Tag Archives: quotient ring

Facts about graded rings and graded ideals

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.