Homework 2 for UTK – M351 – Algebra I Spring 2004, Jochen Denzler, MWF 10:10–11:00, Ayres 111

Problem 3: 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 4: 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 5: 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 6: 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 7: 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 8: Suppose, in a ring, the extra property a · a = a is verified for every a. (We had two examples where this happened: Ex. 3.4 on p. 7 of the book, and the example in the previous problem). 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 9: 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 10: 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 7, 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 11: Show that in a ring with identity that has more than one element, the multiplicative identity is automatically different from the additive identity.

Problem 12: 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 13: 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 10a 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 14: 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 15: Why can a similar substitution of the additive identity not happen?

Problem 16: Here is another example of an ordered ring: Let R be the ring of polynomials a_{0} + a_{1}x + a_{2}x^{2} + … + a_{n}x^{n} with real coefficients. The operations are the usual ones you know from calculus, so R is a subring of C^{0}(ℝ).

Show that the definition below of R^{+} satisfies the axioms for an ordered ring:

(a) If p ∈ R^{+} and q ∈ R^{+}, then p + q ∈ R^{+}

(b) If p ∈ R^{+} and q ∈ R^{+}, then p · q ∈ R^{+}

(c) Exactly one of the following three is true: p = 0, or p ∈ R^{+}, or −p ∈ R^{+}.

\underline{Def:} p ∈ R^{+}, iff in the list of coefficients \{a_{0}, a_{1}, …, a_{n}\} of p := a_{0} + a_{1}x + … + a_{n}x^{n}, the first nonzero number is positive. If all coefficients are 0 or the first nonzero coefficient is negative, then p ∉ R^{+}.

Order the following polynomials from smallest to largest. p_{1} = 1, p_{2} = x^{2}, p_{3} = x − x^{2}, p_{4} = 1 − x, p_{5} = 3x^{3}, p_{6} = x + x^{2}, p_{7} = x^{2} + 25x^{11}, p_{8} = 2x − x^{4}, p_{9} = −x^{4} + 3x^{5}, p_{10} = x^{8}.

Problem 17: The same ring can be ordered in a different way: In this problem define p ∈ R^{+}, iff in the list of coefficients {a_{0}, a_{1}, …, a_{n}} of p := a_{0} + a_{1}x + … + a_{n}x^{n}, the last nonzero number is positive.

Order the same polynomials as previously from smallest to largest, according to the new definition.

Comment: There are many more choices for an ordering; relying on calculus, you could choose any x_{0} ∈ ℝ and say p ∈ R^{+}, provided the first nonzero among p(x_{0}), p′(x_{0}), p′′(x_{0}), ldots, p^{(n)}(x_{0}) is positive. — When we study polynomials in more detail later, we can do the same thing within a pure algebraic framework (not relying on calculus), but the punchline here is just to give you more illustrations of the concept of ordered rings.

Problem 18: Prove: In a commutative ring R, it holds: (ab)^{n} = a^{n}b^{n} for any n ∈ ℕ. and for any a, b ∈ R. (from p. 34 of textbook)

Problem 19: In the ring M_{2}(ℤ) (the ring of 2 × 2 matrices with integer entries), prove that

[ \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ]^{n} = [ \begin{matrix}1 & n \\ 0 & 1 \end{matrix} ]

(from p. 35 of textbook)

Problem 20: Review the definition of ‘multiple’; in the ring P(M) described in problem 7, what is nA for an integer n and a set A ⊆ M?

Format of the answer: If n is  , then nA =  . If n is  , then nA =  .