Homework 4 for UTK – M351 – Algebra I Spring 2004, Jochen Denzler, MWF 10:10–11:00, Ayres 111

Problem 39: Given a commutative ring R with identity, we consider the set Seq(R) consisting of all sequences s = (s_{0}, s_{1}, s_{2}, s_{3}, …) where each s_{i} is an element of R. For instance, with R = ℤ, the following are elements of Seq(ℤ): (0, 1, 4, 9, …), or (1, 0, −1, 0, 1, 0, −1, …). Generally, we will denote by s_{i} the i^{th} entry in the sequence s, where we begin to count entries at number 0. We define the following operations on Seq(R):

The sum a + b of two sequences is defined componentwise: a + b = (a_{0} + b_{0}, a_{1} + b_{1}, a_{2} + b_{2}, …). The Cauchy product of two sequences is defined as follows:

ab = (a_{0}b_{0}, a_{0}b_{1} + a_{1}b_{0}, a_{0}b_{2} + a_{1}b_{1} + a_{2}b_{0}, …)

such that (ab)_{n} = ∑_{i=0}^{n} a_{i}b_{n−i} = a_{0}b_{n} + a_{1}b_{n−1} + … + a_{n−1}b_{1} + a_{n}b_{0}.

(a) Make sure that you understand the definition: To this end, calculate the Cauchy product ab of the sequence a = (1, 1, 1, 1, 1, 1, …) with b = (0, 1, 2, 3, 4, 5, …) in Seq(ℤ). Which number is the the entry (ab)_{30}?

(b) Now show that Seq(R) with these operations is a commutative ring.

We call this ring R[[X]] (The ad-hoc name Seq(R) was just for the set.)

Problem 40: In the ring ℤ[[X]], show that the element a = (1, 1, 1, 1, …) is invertible and give its inverse.

Problem 41: We consider the subset Seq_{0}(R) of Seq(R), consisting of those sequences that have only finitely many non-zero entries. For instance, the sequence (1, 2, 0, −7, 3, 0, 0, 0, 0, …) is in Seq_{0}(ℤ). Such sequences can be written in abbreviated form as finite sequences by omitting the trailing zeros: (1, 2, 0, −7, 3). Show that Seq_{0}(R) is a subring of Seq(R). In particular, to gain sufficient understanding concerning the closure of multiplication, calculate the Cauchy product of (1, 2, 0, −7, 3) and (2, −1, 4).

Problem 42: In the ring Seq_{0}(R), we denote the element (0, 1) as X. Calculate X^{0}, X^{2}, X^{3} etc., and write (1, 2, 0, −7, 3) as a linear combination of powers of X.

Problem 43: From now on, we will take the liberty of writing the elements of ℤ_{n} as 0, 1, 2, …, n − 1, rather than [0], [1], [2], …, [n − 1] when no confusion arises. Calculate (1 + 2X)^{3} in the ring ℤ_{3}[X].

Comments: The usual symbol for the ring Seq_{0}(R) is R[X], and this ring is called the polynomial ring with coefficients in R. Even though we can and will later plug in elements of R for the symbol X, as you would when viewing polynomials as functions of a variable, it is crucial that you do NOT view the ring of polynomials over R as a subring of the ring of functions from R to R. It MAY NOT BE one!!!

The usual symbol for the ring, consisting of the set Seq(R) and the addition and multiplication defined here, is R[[X]], and it is called the “ring of formal power series with coefficients in R”. (Name to be explained in lecture. Just take note here: unlike the power series you may have encountered at the end of Calculus II, you are NOT expected to plug anything in for X here, and therefore no convergence issues arise.) And one of the reasons I introduce this example is to stress the previous remark about polynomial rings, where plugging in ring elements for X is not part of the definition of R[X] either.

Problem 44: In the polynomial ring ℤ_{6}[X], find two polynomials p and q, such that deg(pq) < (deg p) + (deg q). Note that ℤ_{6} is not an integral domain; so the purpose of this problem is to show that the assumption that the coefficient ring be an integral domain is really needed for the degree formula to hold.

Problem 45: In the ring ℤ[X] take the polynomials a = X^{3} + X^{2} + 2X + 1 and b = 2X^{2}. Show that it is not possible to find polynomials q and r in ℤ[X] such that a = bq + r and deg r < deg b. If the coefficients are taken from a field, the euclidean algorithm asserts that such a division with remainder is possible. So this problem serves as an illustration that the requirement that the coefficient ring be a field is really needed for the euclidean algorithm.

Problem 46: In the ring ℚ[X], find a GCD of a = X^{3} − 7X^{2} + 3X + 3 and b = X^{3} − 6X^{2} + X + 7. Also write the GCD thus obtained as a linear combination of a and b.

Problem 47: In the ring ℤ_{13}[X], find a GCD of the “same” polynomials a = X^{3} − 7X^{2} + 3X + 3 and b = X^{3} − 6X^{2} + X + 7, and write the GCD thus obtained as a linear combination of a and b.

I put the word “same” in quotes, because this is an abuse of language. The coefficent −6 in b of problem 46 is the integer −6, whereas in problem 47, the ‘same’ −6 is a shorthand for the element [−6]_{13} = [7]_{13} ∈ ℤ_{13}. But it’s nevertheless common language usage to consider the ‘same’ polynomial in different rings.

Problem 48: In a polynomial ring R[X] (R is a commutative ring with 1), choose two polynomials p_{1}, p_{2}. Consider the set

I⟨p_{1}, p_{2}⟩ := {r_{1}p_{1} + r_{2}p_{2} | r_{1}, r_{2} ∈ R[X]}

of all linear combinations of p_{1} and p_{2}. (This is a set of common interest in algebra, but the notation I have used for it is different from the usual notation.)

Show that I⟨p_{1}, p_{2}⟩ is a subring of R[X] (it may not have a multiplicative identity, though).

Problem 49: Continuing the previous problem, show that I⟨p_{1}, p_{2}⟩ even is an ideal. — “Ideal” is a new concept for you, and here is the definition: A subring S of a commutative ring T is called an ideal if it has the property: For any s ∈ S and any t ∈ T, it holds st ∈ S.

Rmk: The same set of problems 48, 49 could be done with any number of given polynomials p_{1}, p_{2}, p_{3}, …, including the possibility of only a single polynomial.

Problem 50: Give an example of a polynomial in ℚ[X] that is not prime (i.e. can be factored), but has no root in ℚ. What is the smallest degree such a polynomial can have (explain why)?

Problem 51: Show that the polynomial p = X^{2} + X + 1 is irreducible in ℤ_{2}[X].

(Obviously p is not a constant polynomial, but: ) show that the polynomial function ℤ_{2} → ℤ_{2}, x ↦ p(x) is a constant function.

Problem 52: Show that the polynomial p = X^{4} + 1 is irreducible in ℚ[X], but not in ℝ[X] nor in ℂ[X]. Give a complete factorization in ℝ[X], and a complete factorization in ℂ[X].

Also give three different incomplete factorizations (product of two quadratics) in ℂ[X] (for later use).

Problem 53: In the fields ℤ_{p} for p = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, find one solution of the equations x^{2} + 1 = 0, x^{2} − 2 = 0, x^{2} + 2 = 0 each, or conclude that none exists. Basically that’s trial and error, and I have filled in all but three of the “doesn’t exist” cases, and a few of the existence cases, to save you work. Note also that in the example p = 29, to find solutions, I only needed to test 1, 2, 3, …, 14, since 15 ≡ −14, 16 ≡ −13,….

p x^{2} + 1 = 0 x^{2} − 2 = 0 x^{2} + 2 = 0
2  1  0  0
3  dne  dne
5  2  dne  dne
7  dne
11  dne
13  dne

17

19  dne  dne  6
23  dne  dne
29  12  dne  dne

Once this is accomplished, use the information, and wisdom gleaned from the very last part of the previous problem, to factor X^{4}+1 completely in ℤ_{p}[X] for the prime numbers p = 2, 3, 5, 7, 11, 13, 17 (and more of them, if you are bored, or want to get bored).

Background info: a simple result from the theory of quadratic residues (in elementary number theory), or in other terms, a simple argument about groups, which we have alas no time to go into, implies in particular: if p is an odd prime such that there is no element in ℤ_{p} whose square is −1, and also no element whose square is 2, then there does exist an element whose square is −2.

Accepting this fact, you can conclude that at least one of the factorizations of X^{4}+1 into quadratics (in ℚ[X]) found in problem 52 can serve as a model for factorization in ℤ_{p}[X]; in other words: X^{4}+1 can be factored nontrivially in *every* ℤ_{p}[X].

Problem 54: We have seen that the mapping F [X] → Fct(F →F), which assigns to each polynomial the corresponding polynomial function F → F cannot be one-to-one, if the field F contains finitely many elements. (Simply because in this case there are still infinitely many polynomials, but only finitely many functions F → F).

Now show conversely that, if F contains infinitely many elements, then the mapping F [X] → Fct(F →F) is indeed one-to-one.

Problem 55: We have seen that a polynomial of degree n in F [X] can have at most n roots in F (or any extension field of F). This assumed that F be a field. In contrast, consider the polynomial ring ℤ_{25}[X].

How many roots does the polynomial X^{2} have in ℤ_{25}?

Give several essentially different factorizations of X^{2} in ℤ_{25}, thus showing that the unique factorization property may fail in R[X], if R is not a field.

Problem 56: In ℤ_{2}[X], consider the ideal I of all multiples of the irreducible polynomial X^{3} + X + 1. Denoting the equivalence class [X]_{I} in ℤ_{2}[X]/I as j, list all elements of ℤ_{2}[X]/I, and give their multiplication table. In particular, find the inverse of 1 + j in the field ℤ_{2}[X]/I.