Back to the on-screen lesson ·
Every element a product of irreducibles, in only one way up to order and units: the two halves of the definition, Gauss's lemma, and the ring where $6$ factors twice.
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 state both halves of unique factorisation and say what each depends on, place a named ring in the chain from Euclidean to integral domain, compute norms in a quadratic ring and read off which values are unavailable, exhibit two different factorisations of $6$ in $\mathbb{Z}[\sqrt{-5}]$, quote Gauss's lemma for $R[x]$, and prove that an element is irreducible without being prime.
Every argument about integers that says compare the prime factorisations is using unique factorisation, and school never names it as an assumption. The last lesson showed that in a principal ideal domain irreducible elements are prime. This lesson turns that into the uniqueness statement, and then finds the ring where it fails.
A factorisation of a non-zero non-unit is an expression as a product of irreducibles. It is unique up to order and units when any two such expressions have the same number of factors and can be matched so that paired factors are associates. A domain where every such element has a factorisation, unique in that sense, is a UFD. A polynomial over $\mathbb{Z}$ is primitive when its coefficients have no common factor; Gauss's lemma says a product of primitive polynomials is primitive.
The definition. An integral domain $R$ is a unique factorisation domain when every non-zero non-unit can be written as a product of irreducibles, and any two such expressions agree up to the order of the factors and multiplication by units.
The definition has two independent halves, and they fail for different reasons.
Existence follows from the ascending chain condition on principal ideals: if some element had no factorisation, one could strip off irreducible factors for ever, producing a strictly increasing chain $(a) \subset (a_1) \subset (a_2) \subset \cdots$. Every Noetherian domain therefore has existence, which covers everything in this course.
Uniqueness is equivalent, given existence, to every irreducible being prime. If irreducibles are prime, two factorisations can be matched a factor at a time: $p_1$ divides the other product, so it divides some $q_j$, so it is an associate of it, cancel and induct. If some irreducible is not prime, uniqueness fails.
So the whole subject reduces to Euclid's lemma, and that is why the previous lesson's theorem was the important one.
The chain of conditions, complete.
$$\text{Euclidean} \subsetneq \text{PID} \subsetneq \text{UFD} \subsetneq \text{integral domain}$$
with witnesses $\mathbb{Z}\!\left[\frac{1+\sqrt{-19}}{2}\right]$, $\mathbb{Z}[x]$ and $\mathbb{Z}[\sqrt{-5}]$ for the three strict inclusions.
Gauss's lemma and $R[x]$. If $R$ is a UFD then so is $R[x]$. The proof compares factorisations over $R$ with factorisations over the field of fractions, and the bridge is Gauss's lemma: a product of primitive polynomials is primitive. Consequences: $\mathbb{Z}[x]$, $\mathbb{Z}[x, y]$ and $F[x_1, \ldots, x_n]$ are all UFDs, none of them a PID beyond one variable.
What a UFD keeps and what it loses. It keeps greatest common divisors — take the smallest exponent of each irreducible — and the whole apparatus of comparing exponents. It loses Bézout: a gcd need not be a combination, and in $\mathbb{Z}[x]$ the gcd of $2$ and $x$ is $1$ while $(2, x) \ne \mathbb{Z}[x]$.
Unique factorisation is a property some rings have, not a fact about numbers. It follows from a division algorithm, weakens through principal ideals, and fails outright in $\mathbb{Z}[\sqrt{-5}]$ — where irreducible and prime stop being two words for the same thing.
Another way: picture
Think of the irreducibles as a set of coordinate axes and each element as a point with whole-number coordinates: the exponent of each irreducible in its factorisation. Multiplication adds coordinates, divisibility is being below in every coordinate, and the gcd takes the minimum in each. That picture is exactly what unique factorisation supplies. In $\mathbb{Z}[\sqrt{-5}]$ there are no such coordinates, because $6$ sits in two places at once.
Another way: steps
To decide whether a ring is a UFD: 1. If it is Euclidean or a PID, it is — quote the chain. 2. If it is $R[x]$ with $R$ a UFD, it is — quote Gauss. 3. To show it is not, find an element with two genuinely different factorisations, or equivalently an irreducible element that is not prime. 4. The norm is the standard tool: rule out proper factors by showing no element has the required norm. 5. Check the two factorisations do not differ merely by units.
The failure, once more. The norm $a^{2} + 5b^{2}$ takes the values $0, 1, 4, 5, 6, 9, 14, \ldots$ and never $2$ or $3$. So $2$ (norm $4$), $3$ (norm $9$) and $1 \pm \sqrt{-5}$ (norm $6$) are all irreducible, because any proper factor would need norm $2$ or $3$. And
$$6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5}).$$
The units are $\pm 1$ only, so no unit matches $2$ with $1 + \sqrt{-5}$. Uniqueness genuinely fails.
Why this was a historical crisis. Several nineteenth-century proofs of Fermat's last theorem factored $x^{n} + y^{n}$ in $\mathbb{Z}[\zeta_n]$ and assumed unique factorisation there. Lamé announced a proof on that basis in 1847; Kummer had already found that the assumption is false for $n = 23$.
The repair. Kummer's response was to restore uniqueness at the level of ideals rather than elements. In $\mathbb{Z}[\sqrt{-5}]$ set
$$\mathfrak{p} = (2, 1 + \sqrt{-5}), \quad \mathfrak{q} = (3, 1 + \sqrt{-5}), \quad \overline{\mathfrak{q}} = (3, 1 - \sqrt{-5}).$$
Then $(2) = \mathfrak{p}^{2}$, $(3) = \mathfrak{q}\overline{\mathfrak{q}}$, $(1 + \sqrt{-5}) = \mathfrak{p}\mathfrak{q}$ and $(1 - \sqrt{-5}) = \mathfrak{p}\overline{\mathfrak{q}}$. Both factorisations of $6$ become the same product of ideals, $\mathfrak{p}^{2}\mathfrak{q}\overline{\mathfrak{q}}$, and uniqueness is restored.
That is the origin of the word ideal: an ideal number, invented to be the common factor that the ring's elements failed to supply. The measure of how badly a ring fails — the class number — counts the ideals up to principal ones, and equals $1$ exactly when the ring is a PID.
A moral for this course. When a familiar property fails, the useful question is not how do we avoid this ring but what object has the property instead. The answer here rebuilt algebraic number theory.
*Forgetting up to units and order.* $6 = 2 \cdot 3 = 3 \cdot 2 = (-2)(-3)$ is one factorisation. Uniqueness would be false as a literal statement.
Thinking existence is the hard half. It is not: existence follows from a chain condition. Uniqueness is where the content is.
Assuming a UFD has Bézout. It does not. $\gcd(2, x) = 1$ in $\mathbb{Z}[x]$ and $(2, x)$ is not the whole ring.
Assuming a UFD is a PID. $\mathbb{Z}[x]$ is the standing counterexample.
Taking $\mathbb{Z}[\sqrt{-5}]$ to be badly behaved in general. It is a perfectly good Noetherian domain, its ideals factor uniquely, and it is the ring of integers of a number field. What it lacks is unique factorisation of elements.
*Using irreducible and prime interchangeably while proving uniqueness.* That is the circularity to avoid: uniqueness is what makes them interchangeable, so a proof of uniqueness may not assume it.
In $\mathbb{Z}$, $\gcd(2^{3} \cdot 3 \cdot 5, \; 2 \cdot 3^{2} \cdot 7) = 2 \cdot 3 = 6$: take the smaller exponent of each prime.
Minimum of the exponents.
The same works in $\mathbb{Q}[x]$: the gcd of $(x-1)^{2}(x+1)$ and $(x-1)(x+2)$ is $x - 1$.
Same rule, different irreducibles.
Neither computation would be meaningful without uniqueness — with two factorisations available, the exponent of an irreducible is not defined.
What uniqueness is for.
Gauss's lemma: $\mathbb{Z}$ is a UFD, so $\mathbb{Z}[x]$ is one. Every polynomial factors into a content and primitive irreducibles, uniquely.
Unique factorisation holds.
But $(2, x)$ is not principal: a generator would divide $2$ and $x$, hence be a unit, yet the ideal excludes $1$.
Not a PID.
So the arrow PID to UFD does not reverse. The practical loss is Bézout, and it is a real one — much of the theory of $\mathbb{Z}[x]$ works by passing to $\mathbb{Q}[x]$, which is a PID, and coming back with Gauss's lemma.
The gap, and how it is bridged.
$N(3) = 9$, and no element has norm $3$, so $3$ is irreducible.
Irreducible, by the norm.
Now look for a product it divides. $(1 + \sqrt{-5})(1 - \sqrt{-5}) = 6 = 2 \times 3$, so $3$ divides that product.
A product to test.
If $3$ divided $1 \pm \sqrt{-5}$ then $N(3) = 9$ would divide $N(1 \pm \sqrt{-5}) = 6$, and it does not. So $3$ is irreducible and not prime — a second witness in the same ring.
In $\mathbb{Z}[\sqrt{-5}]$ the norm of $a + b\sqrt{-5}$ is $a^{2} + 5b^{2}$. Fill in the norm of each element listed.
| The value of $a$ | The value of $b$ | Its norm | |
|---|---|---|---|
| The element $1$ | 1 | 0 | |
| The element $2$ | 2 | 0 | |
| The element $3$ | 3 | 0 | |
| The element $1 + \sqrt{-5}$ | 1 | 1 | |
| The element $2 + \sqrt{-5}$ | 2 | 1 | |
| The element $\sqrt{-5}$ | 0 | 1 |
Where does $\mathbb{Z}[i]$ sit in the chain Euclidean, principal ideal domain, unique factorisation domain?
In $\mathbb{Z}[\sqrt{-5}]$, how many essentially different factorisations into irreducibles does $6$ have?
Answer:
Select every statement that is true of a unique factorisation domain.
This task has no paper form; do it on a device.
Put in order the steps of proving that a factorisation into irreducibles is unique.
Number the steps in order (write the number in the box):
Build the proof that $2$ is irreducible but not prime in $\mathbb{Z}[\sqrt{-5}]$.
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.
In $\mathbb{Z}[\sqrt{-5}]$ the norm of $a + b\sqrt{-5}$ is $a^{2} + 5b^{2}$. Fill in the norm of each element listed.
| The value of $a$ | The value of $b$ | Its norm | |
|---|---|---|---|
| The element $1$ | 1 | 0 | |
| The element $2$ | 2 | 0 | |
| The element $3$ | 3 | 0 | |
| The element $1 + \sqrt{-5}$ | 1 | 1 | |
| The element $2 + \sqrt{-5}$ | 2 | 1 | |
| The element $\sqrt{-5}$ | 0 | 1 |
You can decide whether a ring has unique factorisation and produce the counterexample when it does not. Say in your own words why uniqueness is the same condition as irreducible elements being prime. Next: the chain condition that supplies the existence half, in its own right.
9. Your turn: is $3$ prime in $\mathbb{Z}[\sqrt{-5}]$?, step 3