TST ALG3 Midsem

All rings are commutative with $1$.

Problem 11

Let $R$ be a PID.

Exercise 290.1(Exercise).

Let $I\ne (0)$ be a proper ideal in $R$. Show that $I=P_1\dots P_n$ for some maximal ideals $P_1,\dots ,P_n$.

Proof.

Suppose$I=(f)$. Since $R$ is a UFD, we can write $f$ as a product of irreducibles $p_1{p}_2\dots p_n$. Each $p_i$ is prime, so each $(p_i)$ is a prime ideal, and hence each $(p_i)$ is maximal. It is now easy to see that $I=(p_1)(p_2)\dots (p_n)$.□

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Exercise 290.2(Exercise).

An ideal $Q$ in $R$ is said to be primary if $ab\in Q$ and $a\notin Q$ implies $b^n\in Q$ for some $n\ge 1$. Show that an ideal $Q$ is primary iff $Q=(f^n)$ for some irreducible element $f$.

Proof.

Let$Q=(g)$ where $g=p_1{p}_2\dots p_n$ for irreducibles $p_1,\dots ,p_n$. Suppose $Q$ is primary. Suppose $p_i$ and $p_j$ are not associates. Let $p_i$ have $l$ associates in $p_1,\dots ,p_n$. Let $a:=p_i^l$ and $b:=p_1{p}_2\dots p_n/p_i^l$. Then, $ab=g\in Q$. However, $p_j$ does not divide $a^k$ for any $k\ge 1$, and $p_i$ does not divide $b^k$ for any $k\ge 1$. So, $g$ does not divide $a^k$ or $b^k$ for any $k\ge 1$, a contradiction. Thus, every pair in $p_1,\dots ,p_n$ is a pair of associates. It follows that $Q=(p_1^n)$.

Suppose $Q=(f^n)$. If $ab\in Q$, $f$ must divide $a$ or $b$, so $a^n\in Q$ or $b^n\in Q$.□

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Exercise 290.3(Exercise).

Suppose $Q_1,\dots ,Q_n$ are distinct primary ideals. Show that $Q_1\dots Q_n=Q_1\cap \cdots \cap Q_n$.

Proof.

It suffices to show that$Q_1,\dots ,Q_n$ are pairwise comaximal. Let $Q_i=(f_i^{n_i})=(f_i{)}^{n_i}$. $(f_i)$ is prime and hence maximal for each $i$. So, $(f_i)$ and $(f_j)$ are comaximal for all $i\ne j$. In any commutative ring, $I+J=R$ implies $I^m+J^k=R$ for all ideals $I,J\subseteq R$ and positive integers $m,k$ (just raise the identity $\alpha +\beta =1$ to the power $m+k$). Thus, it follows that $(f_i{)}^{n_i}$ and $(f_j{)}^{n_j}$ are comaximal for all $i\ne j$.□

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Exercise 290.4(Exercise).

Show that every proper ideal $I$ in $R$ can be expressed as a finite intersection of primary ideals.

Proof.

Write$I=(p_1)(p_2)\dots (p_n)$ where $p_i$ are irreducible, by Exercise 1. Let $q_1,\dots ,q_k$ be the unique elements among $p_1,\dots ,p_n$ up to association. Then, we can write $I=(q_1{)}^{n_1}(p_2{)}^{n_2}\dots (q_k{)}^{n_k}$. Let $Q_i:=(q_i{)}^{n_i}=(q_i^{n_i})$ and note that $Q_i$ is primary by Exercise 2. By Exercise 3, we have $I=Q_1\dots Q_k=Q_1\cap \cdots \cap Q_k$.□

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[!Exercise]
Suppose $I=Q_1\cap \cdots \cap Q_n$, where $Q_i=(f_i^n)$ and $f_i$ is irreducible for all $i=1,\dots ,n$. Let $\sqrt{I}=\{r\in R:r^m\in I\text{ for some integer }m\ge 1\}$. Describe $\sqrt{I}$ in terms of $f_1,\dots ,f_n$.

[!Exercise]
Show that $R/I\cong R/Q_1\times R/Q_2\times \cdots \times R/Q_n$. Is it possible to write $R/I$ as a direct product of fields?

[!Exercise]
For any ideal $I$ in $R$, show that there are only finitely many maximal ideals that contain $I$.