Monthly Archives: December 2010

The polynomial x² – √2 is irreducible over ZZ[√2]

Prove that p(x) = x^2 - \sqrt{2} is irreducible over \mathbb{Z}[\sqrt{2}].


In this previous exercise, we showed that \mathbb{Z}[\sqrt{2}] is a Euclidean domain, and thus a unique factorization domain.

If p(x) is reducible, then it has a root; say \alpha = a+b\sqrt{2}. Evidently, then, we have (a^2 + 2b^2) + 2ab \sqrt{2} = \sqrt{2}. Comparing coefficients, 2ab = 1 for some integers a and b, a contradiction. Thus p(x) is irreducible over \mathbb{Z}[\sqrt{2}].

Exhibit an isomorphism between two finite fields

Let F = \mathbb{Z}/(11) denote the finite field with 11 elements. Prove that F[x]/(x^2+1) and F[y]/(y^2+2y+2) are finite fields containing 121 elements, and prove that they are isomorphic.


First, we need to demonstrate that p(x) = x^2 + 1 and q(y) = y^2 + 2y + 2 are irreducible over F. Note that each, if reducible, must have a root. The first is irreducible because -1 \equiv 10 is not a square mod 11. The second can be seen to have no roots by computing q(a) for each a \in F. Thus F[x]/(x^2+1) and F[y]/(y^2 + 2y + 2) are finite fields containing 121 elements.

Define \varphi : F[x] \rightarrow F[y]/(y^2 + 2y+2) by extending x \mapsto y+1 homomorphically. Now \varphi(x^2 + 1) = (y+1)^2 + 1 = y^2 + 2y + 2 = 0, so that (x^2+1) \subseteq \mathsf{ker}\ \varphi. Now since F[x] is a Euclidean domain and x^2+1 is irreducible, x^2 + 1 is in fact prime and thus (x^2+1) is maximal. Note that \varphi is nontrivial because (for example) \varphi(1) = 1. Thus \mathsf{ker}\ \varphi = (x^2 + 1). By the first isomorphism theorem, we have an injective ring homomorphism \psi : F[x]/(x^2+1) \rightarrow F[y]/(y^2 + 2y + 2); because these sets are finite, \psi is surjective and thus also a ring isomorphism.

As a ring, the complex numbers are isomorphic to RR[x]/(x² + 1)

Prove that the ring of complex numbers is isomorphic to \mathbb{R}[x]/(x^2+1).


Define \varphi : \mathbb{R}[x] \rightarrow \mathbb{C} by embedding coefficients as usual and mapping x to i; this map is a ring homomorphism, and moreover is surjective since \varphi(a + bx) = a+bi.

Note that \varphi(x^2 + 1) = i^2 + 1 = 0, so that x^2+1 \in \mathsf{ker}\ \varphi. Now if x^2+1 is reducible over the reals, then (being of degree 2) it must have a linear factor and thus a root. Suppose \alpha is such a root; that is, \alpha^2 = -1. Note that \alpha^2 is nonnegative for all real numbers, so we have a contradiction. Thus x^2+1 is irreducible over the reals, and thus \mathbb{R}[x]/(x^2+1) is a field. If \mathsf{ker}\ \varphi properly contains (x^2+1), then in fact \mathsf{ker}\ \varphi = \mathbb{R}[x] since fields have no nontrivial ideals and using the lattice isomorphism theorem for rings. So we have \mathsf{ker}\ \varphi = (x^2 + 1), and by the first isomorphism theorem, \mathbb{C} \cong \mathbb{R}[x]/(x^2+1).

Construct finite fields of given orders

Construct fields of orders 9, 49, 8, and 81.


Recall that \mathbb{Z}/(p) is a finite field of order p for primes p.

By this previous exercise, if F is a field and p(x) \in F[x] is irreducible, then F/(p(x)) is a field. By this previous exercise, if F is a finite field of order q and p(x) has positive degree n, then F/(p(x)) has cardinality q^n. Thus it suffices to exhibit irreducible elements of degree 2 over \mathbb{Z}/(3) and \mathbb{Z}/(7), of degree 3 over \mathbb{Z}/(2), and of degree 4 over \mathbb{Z}/(3).

We saw in this previous exercise that x^2 + 1 \in \mathbb{Z}/(3)[x] is irreducible. Thus \mathbb{Z}/(3)[x]/(x^2 + 1) is a finite field of order 9.

Note that the squares mod 7 are 0, 1, 2, and 4. Thus x^2 + 1 is irreducible in \mathbb{Z}/(7)[x], and \mathbb{Z}/(7)[x]/(x^2+1) is a field of order 49.

We saw in this previous exercise that x^3 + x + 1 is irreducible over \mathbb{Z}/(2), so that \mathbb{Z}/(2)[x]/(x^3 + x + 1) is a field of order 8.

Next, we claim that p(x) = x^4 + x + 2 is irreducible over \mathbb{Z}/(3). Since p(0) = p(2) = 2 and p(1) = 1, p(x) has no linear factors. Suppose now that p(x) has a quadratic factor; say x^2 + ax + b. (We can assume that the factor is monic.) Using the long division algorithm for polynomials, we have p(x) = (x^2 + ax + b)(x^2 - ax + (a^2 - b)) + (1 + 2ab - a^3)x + (2 + b^2 - a^2b). Recall that in F[x] (F a field), the quotient and remainder given by the Euclidean algorithm are unique. Thus it must be the case that 1 + 2ab - a^3 = 0 and 2 + b^2 - a^2b = 0 mod 3. We claim that this system of equations has no solution. To see this, note that a and b must both be nonzero, as otherwise we have 1 \equiv 0 or 2 \equiv 0. If a = b, then the second equation gives 2 + a^2 - a^3 \equiv 0; this is easily seen to have no solution in \mathbb{Z}/(3) by checking. Thus a \neq b and both are nonzero; there are only two remaining cases. If (a,b) = (1,2), then the first equation is violated, and if (a,b) = (2,1), then the second equation is violated. Thus p(x) does not have a quadratic factor over \mathbb{Z}/(3) and hence is irreducible there. So \mathbb{Z}[x]/(x^4 + x + 1) is a field of order 81.

Find all the monic irreducible polynomials of degree at most 3 over ZZ/(2) and ZZ/(3)

Find all the monic irreducible polynomials of degree at most 3 over \mathbb{Z}/(2) and \mathbb{Z}/(3).


Note that there are p^k monic polynomials in \mathbb{Z}[x] of degree k, as there are p choices for each of k coefficients.

Over \mathbb{Z}/(2):

All linear polynomials are irreducible; these are x and x+1.

The quadratic polynomials are x^2 = x \cdot x, x^2 + 1 = (x+1)^2, x^2 + x = x(x+1), and p(x) = x^2 + x + 1 which is irreducible since p(0) = p(1) = 1.

The cubic polynomials are x^3 = x \cdot x \cdot x, x^3 + 1 = (x+1)(x^2+x+1), x^3 + x = x(x+1)^2, x^3 + x^2 = x^2(x+1), p(x) = x^3 + x + 1 which is irreducible since p(0) = p(1) = 1, p(x) = x^3 + x^2 + 1 which is irreducible since p(0) = p(1) = 1, x^3 + x^2 + x = x(x^2 + x + 1), and x^3 + x^2 + x + 1 = (x+1)^3.

Thus the irreducible polynomials of degree at most 3 over \mathbb{Z}/(2) are p_1(x) = x, p_2(x) = x+1, p_3(x) = x^2 + x + 1, p_4(x) = x^3 + x + 1, and p_5(x) = x^3 + x^2 + 1.

Over \mathbb{Z}/(3):

All linear polynomials are irreducible. These are x, x+1, and x+2.

The quadratic polynomials are x^2 = x \cdot x, p(x) = x^2 + 1 which is irreducible since p(0) = 1 and p(1) = p(2) = 2, x^2 + 2 = (x+1)(x+2), x^2+x = x(x+1), x^2 + 2x = x(x+2), x^2 + x + 1 = (x+2)^2, p(x) = x^2 + x + 2 which is irreducible since p(0) = p(2) = 2 and p(1) = 1, x^2 + 2x + 1 = (x+1)^2, and p(x) = x^2 + 2x + 2 which is irreducible since p(0) = p(1) = 2 and p(2) = 1.

The cubic polynomials are x^3 = x \cdot x \cdot x, x^3 +1 = (x+1)^3, x^3 + 2 = (x+2)^3, x^3 + x = x(x^2 + 1), x^3 + 2x = x(x+1)(x+2), x^3 + x + 1 = (x+2)(x^2 + 2x + 2), x^3 + x + 2 = (x+1)(x^2 + 2x + 2), p(x) = x^3 + 2x + 1 which is irreducible since p(0) = p(1) = p(2) = 1, p(x) = x^3 + 2x + 2 which is irreducible since p(0) = p(1) = p(2) = 2, x^3 + x^2 = x^2(x+1), x^3 + x^2 + 1 = (x+2)(x^2 + 2x + 2), p(x) = x^3 + x^2 + 2 which is irreducible since p(0) = p(2) = 2 and p(1) = 1, x^3 + x^2 + x = x(x+2)^2, x^3 + x^2 + 2x = x(x^2 + x + 2), x^3 + x^2 + x + 1 = (x+1)(x^2 + 1), p(x) = x^3 + x^2 + x + 2 which is irreducible since p(0) = p(1) = 2 and p(2) = 1, p(x) = x^3 + x^2 + 2x + 1 which is irreducible since p(0) = 1 and p(1) = p(2) = 2, x^3 + x^2 + 2x + 2 = (x+1)^2(x+2), x^3 + 2x^2 = x^2(x+2), p(x) = x^3 + 2x^2 + 1 which is irreducible since p(0) = p(1) = 1 and p(2) = 2, x^3 + 2x^2 + 2 = (x+1)(x^2 + x + 2), x^3 + 2x^2 + x = x(x+1)^2, x^3 + 2x^2 + 2x = x(x^2 + 2x + 2), p(x) = x^3 + 2x^2 + x + 1 which is irreducible since p(0) = p(2) = 1 and p(1) = 2, x^3 + 2x^2 + x + 2 = (x+2)(x^2+1), x^3 + 2x^2 + 2x + 1 = (x+1)(x+2)^2, and p(x) = x^3 + 2x^2 + 2x + 2 which is irreducible since p(0) = 2 and p(1) = p(2) = 1.

Thus the irreducible polynomials of degree at most 3 over \mathbb{Z}/(3) are p_1(x) = x, p_2(x) = x+1, p_3(x) = x+2, p_4(x) = x^2 + 1, p_5(x) = x^2 + x + 2, p_6(x) = x^2 + 2x + 2, p_7(x) = x^3 + 2x + 1, p_8(x) = x^3 + x^2 + 2, p_9(x) = x^3 + x^2 + x + 2, p_{10}(x) = x^3 + x^2 + 2x + 1, p_{11}(x) = x^3 + 2x^2 + 1, p_{12}(x) = x^3 + 2x^2 + x + 1, and p_{13}(x) = x^3 + 2x^2 + 2x + 2.

Prove that a given polynomial over ZZ is irreducible

Let n \geq 1. Prove that p(x) = \prod_{k=1}^n (x-k) + 1 is irreducible over \mathbb{Z} for all n \neq 4.


[There is probably a more direct way to do this.]

We begin with a combinatorial lemma.

Lemma: Let n = 2m \geq 6 be an even integer. Then there does not exist a partition \{A,B\} of [1,n] such that |A| = |B| and \prod A - \prod B = 2. Proof: Note that \mathsf{gcd}(\prod A, \prod B) \in \{1,2\}. If \mathsf{gcd}(\prod A, \prod B) = 1, then since 2 \in [1,n], one of \prod A and \prod B is odd and the other even. Mod 2 we have 1 \equiv 0, a contradiction. Thus \mathsf{gcd}(\prod A, \prod B) = 2. In particular, for each odd prime p, if p \in A then all multiples of p are in A; likewise for B. Also, one of A or B contains only odd numbers except for one, which is maximally divisible by 2^1.

Now we will consider how 2, 3, 4, and 5 might be distributed between A and B. Suppose 3 and 4 are in different classes. Then 3 and 2 are in the same class, and so 6 and 2 are in the same class. But then 4 divides \prod A and \prod B, a contradiction. Thus 3 and 4 are in the same class. Suppose now that 5 is in the same class as 3 and 4. Now the class containing 3, 4, and 5 must contain all multiples of 3 and 5, as well as all multiples of 2 except for one. Since n is even and greater than 5, this class contains m+1 elements, a contradiction since A and B must have the same cardinality. Thus it must be the case that 3 and 4 are in the same class and 5 is in the other class. Then 2 must be in the same class as 5. But now 10 cannot be in either class, since it is both even and a multiple of 5. So n \in \{6,8\}.

Suppose n = 6. We are forced to have 3 and 4 in one class and 5 and 2 in the other; thus 6 is forced to be in the class with 3 and 4 and 1 with 5 and 2. But then the products \prod A and \prod B are 72 and 10; neither difference of these is 2.

Suppose n = 8. Again, the divisibility criteria force the partition \{\{3,4,6,8\}, \{1,2,5,7\}\}, but now \prod A and \prod B are 70 and 576, a contradiction.

Thus no such partition exists. \square

We are now prepared for the main problem.

Let n \geq 1 and p(x) = \prod_{k=1}^n (x-k) + 1. Suppose p(x) is reducible in \mathbb{Z}[x], with p(x) = a(x)b(x). Note that for each k \in [1,n], we have p(x) = a(x)b(x) = 1. Since a,b \in \mathbb{Z}[x], we have a(k),b(k) \in \mathbb{Z} for all such k; thus a(k),b(k) \in \{1,-1\}, and moreover a(k) = b(k). Let A = \{k \in [1,n] \ |\ a(k) = 1\} and B = [1,n] \setminus A. Certainly then b(k) = -1 for all k \in B. By a lemma to this previous exercise, we have a(x) = q(x)\prod_{k \in A} (x-k) + 1 and b(x) = r(x)\prod_{k \in B} (x-k) - 1. Considering degrees and the monicness of a and b, we have q = r = 1. Now p(x) = a(x)b(x) = (\prod_{k \in A} (x-k) - 1)(\prod_{k \in B} (x-k) + 1) = p(x) + \prod_{k \in A} (x-k) - \prod_{k \in B} (x-k) + 2, and thus \prod_{k \in B} (x-k) - \prod_{k \in A} (x-k) = 2. If |A| \neq |B|, then the left hand side of this equation has positive degree while the right is constant, a contradiction. Thus |A| = |B|, and in particular n = 2m is even. Now evaluating at x = 0, we have (-1)^m \prod_{k \in B} k - (-1)^m \prod k = 2, and thus \prod_{k \in B} k - \prod_{k \in A} k = 2 or \prod_{k \in A} k - \prod_{k \in B} k = 2, depending on whether m is even or odd. In either case, no such partition A exists for n \geq 6, so that n \in \{2,4\}.

Note that for n = 2, p(x) = x^2 - 3x + 3. This polynomial is Eisenstein at 3 and thus irreducible. For n = 4, we have p(x) = x^4 - 10x^3 + 35x^2 - 50x + 25. Taking the partition \{\{2,3\},\{1,4\}, we let a(x) = (x-2)(x-3) - 1 = x^2 - 5x + 5 and b(x) = (x-1)(x-4) + 1 = x^2 - 5x + 5; indeed, p(x) = (x^2 -5x + 5)^2 is reducible.

Demonstrate that a given polynomial is irreducible over ZZ

For n \geq 1, show that p(x) = \prod_{k=1}^n (x-k) - 1 is irreducible over \mathbb{Z}.


We begin with a lemma.

Lemma: Let R be a unique factorization domain with field of fractions F. Let t \in R and A \subseteq R, and suppose p(k) = t for each k \in A. Then for some q(x) \in R[x], we have p(x) = q(x) \prod_{k \in A}(x-k) + t. Proof: Each k \in A is a root of p(x) - t, so that \prod_{k \in A}(x-k) is a divisor of p(x) - t. Note that because p(x) - t \in R[x] and \prod_{k \in A}(x-k) \in R[x] is monic, by this previous exercise we may assume q(x) \in R[x]. \square

Now let n \geq 1 and consider p(x) = \prod_{k = 1}^n (x-k)  - 1. Suppose p(x) = a(x)b(x) is reducible; then by Gauss’ lemma we may assume that a(x),b(x) \in \mathbb{Z}[x]. Note that for each 1 \leq k \leq n, p(k) = a(k)b(k) = -1. Since a(k),b(k) \in \mathbb{Z}, then, we have a(k),b(k) \in \{1,-1\} for each such k, and a(k) = -b(k). Let A = \{k \in [1,n] \ |\ a(k) = -1\} and B = [1,n] \setminus A; certainly then b(k) = -1 for all k \in B. By the lemma, we have a(x) = q(x)\prod_{k \in A} (x-k) - 1 and b(x) = r(x) \prod_{k \in B} (x-k) - 1. Considering degrees and the monicness of a and b, we see that in fact q(x) = r(x) = 1.

Now p(x) = a(x)b(x) = (\prod_{k \in A}(x-k) - 1)(\prod_{k \in B} (x-k) - 1) = \prod_{k=1}^n (x-k) - \prod_{k \in A} (x-k) - \prod_{k \in B} (x-k) + 1 = p(x) - (\prod_{k \in A} (x-k) + \prod_{k \in B} (x-k)) + 2. Hence \prod_{k \in A} (x-k) + \prod_{k \in B} (x-k) = 2. Since n \geq 1, the left hand side of this equation has degree at least 1, while the right hand side is constant- a contradiction.

Thus, p(x) is irreducible after all.

In particular this demonstrates that in \mathbb{Z}[x], irreducible polynomials of every postive degree exist.

Prove that a given polynomial is irreducible

Prove that the following polynomials are irreducible in \mathbb{Z}[x]:

  1. p(x) = x^4 - 4x^3 + 6
  2. p(x) = x^6 + 30x^5 - 15x^3 + 6x - 120
  3. p(x) = x^4 + 4x^3 + 6x^2 + 2x + 1
  4. p(x) = \dfrac{(x+2)^p - 2^p}{x}, where p is an odd prime

  1. This polynomial is Eisenstein at 2.
  2. This polynomial is Eisenstein at 3.
  3. With q(x) = x-1, we have (p \circ q)(x) = x^4 -2x + 2, which is Eisenstein at 2. Thus p(x) is irreducible.
  4. Using the Binomial Theorem, we have p(x) = \sum_{k=0}^{p-1} 2^k \binom{p}{k} x^{p-k-1}. Note that p does not divide the leading coefficient \binom{p}{0}, that p does divide each 2^k \binom{p}{k} for 1 \leq k \leq p-1, and that p^2 does not divide the constant coefficient 2^{p-1}\binom{p}{p-1} = 2^{p-1}p. Thus p(x) is Eisenstein at p.

Factor a given polynomial into irreducibles

Factor each of the following polynomials into irreducibles in the stated ring.

  1. p(x) = x^2 + x + 1 in \mathbb{Z}/(2)[x]
  2. p(x) = x^3 + x + 1 in \mathbb{Z}/(3)[x]
  3. p(x) = x^4 + 1 in \mathbb{Z}/(5)[x]
  4. p(x) = x^4 + 10x^2 + 1 in \mathbb{Z}[x]

  1. If p(x) is reducible, then (being of degree 2) it has a linear factor and thus a root in \mathbb{Z}/(2). However, since p(0) = p(1) = 1, no such root exists. Thus p(x) is irreducible in \mathbb{Z}/(2)[x].
  2. If p(x) is reducible, then (being of degree 3) it has a linear factor and thus a root. Indeed, p(1) = 0, so that x+2 divides p(x). Using the long division algorithm for polynomials, we see that p(x) = (x+2)(x^2 + x + 2). Now x+2 is irreducible, and x^2+x+2 is irreducible because it has no roots in \mathbb{Z}/(3). Thus we have completely factored p(x).
  3. Note that, over \mathbb{Z}/(5), p(x) = x^4 - 4 = (x^2 + 2)(x^2 - 2), so that p(x) = (x^2 + 2)(x^2 + 3). Note that the squares in \mathbb{Z}/(5) are 0, 1, and 4; in particular, x^2+2 and x^2 + 3 do not have roots in \mathbb{Z}/(5) and thus are irreducible. So we have completely factored p(x).
  4. Since p(1) = p(-1) = 12, by the rational root theorem p(x) has no roots in \mathbb{Z} and thus no linear factors over \mathbb{Z}. Suppose now that p(x) = (x^2 + ax + b)(x^2 + cx + d) = x^4 + (a+c)x^3 + (ac+b+d)x^2 + (ad+bc)x + bd factors as a product of two quadratics. Comparing coefficients, we have a+c = 0, ac+b+d = 10, ad+bc = 0, and bd = 1. From bd = 1, we have either b = 1 or b = -1. If b = 1, then d = 1, and so ac + 2 = 10. Now -a^2 = 8, a contradiction. If b = -1, then d = -1, and so ac - 2 = 10. Now -a^2 = 12, also a contradiction. So p(x) has no quadratic factors. Thus p(x) is irreducible over \mathbb{Z}.

Exhibit a Bezout domain that is not a PID

Let R = \mathbb{Z} + x\mathbb{Q}[x] \subseteq \mathbb{Q}[x] be the subring consisting of all polynomials whose constant term is an integer. (We studied this ring in this previous exercise.)

  1. Suppose that f,g \in \mathbb{Q}[x] are nonzero. and that x^r is the largest power of x dividing both f and g. Let f_r and g_r be the r-coefficients of f and g, respectively. (One of these must be nonzero.) By this previous exercise, as additive subgroups of \mathbb{Q} we have f_r \mathbb{Z} + g_r \mathbb{Z} = d_r \mathbb{Z} for some d_r \in \mathbb{Q} with d_r \neq 0. Prove that there exists a polynomial d(x) \in \mathbb{Q}[x] that is a greatest common divisor of f(x) and g(x) and whose term of minimal degree is d_rx^r.
  2. Prove that f(x) = d(x)q_f(x) and g(x) = d(x)q_g(x), where q_f,q_g \in R.
  3. Prove that for all f,g \in \mathbb{Q}[x], there exist a,b \in R and a greatest common divisor d of f and g such that d(x) = a(x)f(x) + b(x)g(x).
  4. Conclude that R is a Bezout domain.
  5. Show that R must contain ideals which are not principal. In fact, show that x\mathbb{Q}[x] is an ideal which is not finitely generated.

  1. Let d(x) be a greatest common divisor of f(x) and g(x) in \mathbb{Q}[x]. (We can do this because \mathbb{Q}[x] is a Euclidean domain.) In particular, (f(x),g(x)) = (d(x)), and x^r|d(x). Moreover, if x^{r+1} divides d(x), then x^{r+1} divides both f(x) and g(x), a contradiction. So the term of lowest degree in d(x) is x^r; let d_1 be the coefficient of this term. Then \frac{d_r}{d_1}d(x) is a greatest common divisor of f(x) and g(x) in \mathbb{Q}[x] whose r-coefficient is d_r.
  2. We of course have f(x) = d(x)q_f(x). Comparing r-terms, we have f_r = d_rq_0. Since f_r \in d_r\mathbb{Z}, in fact q_0 \in \mathbb{Z}, so that q_f(x) \in R. Similarly, with g(x) = d(x)q_g(x), q_g(x) \in R, as desired.
  3. Let f,g \in \mathbb{Q}[x] and let d(x) be a greatest common divisor of f and g. Now f(x) = d(x)t(x) and g(x) = d(x)u(x) for some t,u \in \mathbb{Q}[x]. We claim that without loss of generality, t,u \in R and moreover the constant terms of t and u are relatively prime. To see this, let \ell be a least common multiple of the denominators of the constant coefficients of t and u. We have f(x) = \frac{1}{\ell} d(x) \ell t(x) and g(x) = \frac{1}{\ell}d(x) \ell u(x), and since \frac{1}{\ell} is a unit, \frac{1}{\ell}d(x) is a greatest common multiple of f and g, and \ell t(x) and \ell u(x) are in R. Now suppose the constant terms of \ell t(x) and \ell u(x) have greatest common divisor k. Then likewise f(x) = \frac{k}{\ell}d(x) \frac{\ell}{k} t(x) and g(x) = \frac{k}{\ell} d(x) \frac{\ell}{k} u(x), \frac{k}{\ell} d(x) is a greatest common divisor of f and g, and \frac{\ell}{k} t(x), \frac{\ell}{k} u(x) \in R have relatively prime constant terms. Let d,t,u have these properties, and let t_0 and u_0 be the constant coefficients of t and u, respectively. By Bezout’s identity in \mathbb{Z}, we have x,y \in \mathbb{Z} such that xt_0 - yu_0 = 1.

    Now d(x) = \alpha(x)f(x) + \beta(x)g(x) in \mathbb{Q}[x]. Since f(x) = d(x)t(x) and g(x) = d(x)u(x) and \mathbb{Q}[x] is a domain, we have 1 = \alpha(x)t(x) + \beta(x)g(x). Comparing constant coefficients, 1 = \alpha_0t_0 + \beta_0u_0. So xt_0 - yu_0 = \alpha_0t_0 + \beta_0u_0, and rearranging, we have m = \frac{x - \alpha_0}{u_0} = \frac{y + \beta_0}{t_0}. Let a(x) = \alpha(x) + mu(x) and b(x) = \beta(x) - mt(x). Evidently, d(x) = a(x)f(x) + b(x)g(x), and a(x),b(x) \in R.

  4. Let f,g \in R. As elements of \mathbb{Q}[x], by part (3) there is a greatest common divisor d(x) \in \mathbb{Q} and elements a(x),b(x) \in R such that d = af + bg. In particular, d \in R, so that in fact d is a greatest common divisor of f and g in R. Thus fR + gR = dR, and so R is a Bezout domain.
  5. Note that R is a Bezout domain that is not a unique factorization domain, and so by this previous exercise, R cannot be a principal ideal domain. In particular, R must contain ideals which are not finitely generated.

    Consider x\mathbb{Q}[x] \subseteq R; this subring is certainly an ideal. Suppose x \mathbb{Q}[x] = (A) is finitely generated, with all coefficients of elements in A in lowest terms. Note that no element of A has a nonzero constant term. Let \ell be a least common multiple of the denominators of the coefficients of linear terms of elements in A, and let p be a prime not dividing \ell. Now \frac{1}{p}x \in x\mathbb{Q}[x], but we claim that \frac{1}{p}x \notin (A). To see this, note that every element of A has the form \frac{1}{\ell}(k_ix + x^2p_i(x)) with k_i \in \mathbb{Z} and p_i \in \mathbb{Q}[x]. Now there exist elements a_i + xb_i(x) \in R so that \frac{1}{p}x = \sum_i (a_i + xb_i(x))(\frac{1}{\ell}(k_ix + x^2p_i(x))). After some arithmetic, we have \frac{1}{p}x = \frac{1}{\ell} \sum_i k_ia_i x + x^2v_i(x) for appropriate polynomials v_i. Comparing constant coefficients and letting m = \sum_i k_ia_i \in \mathbb{Z}, we have \ell = pm, a contradiction since p does not divide \ell. Thus x\mathbb{Q}[x] is not a finitely generated ideal of R.