MC313: Int.
to Modern Algebra soln Assignment 11 November 14
DA-IICT, B.Tech-MnC, Sem IV Autumn 2024
1. Prove that every field is an integral domain.
soln:
Consider a, b in a field such that ab = 0. If a ̸= 0 then a−1 exists in the field.
∴ a−1 ab = 0 =⇒ b = 0.
So we can’t have any zero divisor in the field.
2. If U is an ideal of R and 1 ∈ U , prove that U = R.
soln:
Certainly U ⊆ R. Consider x ∈ R. Since U is an ideal and 1 ∈ U , we have 1·x = x ∈ U .
∴ R ⊆ U . So we conclude U = R.
3. Prove that any homomorphism of a field is either an isomorphism or takes each element
to 0.
soln:
Let ϕ : F → R be a homomorphism where F is a field. The kernel of this homomor-
phism is an ideal of F . A field can have only two possible ideals {0} or F . If the kernel
is {0} we have an isomorphism. If the kernel is F every element of F is mapped to 0
in R. That proves the statement.
4. If U and V are ideals of R prove that U + V is also an ideal.
soln:
U and V are subgroups of R. ∴ U + V is also a subgroup of R.
Let x ∈ R and w ∈ U + V . Then w = u + v for some u ∈ U, v ∈ V .
∴ xw = x(u + v) = xu + xv.
Now xu ∈ U and xv ∈ V .
∴ xu + xv = xw ∈ U + V .
∴ U + V is an ideal in R.
5. If R is a ring of integers, let U be the ideal consisting of all multiples of 7. Prove that
if V is an ideal of R such that U ⊂ V ⊂ R then either V = R or V = U .
soln:
All subsets of ⟨Z, +⟩ is of the type nZ where n is a positive integer.
Let V = nZ. Since U ⊂ V and 7 ∈ U , n|7.
∴ n = 1 or n = 7.
If n = 7, V = U . If n = 1, V = Z.
6. Consider the ring of quatrnions over Z5 given by a0 + a1 i + aj + a3 k with i2 = j 2 =
k 2 = −1 and ijk = −1 for any cyclic permutation of i, j, k and the tripple product is
1 for any non cyclic permutation of i, j, k. Prove that this ring is not a division ring.
soln:
The ring is a noncommutative ring with unity with 1 being the multiplicative identity.
1
The multiplicative inverse of an element a0 + a1 i + a2 j + a3 k is (a0 − a1 i − a2 j − a3 k) ·
(a − 02 + a21 + a22 + a23 )−1 provided (a − 02 + a21 + a22 + a23 )( mod 5) ̸= 0.
We can see that 1 + 2i will not have an inverse since 12 + 22 = 5, hence (12 + 22 )(
mod 5) = 0. So this ring is not a division ring.
7. If R is a ring with unit element 1 and ϕ : R → R′ is a homomorphism from R onto R′
then prove that ϕ(1) is the unit element of R′ .
soln:
Let r′ ∈ R′ . Let r ∈ R be such that ϕ(r) = r′ .
Then ϕ(1) · r′ = ϕ(1) · ϕ(r) = ϕ(1 · r) = ϕ(r) = r′ .
Similarly r′ · ϕ(1) = r′ .
∴ ϕ(1) is the unit element of R′ .
8. Consider the ring F of continuous real valued functions on the closed interval [0, 1] ⊂ R.
Show that the mapping ϕ : F → R defined as ϕ(f (x)) = f ( 21 ) is a homomorphism.
Show that the kernel of this homomorphism is a maximal ideal of F using the following
theorem:
Theorem 1 Let R be a commutative ring with unity. An ideal M of a ring R is
maximal if and only if the ring R/M is a field.
Try the proof without this theorem, considering an ideal that contains the kernel within
it.
Can you imagine the field F/M ?
soln:
Let us denote the kernel of the homomorphism as the ideal M of F .
By the homomorphism theorem F/M is isomorphic to R.
R is a field under usual addition and multiplication of real numbers.
So F/M is a field.
F is a commutative ring with unity, the constant function f (x) = 1, ∀ x ∈ [0, 1] being
the multiplicative identity in F .
Then the given theorem says that M is a maximal ideal of F .
Alternatively:
Let us denote the kernel of the homomorphism as M .
Let U be an ideal of F such that M ⊆ U ⊆ F .
If U ̸= M then ∃ g(x) ∈ F such that g( 21 ) = α ̸= 0.
Consider f (x) = g(x) − α.
Then f ( 12 ) = 0.
∴ f (x) ∈ M and since M ⊆ U, f (x) ∈ U .
g(x), f (x) ∈ U =⇒ g(x) − f (x) = h(x) = α ∈ U .
∴ 1(x) = α−1 h(x) = 1 ∈ U .
Let k(x) ∈ F .
2
Then k(x) = k(x) · 1(x) ∈ U .
∴ U = F, =⇒ M is maximal in F .
