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$Φ_{p^{r}} (t) = \sum_{k < p} t^{k p^{r - 1}}$ , and $Φ_{p^{r}} (t + 1)$ is Eisenstein at $p$

Statement

Let $p$ be a prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ) and $r \geq 1$ . Then

$Φ_{p^{r}} (t) = \sum_{k = 0}^{p - 1} t^{k p^{r - 1}} = \frac{t^{p^{r}} - 1}{t^{p^{r - 1}} - 1},$

the polynomial $Φ_{p^{r}} (t + 1) \in Z [t]$ satisfies the Eisenstein criterion at $p$ (Eisenstein criterion over the integers), and consequently $Φ_{p^{r}}$ is irreducible in $Q [t]$ (Irreducible and prime elements of an integral domain).

The sum starts at $k = 0$ : its first term is the constant $1$ , and evaluation at $t = 1$ (Evaluation and roots of a polynomial in a commutative target ring) gives $Φ_{p^{r}} (1) = p$ .

Facts & Assumptions

Given: A prime $p$ and an integer $r \geq 1$ ; the cyclotomic polynomials of The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ and the substitution homomorphisms of Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism, under which $t \mapsto t + 1$ is a ring automorphism of $Z [t]$ and of $Q [t]$ with inverse $t \mapsto t - 1$ .

[L1]

For every $n \geq 1$ , $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ with each $Φ_{d}$ monic in $Z [t]$ of degree $φ (d)$ (The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$ , Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L2]

Let $f = a_{n} x^{n} + \dots + a_{0} \in Z [x]$ be primitive with $n \geq 1$ . If a prime $p$ satisfies $p ∤ a_{n}$ , $p ∣ a_{i}$ for every $i < n$ , and $p^{2} ∤ a_{0}$ , then $f$ is irreducible in $Q [x]$ (Eisenstein criterion over the integers).

[L3]

A nonzero integer polynomial is primitive exactly when no prime divides all of its coefficients (Content is the positive common divisor of the coefficients divisible by every common divisor, Content and primitive integer polynomials).

[L4]

For a fixed integer $k \geq 1$ , in a field $K$ of characteristic $p > 0$ one has $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$ in $K [t]$ (In characteristic $p$ the only $p^{k}$ -th root of unity is $1$ , and $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$ ); $Z / p$ is such a field (For every prime $p$ , the two operations on $Z / p$ make it a field, The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise).

[L5]

$φ (p^{k}) = p^{k} - p^{k - 1}$ for every prime $p$ and $k \geq 1$ (For a prime $p$ and $k \geq 1$ , $φ (p^{k}) = p^{k} - p^{k - 1}$ ).

[L6]

Every integer greater than $1$ has a prime divisor (Every integer $n > 1$ has a prime divisor; indeed the least divisor of $n$ that exceeds $1$ is prime); and if a prime divides a finite product of integers it divides one of the factors (If a prime $p$ divides a finite product $\prod_{i < n} a_{i}$ of integers then $p ∣ a_{i}$ for some $i < n$ ; at $n = 0$ the product is $1$ and the hypothesis cannot hold).

[L7]

$R [x]$ is an integral domain when $R$ is (A polynomial ring over an integral domain is an integral domain).

Proof

technique · direct

1.1L6given

The positive divisors of $p^{j}$ are exactly $p^{0}, \dots, p^{j}$ : a positive $d$ dividing $p^{j}$ with $d > 1$ has a prime divisor $q$ by [L6], and $q ∣ p^{j}$ forces $q ∣ p$ by [L6], hence $q = p$ since $p$ is prime and $q > 1$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ , Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ); writing $d = p d^{'}$ gives $d^{'} ∣ p^{j - 1}$ by cancellation, and repeating reduces $d$ to a power of $p$ not exceeding $p^{j}$ .

2.1step 1.1L1

By [L1] at $n = p^{r}$ and at $n = p^{r - 1}$ , using step 1.1, $\prod_{j = 0}^{r} Φ_{p^{j}} = t^{p^{r}} - 1$ and $\prod_{j = 0}^{r - 1} Φ_{p^{j}} = t^{p^{r - 1}} - 1$ ; dividing, $(t^{p^{r - 1}} - 1) Φ_{p^{r}} = t^{p^{r}} - 1$ .

3.1step 2.1L7algebra

With $u : = t^{p^{r - 1}}$ the elementary identity $(u - 1) \sum_{k = 0}^{p - 1} u^{k} = u^{p} - 1$ gives $(t^{p^{r - 1}} - 1) \sum_{k = 0}^{p - 1} t^{k p^{r - 1}} = t^{p^{r}} - 1$ ; comparing with step 2.1 and cancelling the nonzero factor $t^{p^{r - 1}} - 1$ in the integral domain $Z [t]$ ([L7]) yields $Φ_{p^{r}} = \sum_{k = 0}^{p - 1} t^{k p^{r - 1}}$ .

3.2step 2.1L4L7

Reducing step 2.1 modulo $p$ and applying [L4] twice in $(Z / p) [t]$ gives $\overline{Φ_{p^{r}}} (t - 1)^{p^{r - 1}} = (t - 1)^{p^{r}}$ , and cancelling in the integral domain $(Z / p) [t]$ ([L7]) gives $\overline{Φ_{p^{r}}} = (t - 1)^{p^{r} - p^{r - 1}}$ .

4.1step 3.1algebra

Evaluating step 3.1 at $t = 1$ gives $Φ_{p^{r}} (1) = p$ , since the sum has $p$ terms each equal to $1$ ; so the constant term of $Φ_{p^{r}} (t + 1)$ is $p$ , which is divisible by $p$ and not by $p^{2}$ .

4.2step 3.2L1L5given

Substituting $t + 1$ , which commutes with reduction modulo $p$ because both are ring homomorphisms fixing the coefficients appropriately, gives $\overline{Φ_{p^{r}} (t + 1)} = t^{p^{r} - p^{r - 1}} = t^{φ (p^{r})}$ by [L5]. So every coefficient of $Φ_{p^{r}} (t + 1)$ other than the leading one is divisible by $p$ , while the leading coefficient is $1$ because $Φ_{p^{r}} (t + 1)$ is monic of degree $φ (p^{r})$ by [L1] and the substitution being degree preserving.

5.1step 4.1step 4.2L2L3L5

$Φ_{p^{r}} (t + 1)$ is primitive by [L3], no prime dividing its leading coefficient $1$ , and its degree $φ (p^{r})$ is at least $1$ by [L5]; steps 4.1 and 4.2 supply the three Eisenstein conditions at $p$ , so [L2] makes it irreducible in $Q [t]$ .

6.1step 5.1given∎

The substitution $t \mapsto t - 1$ is a ring automorphism of $Q [t]$ carrying $Φ_{p^{r}} (t + 1)$ to $Φ_{p^{r}}$ ; a ring automorphism preserves units and factorisations, so it carries irreducible elements to irreducible elements, and $Φ_{p^{r}}$ is irreducible in $Q [t]$ .

Remarks

The term $k = 0$ is load bearing. Dropping it would change $Φ_{p^{r}} (1)$ from $p$ to $p - 1$ , and the Eisenstein constant-term condition would fail. The highest exponent would be unchanged, so the degree alone would not detect the wrong polynomial.
A self-contained route for prime powers. This gives irreducibility over $Q$ for $n$ a prime power without the general argument of $Φ_{n}$ is irreducible in $Q [t]$ for every $n \geq 1$ , and unlike that argument it exhibits an explicit polynomial to which a named criterion applies. The general theorem covers every $n$ and does not supersede this computation; the companion page works out the case $p = 7$ in $Φ_{7} (t + 1)$ is Eisenstein at seven ↗.

Depends on

Used by

Total ramification at a prime-power cyclotomic level Corollary
A cyclotomic different preview Example
Quadratic Gauss sum at p=5 Example
Φ₁ through Φ₁₂ computed from the divisor recursion Example
Φ₇(t+1) is Eisenstein at seven Example
Prime-power cyclotomic ring, discriminant support and p factor Lemma

Dependency tree · two levels

88 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

K. Conrad, Cyclotomic Extensions (expository blurb), Theorem 5.3 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, v5.10, Lemma 1.42 (standard reference, not scraped)

Φpr(t)=∑k<ptkpr−1Φpr​(t)=∑k<p​tkpr−1, and Φpr(t+1)Φpr​(t+1) is Eisenstein at pp