Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: Literature-sourcedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
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Φ7(t+1) is Eisenstein at seven

Example

The translated seventh cyclotomic polynomial is

Φ7(t+1)=t6+7t5+21t4+35t3+35t2+21t+7.

Its leading coefficient is 1, every other coefficient is divisible by 7, and its constant term 7 is not divisible by 72. So it satisfies Eisenstein's criterion at the prime 7, and therefore Φ7 is irreducible over Q.

Facts & Assumptions

Given: The prime-power cyclotomic formula and the polynomial Φ7.

[L1]

For a prime p and r1, Φpr(t)=k=0p1tkpr1, and Φpr(t+1) is Eisenstein at p (Φpr(t)=k<ptkpr1, and Φpr(t+1) is Eisenstein at p).

[L2]

Eisenstein's criterion: if a prime p divides every non-leading coefficient of a polynomial in Z[t], does not divide the leading coefficient, and p2 does not divide the constant term, then the polynomial is irreducible over Q (Eisenstein criterion over the integers).

Verification

technique · direct
1.1

Applying [L1] at p=7 and r=1 gives Φ7(t)=1+t+t2+t3+t4+t5+t6.

L1
2.1

Therefore Φ7(t+1)=(t+1)71t=t6+7t5+21t4+35t3+35t2+21t+7, by the binomial theorem.

step 1.1algebra
3.1

In the polynomial of step 2.1 the leading coefficient is 1, the remaining coefficients 7,21,35,35,21,7 are all divisible by 7, and the constant term 7 is not divisible by 49; so [L2] applies at the prime 7.

step 2.1L2algebra
4.1

Hence Φ7(t+1) is irreducible over Q, and this is exactly the degree-one prime-power case of [L1].

step 3.1L1

Remarks

  • Why this example matters later. The explicit coefficients are what the counterexample page uses when it says the Eisenstein route already proves irreducibility for prime-power cyclotomic polynomials before the general Dedekind argument is built.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources