
Solution to Exercise 3.8. Obviously if A, B ∈ R, then A − B ∈ R. We are left to prove that
A ∗ B ∈ R for all A, B ∈ R. So, if
( ) ( )
a b
c d
A = and B = ,
2b a
2d c
then
( )
ac + 2bd ad + bc
A ∗ B =
∈ R.
2(ad + bc) ac + 2bd
Thus R is a subring of M 2 (R).
Let F be those matrices from R whose entries are from Q. It is straightforward to check that
F is a commutative subring of M 2 (R). Let A ∈ F \ {0}:
( ) a b
A = .
2b a
If det(A) = a 2 − 2b 2 = 0, then √ 2 ∈ Q, a contradiction. Hence det(A) ≠ 0 for all A ∈ F \ {0}
and so A −1 in M 2 (R). Since the entries of A −1 are obtained from those of A by adding and
multiplying it follows that A −1 ∈ F . Hence every non-zero element of F is a unit and F is a
field.
Let Σ be the set of matrices of the type
( ) a b
A =
2b a
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