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Units, associates, irreducible and prime: the two words that the integers keep in step, the ring where they part company, and the norm that decides both.
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 identify the units of a ring, decide when two elements are associates, state the definitions of irreducible and of prime and keep them apart, compute the norm in a quadratic ring and use it to test irreducibility, exhibit an element that is irreducible and not prime, and prove that every prime element of an integral domain is irreducible.
The first course built rings, integral domains and fields, and factored polynomials over a field. It relied throughout on unique factorisation in $\mathbb{Z}$ and in $F[x]$ without ever asking why either holds. This unit asks. The first step is to separate two words that the integers keep confusingly in step.
In an integral domain $R$: a unit divides $1$; $a$ and $b$ are associates when each divides the other, equivalently $a = ub$ for a unit $u$. A non-zero non-unit is irreducible when every factorisation of it has a unit factor, and prime when dividing a product forces it to divide a factor. In $\mathbb{Z}[\sqrt{d}]$ the norm is $N(a + b\sqrt{d}) = a^{2} - db^{2}$, and it is multiplicative.
Work inside an integral domain $R$. Write $a \mid b$ when $b = ac$ for some $c \in R$.
Units. $u$ is a unit when $uv = 1$ for some $v$. Units divide everything, so they are useless as factors and are excluded from the definitions below. The units of $\mathbb{Z}$ are $\pm 1$; of $\mathbb{Z}[i]$, the four elements $\pm 1, \pm i$; of $F[x]$, the non-zero constants; of a field, everything non-zero.
Associates. $a$ and $b$ are associates when $a \mid b$ and $b \mid a$, which in a domain means $a = ub$ for a unit $u$. Factorisations differing only by units are not counted as different — $6 = 2 \cdot 3 = (-2)(-3)$ is one factorisation, not two.
Irreducible. A non-zero non-unit $a$ is irreducible when
$$a = bc \ \Longrightarrow\ b \text{ or } c \text{ is a unit}.$$
There is no way to break it up except trivially.
Prime. A non-zero non-unit $p$ is prime when
$$p \mid bc \ \Longrightarrow\ p \mid b \ \text{ or } \ p \mid c.$$
That is Euclid's lemma promoted from theorem to definition.
The relation between them. In any integral domain, prime implies irreducible: if $p = bc$ then $p \mid bc$, so $p \mid b$ say, and cancelling gives that $c$ is a unit.
The converse is false in general. In $\mathbb{Z}[\sqrt{-5}]$ the element $2$ is irreducible — no element has norm $2$ — yet $2 \mid (1 + \sqrt{-5})(1 - \sqrt{-5}) = 6$ while dividing neither factor. In $\mathbb{Z}$, in $F[x]$ and in $\mathbb{Z}[i]$ the converse does hold, and that is precisely what unique factorisation amounts to.
The norm. In $\mathbb{Z}[\sqrt{d}]$, $N(a + b\sqrt{d}) = a^{2} - db^{2}$ satisfies $N(zw) = N(z)N(w)$. So a factorisation in the ring forces one in $\mathbb{Z}$, and three things follow immediately: units are the elements of norm $\pm 1$; prime norm implies irreducible; and searching for a factor means searching for an element of a specific, smaller norm — a finite job.
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
Irreducible is a statement about the element on its own: nothing can be cut off it. Prime is a statement about how it behaves towards everything else: it cannot be smuggled into a product in pieces. In the integers these coincide, and the coincidence is so familiar that the two ideas feel like one. In $\mathbb{Z}[\sqrt{-5}]$ they come apart, and suddenly it matters which one a proof was using.
Another way: steps
To decide whether an element of a quadratic ring is irreducible: 1. Compute its norm. 2. If the norm is a prime number, the element is irreducible. 3. Otherwise list the factorisations of the norm into two integers above $1$. 4. For each, look for an element of the ring with the smaller norm; there are finitely many candidates. 5. If one divides the element, it is reducible; if none does, it is irreducible. 6. To test primality instead, look for a product it divides without dividing a factor.
Almost every example in this unit lives in $\mathbb{Z}[\sqrt{d}] = \{a + b\sqrt{d} : a, b \in \mathbb{Z}\}$ for a squarefree integer $d$, so it is worth setting the machinery out once.
The norm. $N(a + b\sqrt{d}) = a^{2} - db^{2}$, which for negative $d$ is positive and equals the square of the complex absolute value. It is multiplicative, which is the only property used.
Units. $u$ is a unit exactly when $N(u) = \pm 1$. For $d < -1$ that leaves only $\pm 1$; for $d = -1$ it gives $\pm 1, \pm i$; for positive $d$ there are infinitely many, generated by a fundamental solution of Pell's equation — $\mathbb{Z}[\sqrt{2}]$ has $1 + \sqrt{2}$ and all its powers.
$\mathbb{Z}[i]$, completely described. Its irreducibles are: $1 + i$, of norm $2$; the two conjugate factors $a \pm bi$ of each rational prime $p \equiv 1 \pmod 4$, where $p = a^{2} + b^{2}$; and each rational prime $p \equiv 3 \pmod 4$, which stays irreducible because $a^{2} + b^{2} = p$ has no solution. Every irreducible here is also prime, and $\mathbb{Z}[i]$ has unique factorisation.
$\mathbb{Z}[\sqrt{-5}]$, where it fails. Norms are $a^{2} + 5b^{2}$, so the possible values are $0, 1, 4, 5, 6, 9, \ldots$ — and never $2$ or $3$. Therefore $2$ and $3$ are irreducible, since a proper factor would need norm $2$ or $3$. So are $1 \pm \sqrt{-5}$, of norm $6$, since a proper factor would need norm $2$ or $3$ again. Yet
$$6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5}),$$
and the units are only $\pm 1$, so the two factorisations are genuinely different. Unique factorisation fails, and with it the equivalence of irreducible and prime.
That single example is the reason the next three lessons exist: to say which rings behave and why.
Treating irreducible and prime as synonyms. They are equivalent in $\mathbb{Z}$, in $F[x]$ and in any PID, and not in general. Every proof should be clear about which it is using.
Forgetting that irreducibility depends on the ring. $5$ is irreducible in $\mathbb{Z}$ and reducible in $\mathbb{Z}[i]$; $x^{2} + 1$ is irreducible over $\mathbb{R}$ and not over $\mathbb{C}$.
Counting factorisations that differ by units as different. $6 = 2 \cdot 3 = (-2)(-3)$ is one factorisation. Uniqueness is always up to units and order.
Calling a unit irreducible. Units are excluded by definition; otherwise $1 = 1 \cdot 1$ would make nonsense of everything.
Using the norm without checking multiplicativity applies. $N(a + b\sqrt{d}) = a^{2} - db^{2}$ is multiplicative; $|a| + |b|$ is not, and no conclusion follows from it.
Assuming a smaller norm means a factor. An element of norm $2$ would be needed to split something of norm $4$; that no such element exists is what proves irreducibility, and its existence would still require checking that it divides.
In $\mathbb{Z}$ the only factorisations are $5 = 1 \cdot 5 = (-1)(-5)$, each with a unit factor.
Irreducible there.
In $\mathbb{Z}[i]$ the norm is $25$, which splits as $5 \times 5$ — so a proper factor would have norm $5$, and $2 + i$ has norm $5$.
A candidate exists.
And indeed $(2 + i)(2 - i) = 4 - i^{2} = 5$. Neither factor is a unit, so $5$ is reducible in $\mathbb{Z}[i]$ — the same number, a different ring, a different answer.
Reducible there.
$N(2) = 4$, so a proper factor would have norm $2$; but $a^{2} + 5b^{2} = 2$ has no integer solution, since $b = 0$ leaves $a^{2} = 2$.
Irreducible.
Now $(1 + \sqrt{-5})(1 - \sqrt{-5}) = 1 + 5 = 6$, which is $2 \times 3$, so $2$ divides the product.
It divides a product.
But $2 \mid 1 \pm \sqrt{-5}$ would force $N(2) = 4$ to divide $N(1 \pm \sqrt{-5}) = 6$, and it does not. So $2$ divides neither factor: irreducible, and not prime.
The two words come apart.
Its norm is $N(3) = 9$, so a proper factor would have to have norm $3$.
Compute the norm.
Solve $a^{2} + b^{2} = 3$ over the integers: $a^{2}$ can only be $0$ or $1$, and neither leaves a square.
No element of norm $3$.
So no proper factor exists and $3$ is irreducible in $\mathbb{Z}[i]$ — which fits the general rule, since $3$ is $3$ modulo $4$. The same test on $13$ finds $N(3 + 2i) = 13$, and $13 = (3+2i)(3-2i)$ is reducible.
Match each word about divisibility in an integral domain to the condition that defines it.
| An element that divides $1$ | Elements that divide each other | A non-unit whose every factorisation has a unit factor | A non-unit that divides a product only by dividing a factor | An element with no divisors at all | |
|---|---|---|---|---|---|
| A unit | |||||
| Two associates | |||||
| An irreducible element | |||||
| A prime element |
What is the norm $N(a + bi) = a^{2} + b^{2}$ of $2 + i$ in $\mathbb{Z}[i]$?
Answer:
Is $2 + i$ irreducible in $\mathbb{Z}[i]$?
Select every statement that is true in an integral domain.
This task has no paper form; do it on a device.
Put in order the steps of deciding whether an element of a quadratic ring is irreducible, using its norm.
Number the steps in order (write the number in the box):
Build the proof that in an integral domain every prime element is irreducible.
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.
Is $3 + 2i$ irreducible in $\mathbb{Z}[i]$?
You can test an element of a quadratic ring for irreducibility with its norm, and say which of irreducible and prime a given argument needs. Say in your own words why the two words agree in the integers and not in $\mathbb{Z}[\sqrt{-5}]$. Next: the rings where a division algorithm makes them agree.
9. Your turn: is $3$ irreducible in $\mathbb{Z}[i]$?, step 3