## Tag Archives: quotient ring

Our goal here is to prove that the quotient of a graded ring by a graded ideal is graded. We will approach this from a slightly more general perspective.

Let $R$ be a commutative ring with 1 and let $S$ be a semigroup. Suppose we have a family $\mathcal{A} = \{A_i\}_S$ of $(R,R)$-bimodules such that $ra = ar$ for all $a$ and $r$. Suppose further that we have a family $\Psi = \{\psi_{i,j} : A_i \times A_j \rightarrow A_{ij}\}_{i,j \in S}$ of $R$-bilinear mappings such that $\psi_{i,jk}(a, \psi_{j,k}(b,c)) = \psi_{ij,k}(\psi_{i,j}(a,b),c)$ for all $i,j,k \in S$ and $a \in A_i$, $b \in A_j$, and $c \in A_k$. Finally, define an operator $\cdot$ on the $R$-module $\bigoplus_S A_i$ by $(a_i) \cdot (b_j) = (\sum_{ij=k} \psi_{i,j}(a_i,b_j))$.

We claim that $\bigoplus_S A_i$ is a ring under this multiplication. It suffices to show that $\cdot$ is associative and distributes over addition. To that end, let $(a_i), (b_j), (c_k) \in \bigoplus_S A_s$.

1. Note the following.
 $\left[ (a_i) \cdot (b_j) \right] \cdot (c_k)$ = $(\sum_{ij = t} \psi_{i,j}(a_i,b_j)) \cdot (c_k)$ = $(\sum_{tk=u} \psi_{t,k}(\sum_{ij=t}(a_i,b_j),c_k))$ = $(\sum_{tk = u} \sum_{ij=t} \psi_{ij,k}(\psi_{i,j}(a_i,b_j),c_k)$ = $(\sum_{it = u} \sum_{jk=t} \psi_{i,jk}(a_i, \psi_{j,k}(b_j,c_k)))$ = $(\sum_{it=u} \psi_{i,t}(a_i, \sum_{jk=t}(b_j,c_k)))$ = $(a_i) \cdot (\sum_{jk=t} \psi_{j,k}(b_j,c_k)$ = $(a_i) \cdot \left[ (b_j) \cdot (c_k) \right]$.

So $\cdot$ is associative.

2. Note the following.
 $(a_i) \cdot \left[ (b_j) + (c_j) \right]$ = $(a_i) \cdot (b_j+c_j)$ = $(\sum_{ij=k} \psi_{i,j}(a_i, b_j+c_j))$ = $(\sum_{ij=k} \psi_{i,j}(a_i,b_j) + \psi_{i,j}(a_i,c_j))$ = $(\sum_{ij=k} \psi_{i,j}(a_i,b_j)) + (\sum_{ij = k} \psi_{i,j}(a_i,c_j))$ = $\left[ (a_i) \cdot (b_j) \right] + \left[ (a_i) \cdot (c_j) \right]$.

So $\cdot$ distributes over $+$ from the left; distributivity from the right is proved similarly.

So $\bigoplus_S A_s$ is a ring under this multiplication; we will call this the $S$-graded sum of the $A_i$, and denote it by $\mathsf{Gr}^S_R(\mathcal{A},\Psi)$.

Now we claim that if $S$ is a monoid with identity $1$, and if $A_1 = R$, and finally if $\psi_{1,k}(r,a) = ra$ and $\psi_{k,1}(a,r) = ar$ for all $k \in S$, $a \in A_k$, and $r \in R$, then in fact $\mathsf{GR}^S_R(\mathcal{A},\Psi)$ is a unital ring and moreover an $R$-algebra via the injection map $R \rightarrow \mathsf{Gr}^S_R(\mathcal{A},\Psi)$. To see this, consider the element $e \in \mathsf{Gr}^S_R(\mathcal{A},\Psi)$ with $e_1 = 1$ and $e_s = 0$ otherwise. Certainly then $e \cdot (a_i) = (a_i) \cdot e = (a_i)$ for all $a$. So $\mathsf{Gr}^S_M(\mathcal{A},\Psi)$ is a unital ring. Moreover, note that since $ra = ar$ for all $r \in R$, we have $R \subseteq Z(\mathsf{Gr}^S_R(\mathcal{A},\Psi)$. In particular, this ring is an $R$-algebra.

Next we prove that this algebra has an appropriate universal property. Suppose $B$ is an $R$-algebra and that we have a family $\varphi_i : A_i \rightarrow B$ of $R$-module homomorphisms such that $\varphi_{ij}(\psi_{i,j}(a,b)) = \varphi_i(a)\varphi_j(b)$ for all $i \in S$ and all $a$ and $b$. Then there exists a unique $R$-algebra homomorphism $\theta : \mathsf{Gr}^S_R(\mathcal{A},\Psi) \rightarrow B$ such that $\theta \circ \iota_i = \varphi_i$ for each $i \in S$. It suffices to show that the unique induced module homomorphism (via the universal property of coproducts of modules) is an $R$-algebra homomorphism. To that end, let $(a_i), (b_j) \in \mathsf{Gr}^S_R(\mathcal{A},\Psi)$. We have $\theta((a_i)(b_j)) = \theta((\sum_{ij=k} \psi_{i,j}(a_i,b_j)))$ $= (\theta \circ \iota_k)(\sum_{ij=k} \psi_{i,j}(a_i,b_j))$ for each $k$. Then $\varphi_k(\sum_{ij=k} \psi_{i,j}(a_i,b_j))$ $= \sum_{ij=k} \varphi_{ij}(\psi_{i,j}(a_i,b_j))$ $= \sum_{ij=k} \varphi_i(a_i)\varphi_j(b_j)$ $= \theta((a_i)) \theta((b_i))$. (Recall that $\theta((a_i)) = \sum \varphi(a_i)$.)

Now suppose we have a family of submodules $\{B_s\}_S$, with $I_s \subseteq B_s$, such that $\psi_{i,j}[B_i,A_j], \psi_{i,j}[A_i,B_j] \subseteq B_{ij}$ for all $i,j \in S$. Certainly $\bigoplus_S B_s \subseteq \mathsf{Gr}^S_R(\mathcal{A},\Psi)$ is an ideal; we will call ideals of this form $S$-graded.

Let $\mathcal{A}$, $\Psi$, and $\{B_i\}$ be as above. For each pair $i,j \in S$, define $\overline{\psi} : A_i/B_i \times A_j/B_j \rightarrow A_{ij}/B_{i,j}$ by $\overline{\psi}_{i,j}(a_i+B_i, a_j+B_j) = \psi_{i,j}(a_i,b_j) + B_{ij}$. We claim that $\overline{\psi}$ is well-defined and $R$-bilinear. To see well-definedness, note that if $a_i - a_i^\prime \in B_i$ and $a_j - a_j^\prime \in B_j$, then by bilinearity we have $\psi_{i,j}(a_i,a_j) - \psi_{i,j}(a_i^\prime,a_j) \in B_{i,j}$ and $\psi_{i,j}(a_i^\prime,a_j) - \psi_{i,j}(a_i^\prime,a_j^\prime) \in B_{i,j}$. Thus $\psi_{i,j}(a_i,a_j) - \psi_{i,j}(a_i^\prime,a_j^\prime) \in B_{i,j}$, and so $\overline{\psi}_{i,j}$ is well-defined. Bilinearity is clear. Moreover, it is straightforward to show that $\overline{\psi}_{ij,k}(\overline{\psi}_{i,j}(\overline{a}_i,\overline{b}_j),\overline{c}_k) = \overline{\psi}_{i,jk}(\overline{a}_i, \overline{\psi}_{j,k}(\overline{b}_j,\overline{c}_k))$. Thus we may construct the $S$-graded sum $\mathsf{Gr}^S_R(\overline{A},\overline{\Psi})$.

Let $\mathcal{B} = \{B_i\}_S$, and for each pair $i,j \in S$, let $\psi|_{i,j}$ denote the restriction of $\psi_{i,j}$ to $B_i \times B_j$. Let $\Psi|_B$ denote the set of these restrictions. Then the ideal $\bigoplus_S B$ is in fact $\mathsf{Gr}^S_R(\mathcal{B},\Psi|_B)$.

We claim that $\mathsf{Gr}^S_R(\mathcal{A},\Psi)/\mathsf{Gr}^S_R(\mathcal{B},\Psi|_B) \cong_R \mathsf{Gr}^S_R(\overline{\mathcal{A}},\overline{\Psi})$. To see this, it suffices to show that the usual module isomorphism $(\bigoplus A_i)/(\bigoplus B_i) \rightarrow \bigoplus A_i/B_i$ given by $\overline{(a_i)} \mapsto (\overline{a_i})$ preserves multiplication. To that end, note that $\overline{(a_i)(b_j)} = \overline{(\sum_{ij=k} \psi_{i,j}(a_i,b_j))}$ $= (\sum_{ij=k} \overline{\psi}_{i,j}(\overline{a}_i,\overline{b}_j))$ $= \overline{(a_i)} \overline{(b_j)}$, as desired.

### Compute a quotient ring

Let $K = \mathbb{Q}(\sqrt{-37})$ with ring of integers $\mathcal{O} = \mathbb{Z}[\sqrt{-37}]$. Consider the ideal $A = (2,1+\sqrt{-37})$. Show that $\mathcal{O}/A \cong \mathbb{Z}/(2)$.

Let $a+b\sqrt{-37} \in \mathcal{O}$, and suppose $a-b \equiv k$ mod 2 where $k \in \{0,1\}$. Now $a+b\sqrt{-37} \equiv k$ mod $A$. In particular, $\mathcal{O}/A = \{\overline{0}, \overline{1}\}$.

Suppose $1 \in A$. Then we have $1 = 2(a+b\sqrt{-37}) + (1+\sqrt{-37})(h+k\sqrt{-37})$ for some integers $a,b,h,k$. Comparing coefficients, we have $2a+h-37k = 1$ and $2b + h + k = 0$. Mod 2, we have $h+k \equiv 1$ and $h+k \equiv 0$, a contradiction. So $1 \not\equiv 0$ mod $A$.

So $\mathcal{O}/A \cong \mathbb{Z}/(2)$.

### Compute a quotient ring

Let $K = \mathbb{Q}(\sqrt{-5})$ with ring of integers $\mathbb{Z}[\sqrt{-5}]$. Consider the ideals $A = (3,1+\sqrt{-5})$ and $B = (2)$. Compute $\mathcal{O}/AB$.

We showed in this previous exercise that $\mathcal{O}/A = \{\overline{0}, \overline{1}, \overline{2}\}$ and that $\mathcal{O}/B = \{\overline{0}, \overline{1}, \overline{\sqrt{-5}}, \overline{1+\sqrt{-5}}\}$.

Clearly $((2), AB) = B$. By the proof of Theorem 9.14, $\mathcal{O}/AB = \{\overline{2\alpha+\beta} \ |\ \alpha \in \{0,1,2\}, \beta \in \{0,1,\sqrt{-5},1+\sqrt{-5}\}\}$, and these cosets are distinct.

### A counterexample regarding quotients of an algebraic integer ring

Let $\mathcal{O}$ be an algebraic integer ring with ideals $A$ and $B$. If $\mathcal{O}/A = \{a_i+A\}_{i=1}^n$ and $\mathcal{O}/B = \{b_i+B\}_{i=1}^m$, is it necessarily the case that $\mathcal{O}/AB = \{a_ib_j\}_{i=1,j=1}^{n,m}$?

No. If $A = B$ is nontrivial, then the set pairwise products of representatives of $\mathcal{O}/A$ is not large enough.

### Compute the discriminant of an ideal in an algebraic integer ring

Let $K = \mathbb{Q}(\sqrt{-5})$ and let $\mathcal{O} = \mathbb{Z}[\sqrt{-5}]$ be the ring of integers in $K$. Let $A = (3,1+\sqrt{-5})$ and $B = (2)$. Exhibit bases for $A$ and $B$, compute $\mathcal{O}/A$ and $\mathcal{O}/B$, and compute the discriminant of $A$ and $B$.

We claim that $\{3, 1+\sqrt{-5}\}$ is a basis for $A$ over $\mathbb{Z}$. It is clear that $(3, 1+\sqrt{-5})_\mathbb{Z} \subseteq A$, and if $\zeta = 3(a+b\sqrt{-5}) + (1+\sqrt{-5})(h+k\sqrt{-5})$, then evidently $\zeta = 3(a-b-2k) + (1 + \sqrt{-5})(3b+k+h)$. So $(3, 1+\sqrt{-5})_\mathbb{Z} = A$. Now if $a,b \in \mathbb{Z}$ and $3a + (1+\sqrt{-5})b = 0$, then $b = 0$, so that $a = 0$. Hence $\{3, 1+\sqrt{-5}\}$ is a basis for $A$ over $\mathbb{Z}$. The discriminant of $A$ is then $\mathsf{det}^2 \left[ \begin{array}{cc} 3 & 1+\sqrt{-5} \\ 3 & 1-\sqrt{-5} \end{array} \right] = -180$.

We now claim that $A$ is proper. To see this, suppose to the contrary that $1 \in A$; then $1 = 3a + (1+\sqrt{-5})b$ for some integers $a$ and $b$; comparing coefficients, we have $b = 0$ and $3a+b = 1$. But then $3a = 1$, which has no integer solutions; thus $A$ is proper.

Now let $a+b\sqrt{-5}$ be arbitrary in $\mathcal{O}$ and let $a-b \equiv k$ mod 3, where $k \in \{0,1,2\}$. Now $a+b\sqrt{-5} = a-b+b(1+\sqrt{-5}) \equiv a-b \equiv k$ mod $A$, so that $\mathcal{O}/A = \{\overline{0}, \overline{1}, \overline{2}\}$. Since $A$ is proper, $0 \not\equiv 1$ and $2 \not\equiv 1$ mod $A$. If $2 \equiv 0$, then we have $3-2 = 1 \in A$, a contradiction; so $2 \not\equiv 0$ mod $A$. Thus $\mathcal{O}/A \cong \mathbb{Z}/(3)$.

We claim now that $\{2,2\sqrt{-5}\}$ is a basis for $B$ over $\mathbb{Z}$. Certainly $(2,2\sqrt{-5})_\mathbb{Z} \subseteq B$, and if $\zeta = 2(a+b\sqrt{-5}) \in (2)$, then $\zeta = 2a + 2\sqrt{-5}b$. So $(2,2\sqrt{-5})_\mathbb{Z} = B$. Moreover, if $2a + 2b\sqrt{-5} = 0$, then $a = b = 0$. So $\{2,2\sqrt{-5}\}$ is a basis for $B$ over $\mathbb{Z}$. The discriminant of $B$ is then $\mathsf{Det}^2 \left[ \begin{array}{cc} 2 & 2\sqrt{-5} \\ 2 & -2\sqrt{-5} \end{array} \right] = -320$.

We claim that $B$ is proper. Indeed, if $1 \in B$, then $1 = 2a$ for some integer $a$, a contradiction.

Now let $a+b\sqrt{-5}$ be arbitrary in $\mathcal{O}$. Say $a \equiv a_0$ and $b \equiv b_0$ mod 2, where $a_0, b_0 \in \{0,1\}$. Then $a+b\sqrt{-5} \equiv a_0 + b_0 \sqrt{-5}$ mod $B$. Thus $\mathcal{O}/B = \{\overline{0}, \overline{1}, \overline{\sqrt{-5}}, \overline{1+\sqrt{-5}}\}$. Since $B$ is proper, we have $0 \not\equiv 1$ and $\sqrt{-5} \not\equiv 1+\sqrt{-5}$ mod $B$. If $\sqrt{-5} \in B$, then we have $1 = 2b$ for some integer $b$, a contradiction. Thus $\sqrt{-5} \not\equiv 0$ and $1+\sqrt{-5} \not\equiv 1$ mod $B$. Likewise, if $1+\sqrt{-5} \in B$, we have a contradiction, so that $1+\sqrt{-5} \not\equiv 0$ and $\sqrt{-5} \not\equiv 1$ mod $B$.

We claim that $\mathcal{O}/B \cong \mathbb{F}_2[x]/(x^2)$. To see this, define $\varphi : \mathbb{F}_2[x] \rightarrow \mathcal{O}/B$ by $x \mapsto 1+\sqrt{-5}$. Certainly $\varphi$ is surjective. Now $\varphi(x^2) = 0$, so that $(x^2) \subseteq \mathsf{ker}\ \varphi$. Now if $p(x) = a_0 + a_1x + p^\prime(x)x^2 \in \mathsf{ker}\ \varphi$, then $\varphi(p(x)) = (a_0+a_1) + a_1\sqrt{-5} \equiv 0$. Thus $a_1 \equiv a_0 \equiv 0$ mod 2. By the first isomorphism theorem for rings, $\mathcal{O}/B \cong \mathbb{F}_2(x)/(x^2)$.

### The tensor product of the quotients of a commutative ring by two ideals is isomorphic to the quotient by their sum

Let $R$ be a commutative ring with ideals $I$ and $J$. Let $R/I$ and $R/J$ be $R$-modules (in fact $(R,R)$-bimodules) in the usual way.

1. Prove that every element of $R/I \otimes_R R/J$ can be written as a simple tensor of the form $(1 + I) \otimes (r + J)$.
2. Prove that $R/I \otimes_R R/J \cong_R R/(I+J)$.

We will prove the first result first for simple tensors; the extension to arbitrary sums of tensors follows by tensor distributivity. Let $(a+I) \otimes (b+J)$ be an arbitrary simple tensor in $R/I \otimes_R R/J$. Now $(a+I) \otimes (b+J) = (1+I)a \otimes (b+J)$ $= (1+I) \otimes a(b+J)$ $= (1+I) \otimes (ab+J)$, as desired.

Now define $\varphi : R/I \times R/J \rightarrow R/(I+J)$ by $(a+I, b+J) \mapsto (a+b) + (I+J)$. First, suppose $a_1-a_2 \in I$ and $b_1-b_2 \in J$. Then $b_1(a_1-a_2) \in I$ and $a_2(b_1-b_2) \in J$, so that $a_1b_1 - a_2b_2 \in I+J$. Thus $\varphi$ is well-defined. It is clear that $\varphi$ is $R$-balanced, and so induces an $R$-module homomorphism $\Phi : R/I \otimes_R R/J \rightarrow R/(I+J)$. Since $\varphi((1+I) \otimes (r+J)) = r + (I+J)$, $\Phi$ is surjective. Now suppose $(1+I) \otimes (r+J)$ is in the kernel of $\Phi$; then $r \in I+J$. Say $r = a+b$ where $a \in I$ and $b \in J$. Then $(1 + I) \otimes (r + J) = (1+I) \otimes (a+J) + (1+I) \otimes (b+J)$ $= (a+I) \otimes (1+J) + (1+I) \otimes (b+J)$ $= 0+0 = 0$, and hence $\Phi$ is an isomorphism.

### Compute ideal quotients in a quotient ring

Let $R = \mathbb{Q}[y,z]$ and let $K = (y^5-z^4)$ be an ideal in $R$. Let bars denote passage to $R/K$. For each of the following pairs of ideals $I$ and $J$, compute the ideal quotient $(\overline{I} : \overline{J})$.

1. $I = (y^3, y^5-z^4)$, $J = (z)$
2. $I = (y^3, z, y^5-z^4)$, $J = (y)$
3. $I = (y, y^3, z, y^5-z^4)$, $J = (1)$

We will let Maple to compute our Gröbner bases for us. Recall from this previous exercise that $(\overline{I} : \overline{J}) = \overline{(I:J)}$. From this previous exercise, we know that if $G = \{g_i\}$ is a Gröbner basis of $I \cap (p)$, then $\{g_i/p\}$ is a Gröbner basis of $(I : (p))$. Throughout, fix the lex order induced by $t > y > z$.

Note also that $\{y^5-z^4\}$ is the reduced Gröbner basis for $K$. Thus if $(a_1, \ldots, a_k) \subseteq R$ is an ideal, then $\overline{A}$ is precisely $(\overline{b_1}, \ldots, \overline{b_k})$, where each $\overline{b_i}$ is simply $a_i$ mod $y^5-z^4$.

1. First we compute $(I : J)$ in $R$. Note that $tI + (1-t)J = (ty^3, ty^5-tz^4, tz-z)$. Using Maple, we see that $G = \{ z^4, y^3z, tz-z, ty^3 \}$ is a Gröbner basis for this ideal. Then $I \cap J = (z^4, y^3z)$, and hence $(I : J) = (z^3, y^3)$. Thus $(\overline{I} : \overline{J}) = (\overline{y^3}, \overline{z^3})$.
2. First we compute $(I : J)$ in $R$. Note that $tI + (1-t)J = (ty^3, tz, ty^5-tz^4, ty-y)$. Using Maple, we see that $G = \{yz, y^3, tz, ty-y \}$ is a Gröbner basis for this ideal. Then $I \cap J = (yz, y^3)$, and hence $(I:J) = (y^2, z)$. Thus $(\overline{I}, \overline{J}) = (\overline{y^2}, \overline{z})$.
3. First we compute $(I : J)$ in $R$. Note that $tI + (1-t)J = (ty, ty^3, tz, ty^5-tz^4, t-1)$. Using Maple, we see that $G = \{ z, y, t-1 \}$ is a Gröbner basis for this ideal. Then $I \cap J = (y,z)$, and hence $(I : J) = (y,z)$. Thus $(\overline{I} : \overline{J}) = (\overline{y}, \overline{z})$.

### A basic property of ideal quotients in a quotient ring

Let $R$ be a ring and let $I$, $J$, and $K$ be ideals in $R$ such that $K \subseteq I$. If $\overline{I}$ and $\overline{J}$ denote the images of $I$ and $J$ in $R/K$ via the natural projection, prove that $\overline{(I:J)} = (\overline{I}:\overline{J})$ where $\overline{(I:J)}$ is the image of the ideal quotient $(I:J)$.

$(\subseteq)$ Suppose $r + K \in \overline{(I:J)}$. Then $r \in (I:J)$, and thus $rJ \subseteq I$. Then $rJ + K \subseteq I + K$, and so $(r+K)(J + K) \subseteq I+K$. So $r+K \in (\overline{I}:\overline{J})$.

$(\supseteq)$ Suppose conversely that $r+K \in (\overline{I}:\overline{J})$. Then $(r+K)(J+K) \subseteq I+K$, and thus $rJ+K \subseteq I+K$. Since $K \subseteq I$, $rJ \subseteq I$. So $r \in (I:J)$, and $r+K \in \overline{(I:J)}$.

### Exhibit a ring in which some element is neither prime nor a product of irreducibles

Let $R = \mathbb{Z} + x\mathbb{Q}[x] \subseteq \mathbb{Q}[x]$ be the set of all rational polynomials whose constant term is an integer.

1. Prove that $R$ is an integral domain and that the units in $R$ are $\pm 1$.
2. Show that the irreducible elements in $R$ are $\pm p$ for prime integers $p$ and the irreducible polynomials $p(x) \in \mathbb{Q}[x]$ whose constant coefficient is $\pm 1$.
3. Show that $x \in R$ cannot be written as a product of irreducibles. and conclude that $R$ is not a unique factorization domain.
4. Show that $x \in R$ is not prime and describe the quotient $R/(x)$.

1. As a subring of the integral domain $\mathbb{Q}[x]$, $R$ cannot contain zero divisors and thus is also an integral domain. Now suppose $a(x)b(x) = 1$ with $a,b \in R$. Computing the degrees of both sides, we see that $a$ and $b$ are constants, and thus integers. So $a, b = \pm 1$. Hence the units in $R$ are $1$ and $-1$.
2. Let $p(x) \in R$ be irreducible. If $\mathsf{deg}(p(x)) = 0$, then $p \in \mathbb{Z}$. Then whenever $ab = p$, without loss of generality $a = \pm 1$. Moreover, computing degrees, $b \in \mathbb{Z}$. Thus $p$ is a prime integer. Suppose now that $p(x)$ has positive degree. Suppose further that $p(x) \in \mathbb{Q}[x]$ is reducible; say $p = ab$ where $a,b \in \mathbb{Q}[x]$. We may also write $p(x) = qp^\prime(x)$, where $q \in \mathbb{Q}$ and $p^\prime(x) \in \mathbb{Z}[x]$. Then $p^\prime = q^{-1}a(x)b(x)$ is reducible in $\mathbb{Q}[x]$. By Gauss’ Lemma, $p^\prime = a^\prime(x)b^\prime(x)$, with $a^\prime,b^\prime \in \mathbb{Z}[x]$. Then $p(x) = qa^\prime(x)b^\prime(x)$. In particular, the constant coefficient $qa_0b_0$ is an integer, where $a_0$ and $b_0$ are integers and $q$ is rational. Suppose $q$ is given in lowest terms and write the denominator of $q$ as $q_aq_b$, where $q_a|a_0$ and $q_b|b_0$. Let $q_n$ be the numerator of $q$. Then $p(x) = (\frac{q_n}{q_a}a(x))(\frac{1}{q_b}b(x))$ is a factorization of $p(x)$ in $R$, and thus $p(x) \in R$ is reducible – a contradiction. So $p(x) \in \mathbb{Q}[x]$ must be irreducible. Now write $p(x) = a + xp^\prime(x)$. If $|a| \neq 1$, then there exists a prime integer $q$ dividing $a$; say $a = qb$. Then $p(x) = q(b + \frac{1}{q}xp^\prime(x))$. Neither of these factors is a unit, so that $p(x)$ is not irreducible after all – a contradiction. Thus $|a| = 1$ as desired.
3. Suppose $x = \prod a_i$ is a product of irreducibles $a_i$. Since the degree of a product is the sum of degrees, exactly one of the $a_i$ is linear – say $a_0$ – and the rest are constants. By part (2), $a_0 = ax \pm 1$. But then $\prod a_i$ has a nonzero constant term, a contradiction. Thus $x$ cannot be written as a product of irreducibles in $R$.
4. Now we show that $x \in R$ is not prime. Note that $2 \cdot \frac{1}{2}x = x$, and both of these factors is in $R$. However, $2$ is not divisible by $x$ (computing degrees). Suppose $\frac{1}{2}x \in (x)$. Then $\frac{1}{2}x = ax$, and since $R$ is an integral domain, $a = \frac{1}{2} \in R$, a contradiction. So $x$ is not prime in $R$.

We claim that $R/(x) = \{ a + bx + (x) \ |\ a \in \mathbb{Z}, b \in \mathbb{Q}, 0 \leq b < 1 \}$. The $(\supseteq)$ direction is clear. To see $(\subseteq)$, suppose $p(x) + (x) \in R/(x)$ with $p(x) = \sum a_i x^i$. All powers of $x$ of exponent at least 2 are in $(x)$. Moreover, if $a_1x$ is the linear coefficient of $p(x)$, we can write $a_1x = a_1^\prime x + kx$ where $k \in \mathbb{Z}$ and $a_1^\prime$ is in the interval $[0,1)$. Since $kx \in (x)$, we have $p(x) + (x) = a_0 + a_1^\prime x + (x)$, as desired. Next we show that these representatives give distinct cosets. Suppose $a_1 + b_1x + (x) = a_2 + b_2x + (x)$; then $(a_1 - a_2) + (b_1 - b_2)x \in (x)$. Since every element of $(x)$ has zero constant term, $a_1 = a_2$. Now $(b_1 - b_2)x \in (x)$, so that $(b_1 - b_2)x = q(x)x$. Computing degrees, in fact $q$ is constant and we have $b_1 - b_2 \in \mathbb{Z}$. Thus $b_1 - b_2 = 0$, so that $b_1 = b_2$.

### Characterize the ideals of ZZ[x]/(2, x³ + 1)

Determine all of the ideals in $\mathbb{Z}[x]/(2, x^3 + 1)$.

Note that $R = \mathbb{Z}[x]/(2,x^3+1) \cong (\mathbb{Z}[x]/(2))/((2,x^3+1)/(2))$ $\cong \mathbb{Z}/(2)[x]/(x^3+1)$. Since $\mathbb{Z}/(2)$ is a field, by this previous exercise, the ideals of $R$ correspond to the divisors of $x^3 + 1$ in $\mathbb{Z}/(2)[x]$. Evidently, $x^3 + 1 = (x+1)(x^2+x+1)$ in this ring. Now $x+1$ is linear, hence irreducible. We claim that $x^2+x+1$ is also irreducible. To see this, note that if $x^2+x+1$ is reducible, then it must factor into linear polynomials. Now if $x^2+x+1 = (ax+b)(cx+d) = acx^2 + (ad + cb)x + bd$. Comparing the leading and constant coefficients, we have $a = b = c = d = 1$. But then $1+1 = 1$, a contradiction. So $x^2+x+1$ is irreducible, and the prime factorization of $x^3+1$ in $\mathbb{Z}/(2)[x]$ is in fact $(x+1)(x^2+x+1)$.

Thus the ideals of $\mathbb{Z}[x]/(2,x^3+1)$ are $(2,1)/(2,x^3+1)$, $(2,x+1)/(2,x^3+1)$, $(2,x^2+x+1)/(2,x^3+1)$, and $0$.