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The primes of a polynomial ring: why having no root settles the question in degree two and three, why it stops in degree four, and what replaces it over the rationals.
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 decide whether a polynomial of low degree is irreducible over a named field, explain why the no-root test is valid only in degree two and three and give the standard counterexample in degree four, apply the rational root test and Eisenstein's criterion, count the monic irreducible quadratics over a finite field, and say why irreducibility depends on the field.
In the integers, every number factors into primes and a prime is one that cannot be split. $F[x]$ has the same structure, and the unsplittable polynomials are called irreducible. The last lesson tied roots to linear factors; this one asks when a polynomial has any factors, which is a larger question as soon as the degree reaches four.
A non-constant $f \in F[x]$ is irreducible over $F$ when it cannot be written as a product of two non-constant polynomials in $F[x]$; otherwise it is reducible. A polynomial is monic when its leading coefficient is $1$. Eisenstein's criterion is a sufficient condition for irreducibility over the rationals, using a prime that divides all but the leading coefficient.
$f$ is irreducible over $F$ when it is non-constant and cannot be written $f = gh$ with $g, h \in F[x]$ both non-constant. The field must be named: $x^{2} + 1$ is irreducible over $\mathbb{R}$ and reducible over $\mathbb{C}$, and $x^{2} - 2$ is irreducible over $\mathbb{Q}$ and reducible over $\mathbb{R}$.
Irreducible polynomials are the primes of $F[x]$: every non-constant polynomial factors into irreducibles, uniquely up to order and constant multiples, by the same argument that gives unique factorisation in $\mathbb{Z}$.
Degree one is always irreducible, since $1$ cannot be split into two positive parts.
Degrees two and three: no root means irreducible. A factorisation into non-constant factors has degrees adding to $2$ or $3$ with both positive, so one factor has degree $1$ — and a linear factor gives a root. Over a finite field the roots can be found by substituting every element, so irreducibility is decidable by hand in these degrees.
Degree four and above: the test fails. The degrees may split $2 + 2$. The standing counterexample is
$$x^{4} + 4 = (x^{2} - 2x + 2)(x^{2} + 2x + 2),$$
which has no rational root and is reducible over $\mathbb{Q}$.
Other tools, for the rationals. The rational root test limits candidate roots to $\pm$(divisor of the constant term) over $\pm$(divisor of the leading one). Eisenstein's criterion: if a prime $p$ divides every coefficient but the leading one, and $p^{2}$ does not divide the constant term, then the polynomial is irreducible over $\mathbb{Q}$. Reduction modulo $p$: if the reduction has the same degree and is irreducible over $\mathbb{Z}_p$, the original is irreducible over $\mathbb{Q}$ — the converse does not hold.
They are plentiful. Over $\mathbb{Z}_p$ there are $p(p-1)/2$ monic irreducible quadratics, and irreducibles exist in every degree. That is what makes the next lesson's construction always available.
Another way: picture
Think of the degree as a length of rod that a factorisation cuts into pieces, each at least one unit long. A rod of length $2$ or $3$ cannot be cut without producing a piece of length $1$, and a piece of length $1$ is a root. A rod of length $4$ can be cut into two pieces of length $2$, with no unit piece anywhere — and that is precisely the case the no-root test cannot see.
Another way: steps
To test a polynomial for irreducibility over a field: 1. Degree $1$: irreducible, stop. 2. Degree $2$ or $3$: look for a root. No root means irreducible. 3. Degree $4$ or more over a finite field: check roots, then check for quadratic factors as well. 4. Over the rationals: try the rational root test, then Eisenstein, then reduction modulo a small prime. 5. Always name the field; the answer depends on it.
Over a finite field, irreducibility can always be settled by search. Over $\mathbb{Q}$ there are infinitely many candidates, so the tools matter.
The rational root test. If $f = a_n x^{n} + \cdots + a_0$ has integer coefficients and $p/q$ is a rational root in lowest terms, then $p \mid a_0$ and $q \mid a_n$. For a monic polynomial the roots are therefore integers dividing the constant term — a short list to check.
Eisenstein's criterion. If a prime $p$ divides $a_0, \ldots, a_{n-1}$ but not $a_n$, and $p^{2} \nmid a_0$, then $f$ is irreducible over $\mathbb{Q}$. Example: $x^{4} + 6x^{2} + 3$ with $p = 3$ — it divides $6$ and $3$, not the leading $1$, and $9 \nmid 3$. Irreducible, with no root ever computed. The criterion is one-directional: failing it says nothing.
Reduction modulo $p$. If $f$ has integer coefficients and its reduction modulo $p$ has the same degree and is irreducible over $\mathbb{Z}_p$, then $f$ is irreducible over $\mathbb{Q}$. It is useful because the reduced question is a finite search. The converse fails badly: $x^{4} + 1$ is irreducible over $\mathbb{Q}$ and reducible modulo every prime, so no amount of reduction proves it irreducible.
A warning about the word reducible: over $\mathbb{Z}[x]$, factoring out a constant such as $2x + 4 = 2(x + 2)$ does not count as reducible in the sense used here, because constants are units in $\mathbb{Q}[x]$. The two rings give different answers and Gauss's lemma relates them; this course works in $F[x]$ for a field $F$ throughout, where constants are units and the ambiguity does not arise.
*Using no root in degree four or more.* $x^{4} + 4$ has no rational root and factors into two quadratics. The test is valid only in degrees two and three.
Omitting the field. Irreducibility is always over something. $x^{2} + 1$ is irreducible over the reals and reducible over the complex numbers.
Thinking failing Eisenstein means reducible. Eisenstein is sufficient and not necessary. Failing it says nothing at all.
Thinking reducible modulo some prime means reducible over the rationals. $x^{4} + 1$ is reducible modulo every prime and irreducible over $\mathbb{Q}$.
Calling a constant factor a factorisation. In $F[x]$ the non-zero constants are units, so $2(x + 2)$ is not a factorisation any more than $2 \times 3$ makes $6$ composite in the rationals.
Is $x^{3} + x + 1$ irreducible over $\mathbb{Z}_2$? Degree $3$, so the no-root test applies.
The right test for the degree.
Substitute: at $0$ the value is $1$; at $1$ it is $1 + 1 + 1 = 1$.
Two elements, both checked.
No root, so no linear factor, so irreducible. The quotient by it is a field with eight elements.
Which is the next lesson.
Is $x^{4} + 4$ irreducible over $\mathbb{Q}$? By the rational root test the candidates are $\pm 1, \pm 2, \pm 4$, and none is a root — every value is positive.
No rational root.
But degree $4$ allows a $2 + 2$ split, so the search must continue. Trying $(x^{2} + ax + b)(x^{2} - ax + c)$ and matching coefficients gives $a = 2$, $b = c = 2$.
Look for quadratic factors too.
So $x^{4} + 4 = (x^{2} - 2x + 2)(x^{2} + 2x + 2)$: reducible, with no root. This is the example to keep.
The test was not enough.
Degree $2$, so the no-root test applies, and there are only two elements to try.
A complete search.
At $0$: $1$. At $1$: $1 + 1 + 1 = 1$ modulo $2$.
Neither is zero.
No root, so irreducible — and it is the only irreducible monic quadratic over this field, since $p(p-1)/2 = 1$.
Over the integers modulo $3$, count the monic quadratics of each kind.
| How many | |
|---|---|
| Monic quadratics altogether | 9 |
| Reducible ones | |
| Irreducible ones |
Is $x^2 + 1$ over the integers modulo $2$ irreducible?
Match each polynomial over the rationals to its factorisation.
| $(x - 1)(x + 1)$ | $(x - 1)(x^{2} + x + 1)$ | $(x - 1)(x + 1)(x^{2} + 1)$ | $(x^{2} - 2x + 2)(x^{2} + 2x + 2)$ | irreducible over the rationals | |
|---|---|---|---|---|---|
| $x^{2} - 1$ | |||||
| $x^{3} - 1$ | |||||
| $x^{4} - 1$ | |||||
| $x^{4} + 4$ |
Select every statement about testing irreducibility that is true.
This task has no paper form; do it on a device.
Over the integers modulo $3$, how many monic irreducible quadratics are there?
Answer:
Build the proof that a polynomial of degree two or three over a field with no root 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.
Over the integers modulo $2$, how many monic irreducible quadratics are there?
Answer:
You can test a polynomial for irreducibility and say which test the degree allows. Say in your own words why a quartic with no root can still factor. Next: quotienting by an irreducible polynomial, which is how every finite field is built.
9. Your turn: is $x^{2} + x + 1$ irreducible over the integers modulo $2$?, step 3