## Tag Archives: maximal ideal

### Factor a given ideal in an algebraic integer ring

Let $K = \mathbb{Q}(\sqrt{-6})$ and let $\mathcal{O} = \mathbb{Z}[\sqrt{-6}]$ be the ring of integers in $K$. Factor the ideals $(2)$ and $(5)$ in $\mathcal{O}$.

We claim that $(2) = (2,\sqrt{-6})^2$. Indeed, $2 = -\sqrt{-6}^2 - 2^2$, so that $(2) \subseteq (2,\sqrt{-6})^2$. The reverse inclusion is clear.

Now we claim that $(2,\sqrt{-6})$ is maximal. By Corollary 9.11, $N((2)) = 4$, and by Theorem 9.14, we have $N((2,\sqrt{-6}))^2 = 4$, so that $N((2,\sqrt{-6})) = 2$. By Corollary 9.15, $(2,\sqrt{-6})$ is a prime ideal.

Thus $(2) = (2,\sqrt{-6})^2$ is the prime factorization of $(2)$ in $\mathcal{O}$.

Now we claim that $(5) = (5,1+2\sqrt{-6})(5,1-2\sqrt{-6})$. Indeed, we have $(5,1+2\sqrt{-6})(5,1-2\sqrt{-6}) = (25,5+10\sqrt{-6}, 5-10\sqrt{-6}, 25)$ $= (25, 5+10\sqrt{-6}, 10)$ $= (5)$.

Next we claim that $Q_1 = (5,1+2\sqrt{-6})$ and $Q_2 = (5,1-2\sqrt{-6})$ are proper. Suppose to the contrary that $1 \in Q_1$; then we have $1 = 5(a+b\sqrt{-6}) + (1+2\sqrt{-6})(h+k\sqrt{-6})$ for some $a,b,h,k \in \mathbb{Z}$. Comparing coefficients, $5a+h-12k = 1$ and $5b+k+2h = 0$, which yields a contradiction mod 5. So $Q_1$ is proper. Likewise, $Q_2$ is proper. In particular, neither $Q_1$ nor $Q_2$ have norm 1 as ideals. Now $25 = N((5)) = N(Q_1)N(Q_2)$, and neither factor is 1, so that $N(Q_1) = N(Q_2) = 5$. By Corollary 9.15, $Q_1$ and $Q_2$ are prime ideals.

Thus $(5) = (5,1+2\sqrt{-6})(5,1-2\sqrt{-6})$ is the prime factorization of $(5)$ in $\mathcal{O}$.

### Facts about the inverse of a maximal ideal in an algebraic integer ring

Let $K$ be an algebraic number field with ring of integers $\mathcal{O}$. Fix an ideal $A \subseteq \mathcal{O}$. Recall that, for an ideal $A \subseteq \mathcal{O}$, $A^{-1} = \{r \in K \ |\ rA \subseteq \mathcal{O}\}$. Prove the following: (1) $A^{-1}$ is an $\mathcal{O}$-submodule of $K$, (2) If $B \subseteq A$ is an ideal, then $A^{-1}B \subseteq \mathcal{O}$ is an ideal, and (3) If $C \subseteq B \subseteq A$ are ideals, then $A^{-1}C \subseteq A^{-1}B$.

Let $x,y \in A^{-1}$ and let $r \in \mathcal{O}$. Now for all $k \in A$, we have $(x+ry)k = xk + ryk \in \mathcal{O}$, since $xk \in \mathcal{O}$ and $y(rk) \in \mathcal{O}$ (because $A$ is an ideal). Since $0 \in A^{-1}$, by the submodule criterion, $A^{-1}$ is an $\mathcal{O}$-submodule of $K$.

Suppose $C \subseteq B \subseteq A$. Note that if $x \in A^{-1}C$, then $x = \sum r_ic_i$ where $r_i \in A^{-1}$ and $c_i \in C$. Certainly then $x \in A^{-1}B$, as desired in (3).

Suppose $B \subseteq A$; by the previous argument, we have $A^{-1}B \subseteq A^{-1}A \subseteq \mathcal{O}$. Suppose $x,y \in A^{-1}B$ and $r \in \mathcal{O}$, with $x = \sum r_ib_i$ and $y = \sum s_ic_i$. Certainly then since $B$ is an ideal we have $x+ry = \sum r_ib_i + \sum s_irc_i \in A^{-1}B$. Since $0 \in A^{-1}B$, by the submodule criterion, $A^{-1}B$ is an ideal in $\mathcal{O}$.

### Over ZZ, if (a,b) is maximal then gcd(a,b) is prime

Let $a,b \in \mathbb{Z}$ such that $(a,b)$ is maximal as an ideal in $\mathbb{Z}$. What can be said about $a$ and $b$?

Since $\mathbb{Z}$ is a principal ideal domain, $(a,b) = (d)$ for some $d$. If $(d)$ is maximal, then it is prime, so that $d$ is a prime element. Moreover, $d$ is a greatest common divisor of $a$ and $b$. So $\mathsf{gcd}(a,b)$ is prime.

### Over an integral domain, every maximal ideal is irreducible

Let $R$ be an integral domain. Prove that every maximal ideal in $R$ is irreducible.

We begin with a lemma. This lemma is actually true (with a minor modification) over a much larger class of rings. However, I am not entirely comfortable with the proof of the general version. I am, however, comfortable with this more specific version. I referred to a discussion at MathForum.org for a proof of the lemma.

Lemma: Let $R$ be an integral domain and let $I \subseteq R$ be an ideal which is finitely generated as a $\mathbb{Z}$-module. If $I^2 = I$, then either $I = 0$ or $I = R$. Proof: Suppose $I \neq 0$. Let $A = \{\alpha_i\}_{i=1}^n$ be a generating set for $A$ over $\mathbb{Z}$; in particular, some $\alpha_i$ is nonzero. Since $I^2 = I$, there exist $u_{i,j} \in A$ such that $\alpha_i = \sum_j u_{i,j} \alpha_j$ for each $i$. In particular, $0 = \sum_j (u_{i,j} - \delta_{i,j})\alpha_j$ for each $i$, where $\delta_{i,j}$ is the Kronecker delta. That is, $[\alpha_1\ \cdots\ \alpha_n]^\mathsf{T}$ is a nontrivial solution to the matrix equation $Mx = 0$, where $M = [u_{i,j} - \delta_{i,j}]$. Thinking of $R$ as embedded in its field $F$ of fractions, we have $\mathsf{det}(M) = 0$. On the other hand, by the Leibniz expansion of $\mathsf{det}(M)$, we have $\mathsf{det}(M) = \beta - 1$, where $\beta \in I$ (using the fact that $I$ is an ideal). In particular, $-1 \in I$, so that $I = R$. $\square$

Now let $M \subseteq R$ be nonzero and maximal. If $M = AB$, then $M \subseteq A$ and $M \subseteq B$, so that $A,B \in \{M, R\}$. If $A = B = M$, then $M^2 = M$. By the lemma, either $M = 0$ (a contradiction) or $M = R$ (also a contradiction). So either $A$ or $B$ is $R$. Since $R$ is a unit in the semigroup of ideals of $R$ under ideal products, $M$ is irreducible.

### Prove that a given ideal is maximal

Prove that $A = (1+\sqrt{-5}, 1-\sqrt{-5})$ is maximal in $\mathcal{O}_K$, where $K = \mathbb{Q}(\sqrt{-5})$.

We will prove this in slightly more generality, beginning with a lemma.

Lemma: Let $D$ be a squarefree integer with $D \equiv 3$ mod 4. Then $\{1+\sqrt{D}, 1-\sqrt{D}\}$ is a basis for $A = (1+\sqrt{D}, 1-\sqrt{D})$ over $\mathbb{Z}$. Proof: Let $\zeta \in A$. Then $\zeta = (a+b\sqrt{D})(1+\sqrt{D}) + (h+k\sqrt{D})(1-\sqrt{D})$ for some $a,b,h,k \in \mathbb{Z}$. Evidently, $\zeta = u(1+\sqrt{D}) + v(1-\sqrt{D})$, where $u = a+b\frac{D+1}{2} - k\frac{D-1}{2}$ and $v = h + b\frac{D-1}{2} - k\frac{D+1}{2}$. So $A = (1+\sqrt{D},1-\sqrt{D})_\mathbb{Z}$. Suppose now that $h(1+\sqrt{D}) + k(1-\sqrt{D}) = 0$; then $h+k = 0$ and $h-k = 0$, so that $h = k = 0$, as desired. $\square$

Let $D$ be a squarefree integer with $D \equiv 3$ mod 4. Now the ring of integers in $K = \mathbb{Q}(\sqrt{D})$ is $\mathcal{O} = \{a+b\sqrt{D} \ |\ a,b \in \mathbb{Z}\}$. We claim that $A = (1+\sqrt{D}, 1-\sqrt{D})$ is maximal in $\mathcal{O}$.

To see this, note that $2 = (1+\sqrt{D}) + (1-\sqrt{D}) \in A$. Now let $a+b\sqrt{D} \in \mathcal{O}$, and note that $a+b\sqrt{D} = a-b+b+b\sqrt{D} = a-b + b(1+\sqrt{D}) \equiv a-b$ mod $A$. In particular, if $a-b \equiv 0$ mod 2, then $a+b\sqrt{D} \equiv 0$ mod $A$, and if $a-b \equiv 1$ mod 2, then $a+b\sqrt{D} \equiv 1$ mod $A$. We claim that these cosets are distinct. Indeed, if $1 \equiv 0$ mod $A$, then $1 \in A$. In this case, by the lemma we have $h,k \in \mathbb{Z}$ such that $h(1+\sqrt{D}) + k(1-\sqrt{D}) = (h+k) + (h-k)\sqrt{D} = 1$. But then $h+k = 1$ and $h-k = 0$, so that $2h = 1$, a contradiction. So $\mathcal{O}/A = \{0+A, 1+A\}$. Now $\mathcal{O}/A$ is a unital ring with $1 \neq 0$ having only two elements, so that $\mathcal{O}/A \cong \mathbb{Z}/(2)$; that is, $\mathcal{O}/A$ is a field, and thus $A$ is a maximal ideal.

The original problem follows since $-5 \equiv 3$ mod 4.

### Irreducible modules over a commutative ring are precisely its maximal ideal quotients

Let $R$ be a commutative ring with 1 and let $M$ be a unital left $R$ module. Prove that $M$ is irreducible if and only if $M \cong_R R/I$ for some maximal ideal $I \subseteq R$. (Where $\cong_R$ means “isomorphic as an $R$-module.”)

Suppose $M$ is irreducible, fix $m \in M$ nonzero, and define $\varphi_m : R \rightarrow M$ by $\varphi_m(r) = r \cdot m$. Note that for all $x,y \in R$ and $r \in R$, $\varphi_m(x + r \cdot y) = (x+r \cdot y) \cdot m = x \cdot m + r \cdot (y \cdot m)$ $= \varphi_m(x) + r \cdot \varphi_m(y)$. So $\varphi_m$ is an $R$-module homomorphism. Now recall that since $m \neq 0$ and $M$ is irreducible, then by this previous exercise, $M = Rm$. In particular, if $b \in M$, then there exists $a \in R$ such that $b = a \cdot m = \varphi_m(a)$. Thus $\varphi_m$ is surjective.

FInally, we claim that $\mathsf{ker}\ \varphi_m$ is a maximal ideal. To that end, let $x + \mathsf{ker}\ \varphi_m$ be nonzero. In particular, $x \cdot m \neq 0$. Since $M$ is irreducible, we have $M = R(x \cdot m)$. In particular, $m = y \cdot (x \cdot m) = (yx) \cdot m$ for some $y \in R$. Then $1 \cdot m - (yx) \cdot m = 0$, so that $1 - yx \in \mathsf{ker}\ \varphi_m$. That is, $(y + \mathsf{ker}\ \varphi_m)(x + \mathsf{ker}\ \varphi_m) = 1 + \mathsf{ker}\ \varphi_m$. So every nonzero element of $R/\mathsf{ker}\ \varphi_m$ has a left inverse. Since $R/\mathsf{ker}\ \varphi_m$ is commutative, $R/\mathsf{ker}\ \varphi_m$ is a field, so that $\mathsf{ker}\ \varphi_m$ is a maximal ideal of $R$.

By the First Isomorphism Theorem, we have $M \cong_R R/\mathsf{ker}\ \varphi_m$, where $\mathsf{ker}\ \varphi_m$ is a maximal ideal.

Conversely, suppose $I \subseteq R$ is a maximal ideal. Now $R/I$ is a field. Let $x + I$ be nonzero. Then there exists $y \in R$ such that $(y+I)(x+I) = (1+I)$. Now let $r+I \in R/I$. Note that $r+I = (ry+I)(x+I)$, so that $R/I = R(x+I)$. That is, $R/I$ is generated (as an $R$-module) by any nonzero element. By this previous exercise, $R/I$ is an irreducible $R$-module.

### Two prime ideals in ZZ[x,y]

Prove that $(x,y)$ and $(2,x,y)$ are prime ideals in $\mathbb{Z}[x,y]$ but that only $(2,x,y)$ is maximal.

In this previous exercise, we saw that $\mathbb{Z}[x,y]/(x,y) \cong \mathbb{Z}$. Since this quotient is a domain but not a field, $(x,y)$ is a prime ideal but not maximal.

On the other hand, we have $\mathbb{Z}[x,y]/(2,x,y) \cong \mathbb{Z}[x][y]/((2,x) + (y)) \cong \mathbb{Z}[x]/((2)+(x)) \cong \mathbb{Z}/(2)$, which is a field. Thus $(2,x,y)$ is maximal and hence prime.

### Characterization of the ideals in a discrete valuation ring

Let $K$ be a field, let $v$ be a discrete valuation on $K$, and let $R$ be the valuation ring of $v$. For each integer $k \geq 0$, define $A_k = \{ r \in R \ |\ v(r) \geq k \} \cup \{0\}$.

1. Prove that $A_k$ is a principal ideal and that $A_{k+1} \subseteq A_k$ for all $k \in \mathbb{N}$.
2. Prove that if $I$ is a nonzero ideal of $R$, then $I = A_k$ for some $k \geq 0$. Deduce that $R$ is a local ring with maximal ideal $A_1$.

1. It is clear that $A_{k+1} \subseteq A_k$ for all $k$.

Now we show that $A_k$ is an ideal. We have $0 \in A_k$, so $A_k$ is nonempty. Now let $a,b \in A_k$; if one or both of $a$ and $b$ is 0, then $a+b \in A_k$. If $a+b = 0$, then $a+b \in A_k$. If $a+b \neq 0$, we have $v(a+b) \geq \min(v(a),v(b)) \geq k$, so that $a+b \in A_k$. Moreover, $v(-a) = v(a)$, so that $-a \in A_k$. Finally, if $r \in R$, we have $v(r) \geq 0$. Then $v(ra) = v(r) + v(a) \geq k$, so that $ra \in A_k$. Since $K$ is commutative, we have that $A_k$ is an ideal of $R$.

We now show that $A_k$ is principal. Choose $\alpha \in A_k$ such that $v(\alpha) = k$; such an element exists because $v$ is surjective. Note that $\alpha^{-1}$ exists in $K$, and that $v(\alpha^{-1}) = -v(\alpha)$. Let $r \in A_k$; then $v(r) \geq k$. Note that $v(\alpha^{-1}r) = v(\alpha^{-1}) + v(r) = v(r) - v(\alpha) \geq 0$, so that $\alpha^{-1}r \in R$. Moreover, $r = \alpha\alpha^{-1}r$. Thus $r \in (\alpha)$, and we have $A_k \subseteq (\alpha)$, and thus $A_k = (\alpha)$. In particular, $A_k$ is generated by any element of valuation $k$.

2. Let $I \subseteq A_k$ be a nonzero ideal of $R$. Let $k$ be minimal among $v(r)$ for $r \in I$, and let $a \in I$ such that $v(a) = k$. In particular, we have $I \subseteq A_k$. Moreover, since $A_k = (a)$, we have $A_k \subseteq I$. Thus $I = A_k$.

From part (1), every proper ideal of $R$ is contained in $A_1$; thus $A_1$ is maximal, and moreover, is the unique maximal ideal of $R$.

### If the set of all nonunits in a commutative ring is an ideal, then it forms the unique maximal ideal

A commutative ring $R$ is called local if it has a unique maximal ideal. Prove that if $R$ is a local ring with a maximal ideal $M$ then every element of $R \setminus M$ is a unit. Prove conversely that if $R$ is a commutative ring with $1 \neq 0$ in which the set of nonunits forms an ideal $M$, then $R$ is a local ring with unique maximal ideal $M$.

Let $u \in R \setminus M$. Consider the ideal $(u)$; if proper, this ideal must be contained in some maximal ideal, and $M$ is the only maximal ideal. Thus $(u) \subseteq M$, a contradiction since $u \notin M$. Thus $(u) = R$, and for some $v \in R$ we have $uv = vu = 1$. Thus $u$ is a unit.

Suppose now that the set $M$ of nonunits in $R$ form an ideal. First we show that $M$ is maximal. If $M \subsetneq I$ for some ideal $I$, then $I$ contains a unit, so that $I = R$. Now we show that $M$ is the unique maximal ideal. Suppose that we have an ideal $N \subseteq R$ such that $N \not\subseteq M$. Then $N$ contains some element $x$ not in $M$, which is a unit. Thus $N = R$. In particular, every proper ideal of $R$ is contained in $M$. Thus $M$ is the unique proper ideal of $R$, and $R$ is local.

### Use Zorn’s Lemma to construct an ideal which maximally does not contain a given finitely generated ideal

Let $R$ be a ring with $1 \neq 0$, and let $A = (a_1,\ldots,a_n)$ be a nonzero finitely generated (proper) ideal of $R$. Prove that there is an ideal in $R$ which is maximal with respect to the property “does not contain $A$“.

Let $\mathcal{C}$ denote the set of all ideals in $R$ which do not contain $I$; this set is partially ordered by inclusion. Moreover $\mathcal{C}$ is nonempty since 0 does not contain $A$. Let $\{C_i\}_\mathbb{N}$ be a chain in $\mathcal{C}$. Now $\bigcup C_i$ is an ideal in $R$ by this previous exercise. Suppose $A \subseteq \bigcup C_i$. Then for each generator $a_k$ of $A$, we have $a_k \in C_{i_k}$ for some $i_k$. Since $\{C_i\}_\mathbb{N}$ is totally ordered and there are only finitely many $a_k$, there exists an inclusion-maximal element $C_t$ such that $a_k \in C_t$ for each $k$; thus $A \subseteq C_t$, a contradiction. Thus $A \nsubseteq \bigcup C_i$, and we have $\bigcup C_i \in \mathcal{C}$. So $\bigcap C_i$ is an upper bound of the chain $\{C_i\}_\mathbb{N}$ in $\mathcal{C}$. Thus every chain in $\mathcal{C}$ has an upper bound, and by Zorn’s lemma there exists a maximal ideal $B$ with respect to the property “does not contain $A$“.