## 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 $R$.