## Tag Archives: prime ideal

### Characterize the irreducible torsion modules over a PID

Let $R$ be a principal ideal domain and let $M$ be a torsion $R$-module. Prove that $M$ is irreducible if and only if $M = Rm$ for some nonzero element $m \in M$ where the annihilator of $m$ in $R$ is a nonzero prime ideal $P$.

Suppose $M$ is an irreducible, torsion $R$-module. By this previous exercise, we have $M \cong_R R/I$ where $I \subseteq R$ is a maximal ideal. Since $R$ is a principal ideal domain, in fact $I = (p)$ for some prime element $p \in R$. Let $\psi : R/(p) \rightarrow M$ be an isomorphism, and let $m = \psi(1 + (p))$. Now $M = Rm$, and moreover $\mathsf{Ann}_R(m) = \mathsf{Ann}_R(1 + (p)) = (p)$.

Conversely, consider the module $M = Rm$, where $\mathsf{Ann}_R(m) = (p)$ is a prime ideal. Define $\psi : R \rightarrow (m)_R$ by $r \mapsto r \cdot m$. Certainly we have $\mathsf{ker}\ \psi = \mathsf{Ann}_R(m) = (p)$. By the First Isomorphism Theorem for modules, we have $\overline{\psi} : R/(p) \rightarrow M$ an isomorphism. Again by this previous exercise, $R/(p)$ is irreducible.

### No prime ideal in an algebraic integer ring has norm 12

Can there exist an algebraic integer ring $\mathcal{O}$ with a prime ideal $A \subseteq \mathcal{O}$ such that $N(A) = 12$?

Recall that if $A$ is prime, then it is maximal, and that $N(A)$ is precisely $|\mathcal{O}/A|$. Since $A$ is maximal, $\mathcal{O}$ is a field. We claim that no field of order 12 exists.

Suppose to the contrary that $F$ is a field of order 12. The characteristic of $F$ is then either 2 or 3, since it must be a prime dividing 12. Moreover, by Cauchy’s Theorem, $F$ contains nonzero elements $\alpha$ and $\beta$ of (additive) order 2 and 3, respectively. That is, $2\alpha = 0$ and $3\beta = 0$, but no smaller multiple of $\alpha$ or $\beta$ is 0. If $F$ has characteristic 2, then $\beta = 0$, and if $F$ has characteristic 3, then $4\alpha = \alpha = 0$, both contradictions.

Since no field of order 12 exists, no prime ideal can have norm 12.

### 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 $\mathcal{O}$.

### Characterize the prime ideals in algebraic integer rings which are also unique factorization domains

Characterize the prime ideals in those algebraic integer rings which are unique factorization domains.

By Theorem 9.4 in TAN, a ring $\mathcal{O}$ of algebraic integers from some field is a unique factorization domain if and only if every ideal is principal.

We claim that in a principal ideal domain, $(\pi)$ is a prime ideal if and only if $\pi$ is an irreducible element. To that end, suppose first that $(\pi)$ is a prime ideal. If $\pi = \alpha\beta$, then $\alpha\beta \in (\pi)$, and so without loss of generality $\alpha \in (\pi)$. So $\alpha = \pi\gamma = \alpha\beta\gamma$, and so $1 = \beta\gamma$; hence $\beta$ is a unit. So $\pi$ is irreducible. Conversely, if $\pi$ is irreducible then it is also prime. If $\alpha\beta \in (\pi)$, then $\pi|\alpha\beta$. Without loss of generality, $\pi|\alpha$, and so $\alpha \in (\pi)$.

### Principal ideals generated by irreducible elements need not be prime

Suppose $\pi$ is an irreducible element in an algebraic integer ring $\mathcal{O}$. Must $(\pi)$ be prime as an ideal in $\mathcal{O}$?

Recall that $1+\sqrt{-5}$ is irreducible in $\mathcal{O}_K$, where $K = \mathbb{Q}(\sqrt{-5})$, since no element in this ring has norm 3. Moreover, $1 - \sqrt{-5} \notin (1+\sqrt{-5})$, as we show. If to the contrary we have $(a+b\sqrt{-5})(1+\sqrt{-5}) = 1 - \sqrt{-5}$, then comparing coefficients we have $a-5b = 1$ and $a+b = -1$, so that $3b = -2$ for some $b \in \mathbb{Z}$, a contradiction. Using this previous exercise, we have $(1+\sqrt{-5}) \subsetneq (1+\sqrt{-5}, 1-\sqrt{-5}) \subsetneq \mathcal{O}$. In particular, $1+\sqrt{-5}$ is irreducible, but $(1+\sqrt{-5})$ is not maximal and thus not prime.

So it need not be the case that $(\pi)$ is prime if $\pi$ is irreducible.

### Characterize the prime monomial ideals in a polynomial ring

Let $F$ be a field and let $R = F[x_1, \ldots, x_t]$. Let $M = (m_i \ |\ i \in I)$ be a monomial ideal (not necessarily finitely generated).

1. Prove that $M$ is prime if and only if $M = (S)$ for some $S \subseteq \{x_1, \ldots, x_t\}$. In particular, there are only finitely many prime monomial ideals, each of which is finitely generated.
2. Prove that $(x_1, \ldots, x_t)$ is the only maximal ideal which is also monomial.

We begin with a lemma.

Lemma: Let $M = (m_i \ |\ i \in I)$ be a monomial ideal. Then we may assume without loss of generality that each $m_i$ is divisibility-minimal with respect to inclusion in $M$; that is, no proper divisor of $m_i$ is in $M$. Proof: Note that each $m_i$ has finite degree, and thus has finitely many divisors. Let $\{n_{i,j}\}_{j \in J_i}$ be the proper divisors of $m_i$ which are minimal with respect to inclusion in $M$. Certainly $(n_{i,j} \ |\ i \in I, j \in J_i) \subseteq M = (m_i \ I\ i \in I) \subseteq (n_{i,j} \ |\ i \in I, j \in J_i)$. Note that if $I$ is finite, then there are only finitely many $n_{i,j}$, and if $I$ is transfinite, then $|\bigcup_{i \in I} J_i| = |I|$. Thus, given a monomial generating set, we may replace it with one whose elements are divisibility minimal without affecting its cardinality (if transfinite) or finiteness (if finite). $\square$

Now let $M = (m_i \ |\ i \in I)$ be a prime monomial ideal, and let $S \subseteq \{x_1, \ldots, x_t\}$ consist of precisely those indeterminates which divide some $m_i$. By the lemma, we may assume that the $m_i$ are divisibiliy minimal- that no proper divisor of $m_i$ is again in $M$. We claim that $M = (S)$. The $(\subseteq)$ direction is clear. Now let $x_k \in S$; then $m_k = x_kq_k$ for some $m_k$. Since $M$ is a prime ideal, either $x_k$ or $q_k$ is in $M$. But since no proper divisor of $m_k$ is in $M$, it follows that $x_k$ is not a proper divisor; instead, $m_k = ux_k$ for some unit $u$. In particular, $x_k \in M$. Thus $(S) \subseteq M$, and we have $M = (S)$.

On the other hand, suppose $S \subseteq \{x_1,\ldots,x_t\}$ and consider the ideal $M = (S)$. Let’s say that (with a reordering if necessary) $S = \{x_1,\ldots,x_k\}$. Now $F[x_1,\ldots,x_t]/(S) \cong F[x_k+1,\ldots,x_t]$, which is a domain (even if $k = t$). So $(S)$ is prime.

Now let $M$ be a maximal ideal which has a monomial generating set. Since $M$ is also prime, by the above argument we have $M = (S)$ for some $S \subseteq \{x_1, \ldots, x_t\}$. Suppose $x_k \notin M$ for some $k$; then $M \subsetneq (M,x_k) \subsetneq R$, contradicting the maximality of $M$. Thus $M = (x_1, \ldots, x_t)$.

### Exhibit an infinite family of prime ideals in R[x,y] when R is an integral domain

Let $R$ be an integral domain and let $a$ and $b$ be relatively prime integers greater than 0. Prove that the ideal $(x^a - y^b)$ is prime in $R[x,y]$.

We will show that $R[x,y]/(x^a - y^b)$ is an integral domain by demonstrating that it is isomorphic to a subring of the integral domain $R[t]$. To that end, define a ring homomorphism $\varphi : R[x,y] \rightarrow R[t]$ by extending $x \mapsto t^b$, $y \mapsto t^a$, $r \mapsto r$ for $r \in R$ homomorphically. Certainly then $\varphi(x^a - y^b) = t^{ab} - t^{ab} = 0$, so that $(x^a - y^b) \in \mathsf{ker}\ \varphi$.

Now let $\alpha \in R[x,y]$, and consider the coset $\overline{\alpha} \in R[x,y]/(x^a-y^b)$. We may eliminate powers of $\overline{y}$ greater than $b-1$ in favor of powers of $\overline{x}$ in the representative $\alpha$, so that in fact $\overline{\alpha} = \overline{\sum_{i=0}^{b-1} \alpha_i y^i}$, where $\alpha_i \in R[x]$. Thus every element of $R$ has the form $\alpha = \left( \sum_{i=0}^{b-1} \alpha_i y^i\right) + \alpha^\prime$, where $\alpha^\prime \in (x^a-y^b)$. Suppose now that $\alpha \in \mathsf{ker}\ \varphi$. Then we have $\varphi(\alpha) = \sum_{i=0}^{b-1} \varphi(\alpha_i) t^{ai} = 0$. Now letting $\alpha_i(x) = \sum_{j=0}^m r_{i,j} x^j$, we have $\varphi(\alpha) = \sum_{i=0}^{b-1} \sum_{j=0}^m r_{i,j} t^{bj + ai}$.

We claim that the exponents $bj + ai$ are pairwise distinct. To see this, suppose that $bj + ai = bj^\prime + ai\prime$. Mod $b$, we have $ai \equiv ai^\prime$. Since $a$ is relatively prime to $b$, it is a unit mod $b$. Thus $i \equiv i^\prime \mod b$. However, since $0 \leq i,i^\prime < b$, this yields $i = i^\prime$. Then $bj = bj^\prime$, so that $j = j^\prime$. Thus we have $r_{i,j} = 0$ for all $i,j$, so $\alpha_i = 0$ for all $i$, and thus $\alpha \in (x^a - y^b)$.

So $\mathsf{ker}\ \varphi = (x^a - y^b)$, and by the first isomorphism theorem for rings, $R[x,y]/(x^a - y^b)$ is isomorphic to a subring of the integral domain $R[t]$. Thus $R[x,y]/(x^a - y^b)$ is an integral domain, and so the ideal $(x^a - y^b)$ is prime.

### Exhibit a prime ideal whose square is not primary

Let $R = \mathbb{Q}[x,y,z]$ and let bars denote passage to $\mathbb{Q}[x,y,z]/(xy - z^2)$. Prove that $\overline{P} = (\overline{x},\overline{z})$ is a prime ideal. Show that $\overline{xy} \in \overline{P}$ but that no power of $\overline{y}$ is in $\overline{P}^2$; that is, $\overline{P}$ is a prime ideal whose square is not primary.

Note that $(R/(xy - z^2)/((x,z,xy-z^2)/(xy-z^2)) \cong R/(x,z,xy-z^2) = R/(x,z) \cong \mathbb{Q}[y]$ is a domain, so that $\overline{P}$ is a prime ideal.

We have $\overline{x}\overline{y} = \overline{z^2} \in \overline{P}$. Now suppose $\overline{y}^n \in \overline{P}$. That is, $y^n + (xy - z^2) \in (x,z,xy-z^2)/(xy - z^2) = (x,z)/(xy-z^2)$. But then $y^n \in (x,z)$; this is a contradiction considering the $x$ (or $z$) degrees. So no power of $\overline{y}$ is in $\overline{P}$.

### Exhibit a primary ideal which is not a power of a prime ideal

Show that the radical of the ideal $I = (x,y^2)$ in $\mathbb{Q}[x,y]$ is $(x,y)$. Deduce that $I$ is a primary ideal that is not a power of a prime ideal.

We discussed some basic properties of radical ideals and primary ideals here and here, respectively.

Recall that $\mathsf{rad}(I) = \{ r \in R \ |\ r^n \in I\ \mathrm{for\ some}\ n \in \mathbb{N}^+ \}$. Certainly we have $(x,y) \subseteq \mathsf{rad}\ (x,y^2)$. Recall that $(x,y) \subseteq \mathbb{Q}[x,y]$ is maximal; thus $\mathsf{rad}\ (x,y^2) \subseteq \mathsf{Jac}\ (x,y^2) \subseteq (x,y)$, using this prior exercise about Jacobson radicals. Thus $\mathsf{rad}\ (x,y^2) = (x,y)$.

Next we show that $(x,y^2)$ is primary; recall that an ideal $P$ is called primary if whenever $ab \in P$ and $a \notin P$, then $b^n \in P$ for some $n \geq 1$. We saw previously that an ideal $P \subseteq R$ is primary if and only if every zero divisor in $R/P$ is nilpotent. Note that $\mathbb{Q}[x,y]/(x,y^2) \cong \mathbb{Q}[y]/(y^2)$. Let $\alpha(y) + (y^2) \in \mathbb{Q}[y]/(y^2)$ be a zero divisor and say $\alpha(y) = \alpha_0 + \alpha_1y$; then we have $\alpha(y)\beta(y) \in (y^2)$ for some $\beta \notin (y^2)$. Without loss of generality, let $\beta(y) = \beta_0 + \beta_1y$. Now $\alpha_0\beta_0 + (\alpha_0\beta_1 + \alpha_1\beta_0)y = 0$, and thus $\alpha_0\beta_0 = 0$ and $\alpha_0\beta_1 + \alpha_1\beta_0 = 0$. Suppose $\alpha_0 \neq 0$; then $\beta_0 = 0$ and so $\beta_1 = 0$. But then $\beta \in (y^2)$, a contradiction. Thus $\alpha_0 = 0$. Thus $\alpha^2 \in (y^2)$, so that $\alpha$ is nilpotent. So $(y^2)$ is primary in $\mathbb{Q}[y]$, and thus $(x,y^2)$ is primary in $\mathbb{Q}[x,y]$.

Now suppose $(x,y^2) = P^n$, where $n \geq 1$ and $P$ is a prime ideal. Now $P$ cannot contain a constant, as in this case we have $(x,y^2) = \mathbb{Q}[x,y]$, a contradiction. So every element of $P$ has degree at least 1, and thus every element of $P^n$ has degree at least $n$. Since $x \in P^n$, we have $n = 1$. But then $(x,y^2) = P$ is prime; this to is a contradiction since $\mathbb{Q}[x,y]/(x,y^2) \cong \mathbb{Q}[y]/(y^2)$ contains a zero divisor (namely $y + (y^2)$) and thus is not an integral domain. So $(x,y^2)$ is not a power of any prime ideal.

### Exhibit a ring with infinitely many minimal prime ideals

Prove that the ring $\mathbb{Z}[x_1,x_2,\ldots]/(x_1x_2,x_3x_4,x_5x_6,\ldots)$ has infinitely many $\subseteq$-minimal prime ideals.

Let $R = \mathbb{Z}[x_1,x_2,\ldots]$ and let $K = (x_1x_2,x_3x_4,\ldots)$.

Let $X = \{ \{2k+1,2k+2\} \ |\ k \in \mathbb{N} \}$, and let $Y$ denote the set of all choice functions on $X$. (That is, each element of $Y$ is a function that, given $k \in \mathbb{N}$, chooses one element from the pair $\{2k+1,2k+2\}$.) For each $\lambda \in Y$, define $I_\lambda = (\lambda(0),\lambda(1),\ldots)$. For example, one such ideal is that generated by all the $x_i$ with odd indices. Certainly there are infinitely many such ideals, each distinct. Also, we have that $K = (x_1x_2,x_3x_4,\ldots) \subseteq I_\lambda$ for each $\lambda$, since each $x_ix_{i+1}$ is in $I_\lambda$ by construction.

For each $\lambda$, by the third isomorphism theorem for rings we have $(R/K)/(I_\lambda/K) \cong R/I_\lambda$, and $R/I_\lambda$ is isomorphic to $R$. Since $R$ is an integral domain, $I_\lambda/K$ is a prime ideal of $R/K$.

We claim also that each $I_\lambda$ is minimal. To see this, suppose $J/K \subseteq I_\lambda/K$ is a prime ideal. Let $(i,i+1)$ be a pair (with $i$ odd) such that (without loss of generality) $x_i \in I$. Now $x_ix_{i+1} \in K \subseteq J$. Since $J$ is prime and $x_{i+1} \notin J$ (as $x_{i+1} \notin I$), we have $x_i \in J$– a contradiction. So in fact $J = I_\lambda$, and hence $J/K = I_\lambda/K$.

Note that $R/K$ is not an integral domain (since $(x_1+K)(x_2+K) = 0$), so that the zero ideal is not prime. Thus each $I_\lambda$ is an inclusion-minimal prime ideal.