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Every ideal generated by one element: the greatest common divisor as that generator, the standard ideals that need two, and the theorem that makes irreducible and prime the same word.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to find the single generator of an ideal of the integers, decide whether an ideal is prime, maximal or neither by forming the quotient, show that a named ideal needs two generators, state why every non-zero prime ideal of a principal ideal domain is maximal, place the standard counterexamples in the chain of conditions, and prove that irreducible implies prime in a principal ideal domain.
The last lesson proved that a division algorithm forces every ideal to be generated by one element. That conclusion turns out to be more useful than the algorithm itself: it is what the proofs actually use, it holds in rings with no algorithm at all, and it is stable under constructions that a size function is not. So it is promoted here from a consequence to a hypothesis.
An ideal is principal when $I = (a) = \{ra : r \in R\}$ for a single $a$. A principal ideal domain is an integral domain in which every ideal is principal. An ideal $P$ is prime when $R/P$ is a domain, and maximal when $R/P$ is a field. Bézout's identity is the statement $\gcd(a, b) = ax + by$, which in ideal language says $(a, b) = (\gcd(a,b))$.
The definition. An integral domain $R$ is a principal ideal domain when every ideal of $R$ has the form $(a)$ for a single element $a$.
Examples. $\mathbb{Z}$; $F[x]$ for a field $F$; $\mathbb{Z}[i]$; and $\mathbb{Z}\!\left[\frac{1 + \sqrt{-19}}{2}\right]$, which is a PID with no division algorithm. Non-examples. $\mathbb{Z}[x]$, where $(2, x)$ needs two; $F[x, y]$, where $(x, y)$ needs two; $\mathbb{Z}[\sqrt{-5}]$, where $(2, 1 + \sqrt{-5})$ needs two.
Greatest common divisors. In a PID the ideal $(a, b)$ is $(d)$ for some $d$. Then $d \mid a$ and $d \mid b$; and any common divisor $c$ divides everything in $(a,b)$, hence divides $d$. So $d$ is a greatest common divisor, and since $d \in (a, b)$ it can be written $d = ax + by$. Bézout's identity holds in every PID, with no algorithm required to find the coefficients.
The theorem that matters: irreducible implies prime. Suppose $p$ is irreducible, $p \mid ab$ and $p \nmid a$. The ideal $(p, a)$ is $(d)$ with $d \mid p$; irreducibility makes $d$ a unit or an associate of $p$, and it is not the latter because $d \mid a$. So $(p, a) = R$, giving $1 = px + ay$; multiplying by $b$ gives $b = pbx + aby$, and $p$ divides both terms.
That is the exact point at which the two words of the previous lesson merge, and it is why the next lesson can prove unique factorisation.
Non-zero primes are maximal. If $(p) \subseteq (a) \subseteq R$ then $a \mid p$, so $a$ is a unit or an associate of $p$ — there is no room in between. So in a PID the ideals $(p)$ for $p$ irreducible are exactly the non-zero maximal ideals, and $R/(p)$ is a field. That is how finite fields are built: $\mathbb{F}_p[x]/(f)$ for $f$ irreducible.
Ascending chains stop. In a PID no strictly increasing chain of ideals can run for ever, which is the Noetherian condition and is what makes every element factors into irreducibles true at all. That is two lessons away.
Another way: picture
In $\mathbb{Z}$, an ideal is a set of multiples of one number, and containing another ideal means dividing. So the lattice of ideals is the divisibility lattice upside down: bigger ideal, smaller generator. A PID is a ring where that picture is accurate — every ideal is the multiples of something — and the failures are rings where an ideal is genuinely built from two directions at once and has no single number behind it.
Another way: steps
To show a ring is a PID: 1. Easiest route: exhibit a size function and apply the last lesson. 2. Otherwise, take an arbitrary ideal and produce a generator directly. To show it is not: 3. Find two elements whose only common divisors are units. 4. Check the ideal they generate is not the whole ring — usually by a congruence the ideal's elements all satisfy. 5. Conclude that no single generator exists.
Take $R = \mathbb{Z}[x]$ and $I = (2, x)$, the polynomials of the form $2f(x) + xg(x)$.
What is in it. Exactly the polynomials with even constant term. Any $2f + xg$ has constant term $2f(0)$, which is even; conversely a polynomial with even constant term is its constant term plus $x$ times something.
Why it is not principal. Suppose $I = (h)$. Then $h \mid 2$, so $h$ is $\pm 1$ or $\pm 2$ (degree zero, since degrees add). And $h \mid x$, which rules out $\pm 2$ because $x/2 \notin \mathbb{Z}[x]$. So $h = \pm 1$ and $I = R$ — but $1 \notin I$, since its constant term is odd. Contradiction.
What this shows about the chain. $\mathbb{Z}[x]$ is a unique factorisation domain, by Gauss's lemma. So the arrow PID $\Rightarrow$ UFD does not reverse, and $(2, x)$ is the witness.
Two more facts worth carrying.
The same shape elsewhere. $(x, y)$ in $F[x,y]$ and $(2, 1 + \sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$ fail for the same reason: two elements sharing no non-unit factor, generating something smaller than the whole ring. In a Euclidean domain that is impossible, because the algorithm manufactures the combination that equals their greatest common divisor.
*Thinking principal means generated by a prime.* It means generated by one element, whatever that element is. $(6)$ is principal in $\mathbb{Z}$.
Assuming every PID is Euclidean. The implication runs one way only, with a genuine counterexample.
Assuming $R$ a PID makes $R[x]$ one. It does not: $\mathbb{Z}$ is a PID and $\mathbb{Z}[x]$ is not. What does survive is unique factorisation.
*Reading every prime ideal is maximal without the word non-zero.* The zero ideal is prime in any domain and is maximal only in a field.
Concluding a ring is not a PID from one two-generator description. $(4, 6) = (2)$ in $\mathbb{Z}$ — the ideal was merely described wastefully. The failure has to be proved, by showing no single generator can exist.
Forgetting the domain condition. Principal ideal ring without domain is a genuinely different and weaker notion, and the theorems here use the absence of zero divisors.
Let $I \ne 0$ be an ideal and take $g \in I$ non-zero of least degree.
A least element exists because degrees are non-negative integers.
For any $f \in I$, divide: $f = qg + r$ with $r = 0$ or $\deg r < \deg g$. But $r = f - qg \in I$.
The remainder stays inside.
A non-zero $r$ would beat $g$ for degree, so $r = 0$ and $I = (g)$. The generator is the monic polynomial of least degree in $I$.
Principal.
In $\mathbb{Q}[x]$, take $(x^{2} + 1)$. The generator is irreducible over $\mathbb{Q}$, since it has no rational root and has degree $2$.
An irreducible generator.
In a PID, $(p)$ for $p$ irreducible is maximal, so the quotient is a field.
Maximal.
And indeed $\mathbb{Q}[x]/(x^{2}+1) \cong \mathbb{Q}(i)$. The same construction over $\mathbb{F}_2$ with $x^{2} + x + 1$ builds the field of four elements.
A field, built to order.
Suppose $(x, y) = (h)$. Then $h \mid x$ and $h \mid y$.
A single generator would divide both.
Degrees force $h$ to be a constant or an associate of $x$; an associate of $x$ cannot divide $y$, so $h$ is a non-zero constant, hence a unit.
So the ideal would be everything.
But $(x, y)$ consists of the polynomials with zero constant term, and $1$ is not among them. So the ideal is not principal and $\mathbb{Q}[x,y]$ is not a PID — though it is a unique factorisation domain.
Each row gives two integers. Fill in the single positive integer that generates the ideal they generate together.
| First element | Second element | The single generator | |
|---|---|---|---|
| The first ideal | 4 | 6 | |
| The second ideal | 9 | 12 | |
| The third ideal | 8 | 12 | |
| The fourth ideal | 5 | 7 | |
| The fifth ideal | 18 | 24 |
What kind of ideal is $(x^{2} + 1)$ in $\mathbb{R}[x]$?
An ideal of a principal ideal domain is handed over with $7$ generators. At most how many does it actually need?
Answer:
Select every statement that is true of a principal ideal domain.
This task has no paper form; do it on a device.
Match each ring to an ideal of it that cannot be generated by one element — or, in one case, to the fact that none exists.
| $(2, x)$ | $(x, y)$ | $(2, 1 + \sqrt{-5})$ | Every ideal is principal | $(x^{2}, x)$ | |
|---|---|---|---|---|---|
| $\mathbb{Z}[x]$ | |||||
| $F[x, y]$ over a field $F$ | |||||
| $\mathbb{Z}[\sqrt{-5}]$ | |||||
| $\mathbb{Z}$ |
Build the proof that in a principal ideal domain every irreducible element is prime.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
An ideal of a principal ideal domain is handed over with $5$ generators. At most how many does it actually need?
Answer:
You can find the generator of an ideal in a principal ideal domain and show that a given ideal elsewhere needs two. Say in your own words why $(2, x)$ is not principal in $\mathbb{Z}[x]$. Next: what having irreducible equal to prime buys — unique factorisation.
9. Your turn: is $(x, y)$ principal in $\mathbb{Q}[x, y]$?, step 3