Principal Ideal Domains (PID)

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2020)
• Priority tier: T4
• Marks (count): 10 (1)
• Average solve time: ~12 min
• Difficulty mix: medium 1
• Section: A | Dominant type: proof

Why This Chapter Matters

PIDs occupy the middle rung of the hierarchy Euclidean domain $\Rightarrow$ PID $\Rightarrow$ UFD $\Rightarrow$ integral domain. UPSC 2020 targeted the most accessible node of this hierarchy: proving that $\mathbb{Z}$ is a PID. The proof is short and structured — every ideal of $\mathbb{Z}$ is of the form $n\mathbb{Z}$ by the well-ordering principle — making this one of the highest-return proofs in the algebra syllabus.

Minimum Theory

Ideals

Let $R$ be a commutative ring with unity. A non-empty subset $I \subseteq R$ is an ideal if:

1. $a, b \in I \Rightarrow a - b \in I$ (additive subgroup).
2. $a \in I,\, r \in R \Rightarrow ra \in I$ (absorption under multiplication).

A principal ideal generated by $a \in R$ is $\langle a \rangle = (a) = \{ ra : r \in R \}.$ This is the smallest ideal containing $a$.

Integral Domain

A commutative ring with unity is an integral domain if it has no zero divisors: $ab = 0 \Rightarrow a = 0$ or $b = 0$.

Principal Ideal Domain

A principal ideal domain (PID) is an integral domain in which every ideal is principal.

The Hierarchy

$\text{Euclidean domain} \Rightarrow \text{PID} \Rightarrow \text{UFD} \Rightarrow \text{Integral domain}$

None of these implications reverses in general:

• $\mathbb{Z}[\frac{1+\sqrt{-19}}{2}]$ is a PID but not Euclidean (under any norm).
• $\mathbb{Z}[x]$ is a UFD but not a PID (the ideal $\langle 2, x \rangle$ is not principal).

Standard PIDs

$\mathbb{Z}$ is a PID: The Proof

Theorem. $\mathbb{Z}$ is a principal ideal domain.

Proof.

First, $\mathbb{Z}$ is an integral domain (standard fact: $\mathbb{Z}$ is a commutative ring with unity and has no zero divisors, since $\mathbb{Z}$ is embedded in $\mathbb{R}$).

Let $I$ be any ideal of $\mathbb{Z}$.

Case 1: $I = \{0\} = \langle 0 \rangle$. Principal.

Case 2: $I \ne \{0\}$. Then $I$ contains a non-zero element; since $I$ is closed under negation, $I$ contains a positive element. By the well-ordering principle, $I$ has a least positive element; call it $n > 0$.

We claim $I = \langle n \rangle = n\mathbb{Z}$.

$(\supseteq)$: Since $n \in I$ and $I$ is an ideal, $kn \in I$ for all $k \in \mathbb{Z}$, so $n\mathbb{Z} \subseteq I$.

$(\subseteq)$: Let $a \in I$. By the division algorithm, $a = qn + r$ with $0 \le r < n$. Then $r = a - qn \in I$ (since $a \in I$ and $qn \in I$). Since $0 \le r < n$ and $n$ is the least positive element of $I$, we must have $r = 0$. So $a = qn \in n\mathbb{Z}$.

Hence $I = n\mathbb{Z} = \langle n \rangle$, which is principal.

Since every ideal is principal, $\mathbb{Z}$ is a PID. $\blacksquare$

$\mathbb{Z}[x]$ is NOT a PID

The ideal $I = \langle 2, x \rangle = \{ 2f(x) + xg(x) : f, g \in \mathbb{Z}[x] \}$ (the set of polynomials in $\mathbb{Z}[x]$ with even constant term) is not principal:

Suppose $I = \langle p(x) \rangle$. Then $p(x) \mid 2$ and $p(x) \mid x$ in $\mathbb{Z}[x]$. Since $p(x) \mid 2$, $p$ is a constant $\pm 1$ or $\pm 2$. Since $p(x) \mid x$, the constant $p$ must equal $\pm 1$. But then $\langle p(x) \rangle = \mathbb{Z}[x] \ni 1$, yet $1 \notin I$ (the constant term of $1$ is $1$, which is odd). Contradiction.

Question Archetypes

Z-is-PID (1 question; 2020)

Recognition Cues

• “Prove that $\mathbb{Z}$ is a principal ideal domain”
• “Show that every ideal of $\mathbb{Z}$ is of the form $n\mathbb{Z}$”
• 10 marks: state the PID definition, then prove it for $\mathbb{Z}$

Solution Template

1. State definition of PID (integral domain + every ideal principal).
2. Assert $\mathbb{Z}$ is an integral domain (no zero divisors, commutative, has unity).
3. Let $I$ be any ideal. Handle $I = \{0\}$ separately.
4. For $I \ne \{0\}$: use well-ordering to find the least positive element $n$.
5. Show $I = n\mathbb{Z}$ using division algorithm.
6. Conclude.

Worked Example

2020 Paper 2, 2020-P2-Q2a (10 marks)

Prove that $\mathbb{Z}$, the ring of integers, is a principal ideal domain.

Solution.

Definition. A PID is an integral domain in which every ideal is principal.

Step 1: $\mathbb{Z}$ is an integral domain. $\mathbb{Z}$ is a commutative ring with unity $1$. If $ab = 0$ in $\mathbb{Z}$ then, since $\mathbb{Z} \subset \mathbb{R}$ and $\mathbb{R}$ has no zero divisors, $a = 0$ or $b = 0$. Hence $\mathbb{Z}$ is an integral domain.

Step 2: Every ideal of $\mathbb{Z}$ is principal. Let $I$ be an ideal of $\mathbb{Z}$.

If $I = \{0\}$, then $I = \langle 0 \rangle$ is principal.

Otherwise, $I$ contains a non-zero integer. Since $a \in I \Rightarrow -a \in I$, the set $I \cap \mathbb{Z}_{>0}$ is non-empty. By the well-ordering principle, $I \cap \mathbb{Z}_{>0}$ has a least element; call it $n$.

Claim: $I = n\mathbb{Z}$.

$(\supseteq)$: $n \in I$ and $I$ is an ideal, so for any $k \in \mathbb{Z}$, $kn \in I$. Thus $n\mathbb{Z} \subseteq I$.

$(\subseteq)$: Let $a \in I$. By the division algorithm in $\mathbb{Z}$, write $a = qn + r$ with $q, r \in \mathbb{Z}$ and $0 \le r < n$. Then $r = a - qn \in I$ (since $a \in I$ and $qn \in I$). Since $0 \le r < n$ and $n$ is the smallest positive element of $I$, we must have $r = 0$. Therefore $a = qn \in n\mathbb{Z}$, so $I \subseteq n\mathbb{Z}$.

Hence $I = n\mathbb{Z} = \langle n \rangle$ is principal.

Conclusion. Since every ideal of $\mathbb{Z}$ is principal and $\mathbb{Z}$ is an integral domain, $\mathbb{Z}$ is a PID.

$\boxed{\mathbb{Z} \text{ is a PID}}\;\blacksquare$

Common Traps

• Forgetting to verify that $\mathbb{Z}$ is an integral domain before claiming it is a PID.
• Forgetting to handle the case $I = \{0\}$ separately.
• In the division step, not justifying why $r = 0$ — the minimality of $n$ is the key.
• Confusing PID with UFD: $\mathbb{Z}[x]$ is a UFD but not a PID; stating ”$\mathbb{Z}[x]$ is a PID” is a serious error.

Marks-Aware Writing

10 marks (Section A):

• Definition of PID stated correctly (1–2 marks).
• $\mathbb{Z}$ is an integral domain (1–2 marks).
• Case $I = \{0\}$ handled (1 mark).
• Well-ordering to find $n$ (1–2 marks).
• Division algorithm argument for $I \subseteq n\mathbb{Z}$ (2–3 marks).
• Conclusion (1 mark).

Practice Set

Only one historical question on this atom (shown above).