Mathematics 351 – Abstract Algebra I Solutions for Midterm Exam 2 – November 14, 2007

I. Terminology.

A) Let G be a group and n be an integer. What does it mean to say that |g| = n for g ∈ G?

Solution: |g| is the order of the element g. This is, by definition, the smallest positive integer such that g^{n} = e (the identity element in G).

B) Let I be an ideal in a ring R. What does it mean to say that a ≡ b mod I?

Solution: This equivalent to saying a − b ∈ I.

C) Give an example of an ideal in a ring R that is prime but not maximal.

Solution: One such example is the ideal P = {(a, 0) : a ∈ ℤ} in R = ℤ × ℤ. The quotient ring R/P is isomorphic to ℤ, which is an integral domain but not a field. Hence P is a prime ideal, but not a maximal ideal.

II. Let R be the ring of functions f : ℝ → ℝ that have derivatives of all orders. Show that I = {f ∈ R : f(0) = f′(0) = 0} is an ideal in R.

Solution: I is clearly nonempty since it contains the zero function, f(x) = x^{2}, etc. Let f, g ∈ I so f(0) = f′(0) = 0 and g(0) = g′(0) = 0. Then (f − g)(0) = f(0) − g(0) = 0 and (f − g)′(0) = f′(0) − g′(0) = 0 − 0 = 0 (by the sum rule for derivatives). Hence f − g ∈ I. Next, let f ∈ I, and let g be any function in R. Since R is a commutative ring, we only need to check that fg ∈ I. But (fg)(0) = f(0)g(0) = 0 · g(0) = 0. Moreover, by the product rule for derivatives,

(fg)′(0) = f′(0)g(0) + f(0)g′(0) = 0 · g(0) + 0 · g′(0) = 0.

Hence fg ∈ I.

III. All parts of this problem refer to p(x) = x^{2} + x in ℤ_{2}[x].

A) How many distinct cosets of the ideal (p(x)) are there in ℤ_{2}[x]?

Solution: Since ℤ_{2} is a field, we have a division algorithm in the polynomial ring ℤ_{2}[x], and by our general results in this situation, the distinct cosets of (p(x)) are in one-to-one correspondence with the possible remainders on division by p(x). Since p(x) has degree 2, these remainders are all the polynomials of degree 1 or less in ℤ_{2}[x] (and the zero polynomial): 0, 1, x, x + 1. There are 4 distinct cosets:

0 + (p(x)), 1 + (p(x)), x + (p(x)), x + 1 + (p(x)).

B) Construct the coset addition and multiplication tables for the quotient ring ℤ_{2}[x]/(p(x)).

Solution: The addition table (omitting the +(p(x)) in the description of the cosets for simplicity):

\begin{matrix}+ & 0 & 1 & x & x + 1 \\ 0 & 0 & 1 & x & x + 1 \\ 1 & 1 & 0 & x + 1 & x \\ x & x & x + 1 & 0 & 1 \\ x + 1 & x + 1 & x & 1 & 0 \end{matrix}

The multiplication table is:

\begin{matrix}· & 0 & 1 & x & x + 1 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & x & x + 1 \\ x & 0 & x & x & 0 \\ x + 1 & 0 & x + 1 & 0 & x + 1 \end{matrix}

C) State and prove the First Isomorphism Theorem for rings.

Solution: The First Isomorphism states that if f : R → S is a ring homomorphism, then im(f) ≃ R/ ker(f). To prove this, we write K = ker(f) and consider the mapping ϕ : R/K → im(f) defined by ϕ(a + K) = f(a). This is well-defined since if a + K = b + K, then a − b ∈ K, so f(a − b) = 0, which shows f(a) = f(b). Hence ϕ(a + K) = ϕ(b + K). Next, we show that ϕ is a ring homomorphism. Let a + K and b + K be elements of R/K. Then ϕ((a + K) + (b + K)) = ϕ((a + b) + K) = f(a + b) = f(a) + f(b) = ϕ(a + K) + ϕ(b + K). Similarly, ϕ((a + K) · (b + K)) = ϕ((a · b) + K) = f(a · b) = f(a) · f(b) = ϕ(a + K) · ϕ(b + K). (We are using the definition of the coset sum and product operations and the fact that f is a ring homomorphism here. Now, ϕ is onto im(f) by definition. Every element of im(f) is of the form f(a) for some a ∈ R. Hence f(a) = ϕ(a + K). Finally, ϕ is one-to-one since if ϕ(a + K) = ϕ(b + K), then f(a) = f(b) which implies f(a − b) = 0 so a − b ∈ K, and hence a + K = b + K.

D) Let ℤ_{2} × ℤ_{2} = {(a, b) : a, b ∈ ℤ_{2}}, a ring under the component-wise sum and product operations. Show that f : ℤ_{2}[x] → ℤ_{2} × ℤ_{2} defined by f(g(x)) = (g(0), g(1)) is a ring homomorphism and determine ker(f).

Solution: f is a ring homomorphism because

f(g(x)+h(x)) = (g(0)+h(0), g(1)+h(1)) = (g(0), g(1))+(h(0), h(1)) = f(g(x))+f(h(x))

and

f(g(x) · h(x)) = (g(0) · h(0), g(1) · h(1)) = (g(0), g(1)) · (h(0), h(1)) = f(g(x)) · f(h(x)).

The kernel of f consists of all polynomials g(x) ∈ ℤ_{2}[x] such that g(0) = 0 and g(1) = 0. This means that x and (x + 1) must divide g, so g(x) ∈ (p(x)) = (x(x + 1)). On the other hand every polynomial in (p(x)) is clearly in ker(f) since such polynomials are zero at x = 0 and x = 1. Hence ker(f) = (p(x)).

E) Deduce that ℤ_{2}[x]/(p(x)) is isomorphic as a ring to ℤ_{2} × ℤ_{2}.

Solution: This follows from part D by the First Isomorphism Theorem. Let f be as in part D. The image of f is all of ℤ_{2} × ℤ_{2}, so that is isomorphic to ℤ_{2}[x]/ ker(f) and ker(f) = (p(x)). Comment: It is also possible to see this by comparing the addition and multiplication tables for the two rings. But it is necessary then to be somewhat careful in defining the correspondence between the elements. For example, the “obvious” mapping

0 + (p(x)) ↦ (0, 0)
1 + (p(x)) ↦ (1, 0)
x + (p(x)) ↦ (0, 1)
x + 1 + (p(x)) ↦ (1, 1)

is an isomorphism of the additive groups of the two rings, but not an isomorphism of rings – note that the multiplicative identity in ℤ_{2}[x]/(p(x)) is 1 + (p(x)), but the multiplicative identity in ℤ_{2} × ℤ_{2} is (1, 1). A ring isomorphism would have to map the multiplicative identity to the multiplicative identity, so a correct mapping to get a ring isomorphism here is:

0 + (p(x)) ↦ (0, 0)
1 + (p(x)) ↦ (1, 1)
x + (p(x)) ↦ (0, 1)
x + 1 + (p(x)) ↦ (1, 0).

This is the mapping obtained from f from part D by the proof of the First Isomorphism Theorem(!) There is also another correct way to do this by mapping

0 + (p(x)) ↦ (0, 0)
1 + (p(x)) ↦ (1, 1)
x + (p(x)) ↦ (1, 0)
x + 1 + (p(x)) ↦ (0, 1).

IV. Let G = {4, 8, 12, 16} ⊂ ℤ_{20}. Is G a group under multiplication mod 20? If so, say why and determine if G is cyclic. If not, identify which of the group axioms fail.

Solution: G is a group under multiplication mod 20, as can be seen from the operation table

\begin{matrix}· & 4 & 8 & 12 & 16 \\ 4 & 16 & 12 & 8 & 4 \\ 8 & 12 & 4 & 16 & 8 \\ 12 & 8 & 16 & 4 & 12 \\ 16 & 4 & 8 & 12 & 16 \end{matrix}

We see 16 is an identity for multiplication, every element has an inverse (that is, for each u, there is some v such that uv = 16), and associativity follows from associativity of multiplication in ℤ_{20}. The group G is cyclic, since the powers of 8 generate all the elements: 8^{1} = 8, 8^{2} = 4, 8^{3} = 12, 8^{4} = 16.

V.

A) State Lagrange’s Theorem for groups.

Solution: Lagrange’s Theorem says that if G is a finite group and H is a subgroup of G, then |H| divides |G|. In particular, |G| = [G : H]|H|.

B) Let G be a group of order 60. Is it possible for G to contain elements a, b with |a| = 3 and aba^{−1} = b^{4}? Why or why not?

Solution: There certainly can be such elements. Note for instance that a cyclic group of order 60 generated by c will contain elements a = c^{20} and b = c^{40}, both of order 3. Then aba^{−1} = baa^{−1} = b = b^{4} since a and b commute and b also has order 3.