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 $I=(p_1)(p_2)\dots (p_n)$.□

389f77

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 $b^n\in Q$.□

cf1276

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 $i\ne j$.□

7bd246

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 Exr 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 Exr 2. By Exr 3, we have $I=Q_1\dots Q_k=Q_1\cap \cdots \cap Q_k$.□

bb9980

Note that the manner in which we constructed the primary decomposition here ensures that it is irredundant, i.e, all the primary ideals are distinct and removing any one of them changes the intersection.

[!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 $I$.