Homework 2 for UTK – M351 – Algebra I Spring 2007, Jochen Denzler, MWF 10:10–11:00, Ayres 205

Problem 11: Explore and discover…

Check each field axiom for validity in the following examples (X, +, ·), and if not all are verifiied, decide whether they are rings, commutative rings, with or without unity. Here X is

(a) The set of integers, ℤ with the usual meanings of + and · (also in the following examples with sets of numbers).

(a1) The set of odd integers

(a2) The set of even integers

(b) The set of rationals, ℚ

(c) The set of 2 × 2 matrices with real entries (where + and · denote addition and multiplication of matrices)

(c1) The set of 2 × 2 matrices of the form [ \begin{matrix}a & b \\ 0 & 0 \end{matrix} ].

(d) The set P of polynomials in the variable x, with rational coefficients.

(e) The set {E, O}, where the following rules define + and · : E + E = E, E + O = O + E = O, O + O = E. E · E = E, E · O = O · E = E, O · O = O.

(f) You invent this problem: The set {Z, I, T } which made a commutative group with an operation + in Example&Hwk #5. Together with definitions for a multiplication · that makes the set at least a ring. Check any further field axioms for validity in this example.

(g) Same task as before; but this time we want a set of four elements. You should now focus on which axioms we are losing, as compared to (e) and (f). Any hunch what feature of the numbers 2, 3, 4 is decisive?

Problem 12: Let R be any ring (with operations + and ·). Define the matrix ring M_{n}(R) as the set of all n × n matrices whose entries are in R. The addition will be componentwise, and the multiplication will also be defined as in the usual matrix algebra course: (AB)_{ik} = ∑_{j=1}^{n} A_{ij}B_{jk}.

Show that M_{n}(R) is a ring, and show that it has an identity provided R has.

Note: You should be able to handle the ∑ notation. If not, you may ask for help. I will accept a solution that only takes care of n = 2. But at least the stronger half of the students should attempt to do it for general n using the sum notation, or possibly a three-dots-substitute for ∑. Be aware that the ∑ notation for general n is shorter than the pedestrian way for n = 2 only!

Problem 13: Let R be a ring (with operations + and ·). We define operations on R × R as follows:

(x, y) + (u, v) := (x + u, y + v) , (x, y) · (u, v) := (xu − yv, xv + yu)

Here, as usual, a − b stands for a plus the additive inverse of b.

Show that this defines a ring. We are going to denote R × R, when adorned with these operations, as R[i]. (This is admittedly a strange name as of yet).

Problem 14: Continuing the previous problem, show that R[i] has an identity, if R has. Show also that R[i] is commutative, if R is.

Assume that R is a field. Must R[i] necessarily be a field? If not, what condition must be satisfied in R to guarantee that R[i] is a field? Some may find it convenient to attempt #6 before this second part of #5; try it in case you have difficulties at this moment.

Problem 15: Continuing the previous problem, let R be a commutative ring with identity 1. In R[i], we’ll denote the element (0, 1) with the special symbol i. (You start getting an idea where R[i] got its name from.) Calculate i · i (too easy…).

I claim that, for the case R = ℝ, the field of real numbers, you should be at least vaguely familiar with ℝ[i] under a different name. Which one? Set up a complete translation dictionary (it has only a few lines) that translates the notation set up in Problem 4 into the more familiar one.

Show that ℝ[i] is a field.

Problem 16: I claimed in class that the power set P(M) (which is the set of all subsets of M), together with the operations A + B := (A \ B) ∪ (B \ A) and A · B := A ∩ B is a commutative ring with identity. Prove the distributive law (as far as not done in class yet) and the associativity for + .

Problem 17: Suppose, in a ring, the extra property a · a = a is verified for every a. (The previous problem is an example where this happens.) Show generally, that a ring satisfying that extra property is automatically commutative: Since this is a bit tricky, I give you the steps (the steps how I did it; I wouldn’t claim with certainty that there cannot be another, shorter way):

(a) Show that b + b = 0 for every b. You do this by calculating (b + b) · (b + b) in two different ways.

(b) Show that bcb = cbc for every b, c. You do this by calculating (b · c − c · b) · (b · c − c · b) in two different ways.

(c) Conclude b · c = c · b from part (b) by appropriate multiplications and by again using a · a = a.

Each step needs to be justified by explicit reference to the ring axioms (or to consequences thereof that were proved in class).

Problem 18: Show: A ring with exactly 3 elements, {0, a, b} must be commutative. Hint: First show a + a = b.

Problem 19: In the ring ℤ, find the gcd of 43728 and 15360 (‘the’ gcd: so make it a positive number), and express this gcd in the form 43728k + 15360ℓ with integers k, ℓ.

Problem 20: In this problem, we’ll see that the division algorithm can be mimicked in the ring ℤ[i], which consists of the numbers a + bi where a, b ∈ ℤ and i is the imaginary unit. You may view this ring either as a subring of ℂ, or as an instance of the class of rings constructed in Problem 5.

Given a = a_{1} + a_{2}i ∈ ℤ[i] and b = b_{1} + b_{2}i ∈ ℤ[i] with b ≠ 0, we want to find q = q_{1} + q_{2}i ∈ ℤ[i] and r = r_{1} + r_{2}i ∈ ℤ[i] such that a = qb + r and r “smaller” than b. We cannot require “0 ≤ r < b” because we do not have an order in the ring ℤ[i]; a statement “0 ≤ r < b” would be meaningless. Instead we will use the absolute value of complex numbers and require that |r| is smaller than |b|, or, equivalently: r_{1}^{2} + r_{2}^{2} < b_{1}^{2} + b_{2}^{2}.

Given a = a_{1} + a_{2}i ∈ ℤ[i] and b = b_{1} + b_{2}i ∈ ℤ[i] \ {0}, let ϑ = ϑ_{1} + iϑ_{2} ∈ ℂ be the exact quotient ϑ = a/b. Let q_{1} be an integer closest possible to ϑ_{1} (there may be several equally good choices) and let q_{2} be an integer closest possible to ϑ_{2}. Let r be the remainder making a = qb + r true

(a) To make sure you understand the principle, find q and r according to the prescription of the preceding paragraphs in the case a = 517 + 213i, b = 11 + 25i. Check that r_{1}^{2} + r_{2}^{2} is indeed less than b_{1}^{2} + b_{2}^{2}.

(b) Write out explicitly what a = ϑb means for a_{1}, a_{2}, b_{1}, b_{2} and ϑ_{1}, ϑ_{2}. — Write out explicitly what a = qb + r means for a_{1}, a_{2}, b_{1}, b_{2}, q_{1}, q_{2}, r_{1}, r_{2}. — What does your prescription about the choice of q imply about the size of q_{1} − ϑ_{1}, q_{2} − ϑ_{2}?

(c) Express r_{1} and r_{2} in terms of b_{1}, b_{2}, q_{1} − ϑ_{1}, q_{2} − ϑ_{2} and conclude that r_{1}^{2} + r_{2}^{2} < b_{1}^{2} + b_{2}^{2}.

Problem 21: In many rings that are not fields, it can happen that ab = 0 for certain a ≠ 0 and b ≠ 0. The next problem gives a whole lot of examples, this one wants you merely to show:

In any ring, if ab = 0, but a ≠ 0 and b ≠ 0, then neither a nor b has a multiplicative inverse.

(Comment: Therefore, in fields this phenomenon ab = 0 with a ≠ 0 and b ≠ 0 cannot happen, because there, all nonzero elements have multiplicative inverses. The phenomenon also does not occur in the ring ℤ, or, for that matter, in any ring that is subring of a field.)

Problem 22: 2pts each for (a), (b), (c) ∪ (d), (e) Let me introduce a name: In a ring, whenever a ≠ 0 and b ≠ 0 satisfy ab = 0, then a and b are called zero divisors. In this problem, you’ll find zero divisors in various rings:

(a) The ring C^{0}[0, 1] of continuous, real-valued functions on the interval [0, 1], with the usual addition and multiplication of functions. (The proof of the ring properties is straightforward, you are not required to write it out here.) Find a pair of zero divisors. If you find this difficult, then the most likely source of your difficulty is that you are shying away from piecewise defined functions.

(b) In the ring M_{2}(ℤ) = ℤ^{2×2} of 2 × 2 matrices with integer entries, find a pair of zero divisors.

(c) In the direct sum ℤ ⊕ ℤ, find a pair of zero divisors.

(d) In the ring P(M) described in Problem 16, where M = {□, ♢, ⋆, △}, find a pair of zero divisors.

(e) Bonus problem: How many pairs of zero divisors does the commutative ring in (d) have, not counting pairs (A, B) and (B, A) as different?

Problem 23: Show that in a ring with identity that has more than one element, the multiplicative identity is automatically different from the additive identity.

Problem 24: In a ring with identity (not necessarily commutative!), assume that the elements a and b each have a multiplicative inverse; we’ll call them a^{−1} and b^{−1} respectively. Show that ab has a multiplicative inverse as well, and give a ‘formula’ for it, in terms of a^{−1} and b^{−1}.

Problem 25: Let A be any subset of [0, 1] (think of finitely many numbers between 0 and 1). Within the ring C^{0}[0, 1] (defined in 22a above), consider the set

C_{A}^{0}[0, 1] := {f | f(x) = 0 for all x ∈ A}

Show that C_{A}^{0}[0, 1] is a subring of C^{0}[0, 1]. (Comment: The name C_{A}^{0}[0, 1] is an ad-hoc name given for this problem, unlike the name C^{0}[0, 1], which is generally understood in the mathematical community.)

Problem 26: Warning / Surprise: If R is a ring with identity 1_{R} and S is a subring not containing the element 1_{R}, then S might still have an identity 1_{S} different from 1_{R}. In that case, by the uniqueness of the identity, 1_{S} could not serve as a multiplicative identity in R. In this problem, you’ll see two examples:

(a) Take the ring ℤ⊕ℤ. Give its multiplicative identity. Show that the ring ℤ⊕{0} = {(a, 0) | a ∈ ℤ} is a subring of ℤ ⊕ ℤ. Show that it does have a multiplicative identity, and exhibit it.

(b) In the ring P(M), where M = {□, ♢, ⋆, △}, what is the multiplicative identity? Show that P(N), where N = {□, ⋆, △}, is a subring. What is its multiplicative identity?

Problem 27: Why can a similar substitution of the additive identity not happen?

Problem 28: Divisibility by 11: To find the remainder of a number when divided by 11, for an integer given in decimal notation, the following rule can be used with the digits: Add the digits from right to left, with alternating sign. Add/subtract multiples of 11 as needed or desired. The result (between 0 and 10) is the remainder of the given integer upon division by 11.

Example: a = 357123946803; We calculate c = 3 − 0 + 8 − 6 + 4 − 9 + 3 − 2 + 1 − 7 + 5 − 3 = −3 Add 11 to get 8 (between 0 and 10): The remainder of a when divided by 11 is therefore 8.

Prove this rule by writing up a claculation in the ring ℤ_{11}

Problem 29: Given an integer a, let Q(a) be the sum of its digits. E.g., Q(37491) = 3 + 7 + 4 + 9 + 1 = 24. What is

Q(Q(Q(4444^{4444}))) ?

To answer the problem, give a rough estimate how large the number could be at most, and use a calculation in ℤ_{9} as a second piece of information.

Problem 30: Show that 13 (which is a prime in ℤ of course) is not irreducible in the ring ℤ[i]. In other words, find integers a, b, c, d such that (a+bi)(c+di) = 13, but neither of the numbers a+bi, c+di should be 1, −1, i or −i.

Hint: such numbers are easier to guess (and finding one solution is good enough) than to find systematically; see if you can make c + di = a − bi.

Problem 31: Let’s try the ring ℤ[\sqrt{−5}] for a change: another subring of ℂ; it consists of all the numbers a+b\sqrt{−5} with a, b ∈ ℤ.

First show that the only numbers dividing the identity 1 in this ring are +1 and −1: you have to find all integers a, b, c, d such that (a + b\sqrt{−5})(c + d\sqrt{−5}) = 1.

Now show that 3 has no divisors but ±3 and ±1 of 3 in this ring. Show the same for the numbers 2 and 1 ± \sqrt{−5}. In other words, all of these numbers are irreducible in the ring ℤ[\sqrt{−5}].

Hint: The task to find all integers a, b, c, d such that (a + b\sqrt{−5})(c + d\sqrt{−5}) = 1 (or 3 etc) is simplified a lot if you first multiply this equation with its complex conjugate. If you still get stuck, hand it in as pingpong hwk.

Problem 32: Show that in the ring ℤ[\sqrt{−5}], the number 6 can be written as a product of irreducible factors in two essentially different ways. (Refer to previous problem for raw material).