I studied the cyclotomic extension using Fraleigh's text.

To prove that Galois group of the $n$th cyclotomic extension has order $\phi(n)$( $\phi$ is the Euler's phi function.), the writer assumed, without proof, that $n$th cyclotomic polynomial $\Phi_n(x)$ is irreducible over $\mathbb{Q}$.

I know that for n=p, p is prime, $\Phi_n(x)$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion.

But I don't know how $\Phi_n(x)$ is irreducible over $\mathbb{Q}$ when n is not prime.

  • 3
    $\begingroup$ see paramanands.blogspot.com/2009/12/… Its the famous proof by Gauss. $\endgroup$
    – Paramanand Singh
    Commented Oct 20, 2013 at 7:52
  • $\begingroup$ It is hard to see... May I carry that in this website as an answer? $\endgroup$
    – NNNN
    Commented Oct 20, 2013 at 7:58
  • 1
    $\begingroup$ Do you want me to copy from blog the entire proof and paste it here as an answer? $\endgroup$
    – Paramanand Singh
    Commented Oct 20, 2013 at 8:02
  • $\begingroup$ Yeah... Because the post is hard to see.. Would you give me the permission? $\endgroup$
    – NNNN
    Commented Oct 20, 2013 at 8:06
  • 1
    $\begingroup$ I am posting that as an answer. Dont worry $\endgroup$
    – Paramanand Singh
    Commented Oct 20, 2013 at 8:10

8 Answers 8


The proof which follows is the one provided by Gauss and it uses modular arithmetic in very ingenious way. We will summarize the results needed as follows:

1) For a given prime $ p$, the numbers $ 0, 1, 2, \ldots, (p - 1)$ form a finite field under the operations of addition and multiplication modulo $ p$.

2) Since these numbers form a field, say $ F_{p}$, we can talk about polynomials $ f(z)$ whose coefficients are in $ F_{p}$. The set of all such polynomials, say $ F_{p}[z]$, has the unique factorization property i.e. any such polynomial can be factored as a product of irreducible polynomials in $ F_{p}[z]$ in a unique way apart from the order of the factors. The proof is same as that used for normal polynomials with rational coefficients.

3) If $ f(z)$ is a polynomial in $ F_{p}[z]$ then $ \{f(z)\}^{p} = f(z^{p})$. This is true for constant polynomials by Fermat's theorem which says that $ a^{p} \equiv a\,\,\text{mod} (p)$. For higher degree polynomials this is achieved by induction by writing $ f(z) = az^{n} + g(z)$ and using binomial theorem to raise both sides to power $ p$. In so doing we only need to note that the binomial coefficients involved are divisible by $ p$.

The proof of irreducibility of $ \Phi_{n}(z)$ is done in two stages:

Stage 1:

Let $ \zeta$ be a primitive $ n^{th}$ root of unity and let $ f(z)$ be its minimal polynomial i.e. $ f(z)$ is monic (leading coefficient $ 1$), has rational coefficients and irreducible and $ f(\zeta) = 0$. Since $ \zeta$ is also a root of $ z^{n} - 1 = 0$, it follows that $ f(z)$ divides $ (z^{n} - 1)$ and by Gauss Lemma $ f(z)$ has integer coefficients as well. We now establish the following:

If $ p$ is any prime which does not divide $ n$ then $ \zeta^{p}$ is a root of $ f(z) = 0$.

Proof: Since $ \zeta$ is also a root of $ \Phi_{n}(z) = 0$ it follows that $ f(z)$ divides $ \Phi_{n}(z)$. Thus we have $ \Phi_{n}(z) = f(z)g(z)$ where $ g(z)$ is also monic and has integer coefficients (by Gauss Lemma). Since $ p$ is coprime to $ n$ it follows that $ \zeta^{p}$ is also a primitive $ n^{th}$ root. And therefore $ \Phi_{n}(\zeta^{p}) = 0$.

Assuming that $ \zeta^{p}$ is not a root of $ f(z) = 0$ (otherwise there is nothing to prove), we see that it must be a root of $ g(z) = 0$. Therefore $ \zeta$ is a root of $ g(z^{p}) = 0$. Since $ f(z)$ is the minimal polynomial of $ \zeta$, it follows that $ f(z)$ divides $ g(z^{p})$ so that $ g(z^{p}) = f(z)h(z)$ where $ h(z)$ is monic with integer coefficients. Also since $ \Phi_{n}(z)$ is a factor of $ (z^n - 1)$ so that we have $ z^{n} - 1 = \Phi_{n}(z)d(z)$ where $ d(z)$ is again monic with integer coefficients. We thus have the following equations: $$z^{n} - 1 = f(z)g(z)d(z)\tag{1}$$ $$g(z^{p}) = f(z)h(z)\tag{2}$$ We now apply the modulo $ p$ operation to each of the equations above, i.e. we replace each coefficient in the polynomials involved with its remainder when it is divided by $ p$. The resulting polynomials are all in $ F_{p}[z]$ and we will use the same letters to denote them. So the above equations are now to be interpreted as relations between some polynomials in $ F_{p}[z]$. The equation $(2)$ can now be equivalently written as $$\{g(z)\}^{p} = f(z)h(z)\tag{3}$$ Let $ k(z)$ in $ F_{p}[z]$ be an irreducible factor of $ f(z)$. Then from the above equation $(3)$, $ k(z)$ divides $ \{g(z)\}^{p}$ and so divides $ g(z)$. Thus from equation $(1)$ the polynomial $ \{k(z)\}^{2}$ divides $ (z^{n} - 1)$. Thus $ (z^n - 1)$ has repeated factors and therefore $ (z^{n} - 1)$ and its derivative $ nz^{n - 1}$ must have a common factor. Since $ n$ is coprime to $ p$, therefore the derivative $ nz^{n - 1}$ is non-zero polynomial and it clearly does not have any common factor with $ (z^{n} - 1)$.

We have reached a contradiction and therefore the initial assumption that $ \zeta^{p}$ is not the root of $ f(z) = 0$ is wrong. The result is now proved.

Stage 2:

We have thus established that if $ f(z)$ is the minimal polynomial for any primitive $ n^{th}$ root of unity then for any prime $ p$ not dividing $ n$, $ \zeta^{p}$ (which is again a primitive $ n^{th}$ root) is also a root of $ f(z) = 0$. And since $ f(z)$ is irreducible and monic, it will act as minimal polynomial for the primitive root $ \zeta^{p}$.

The same logic can be applied repeatedly and we will get the result that $ f(z)$ is the minimal polynomial for $ \zeta^{p_{1}p_{2}\ldots p_{m}}$ where $ p_{1}, p_{2}, \ldots, p_{m}$ are any primes not dividing $ n$. It follows that $ \zeta^{k}$ where $ k$ is coprime to $ n$ is also a root of $ f(z)$. Thus all the primitive $ n^{th}$ roots of unity are roots of $ f(z) = 0$. Hence $ \Phi_{n}(z)$ divides $ f(z)$. Since $ f(z)$ is irreducible it follows that $ f(z) = \Phi_{n}(z)$ (both $ f(z)$ and $ \Phi_{n}(z)$ are monic). We thus have established that $ \Phi_{n}(z)$ is irreducible.

Update: User Vik78 points out in comments that Gauss had proved the irreducibility of $\Phi_{n}(z)$ for prime values of $n$ only and it was Dedekind who established it for non-prime values of $n$. Note that the problem is far simpler when $n$ is prime (one easy proof is via Eisenstein's criterion of irreducibility) and the crux of the argument presented above is essentially due to Gauss. I came to know of this proof from the wonderful book Galois Theory of Algebraic Equations by Jean Pierre Tignol.

Tignol mentions in his book that Gauss proved the irreducibility of $\Phi_{n}(z)$ over $\mathbb{Q}$ for non-prime values of $n$ in 1808 (see section 12.6 titled Irreducibility of cyclotomic polynomials on page 196 in the book mentioned in last paragraph). Dedekind on the other hand proved an even more powerful theorem:

Theorem: Let $\zeta_{m}$ denote a primitive $m^{\text{th}}$ root of unity. If $m, n$ are relatively prime then $\Phi_{n}(z)$ is irreducible over the field $\mathbb{Q}(\zeta_{m})$.

And Gauss used this result implicitly (thinking that it could be proved in similar manner as irreducibility of $\Phi_{n}(z)$ over $\mathbb{Q}$) in obtaining solution to $\Phi_{n}(z) = 0$ via radicals.

  • $\begingroup$ Thanks a lot!! By the way, I have a question. Obvously, $\Phi_n(x)$ divide $x^n -1$. However, $x^{15} -1$ can be factored as $(x^5-1)(x^{10}+x^5+1)$. But $\Phi_{15}(x)$ is a polynomial of degree 8. Then does the $\Phi_{15}(x)$ divide $x^{10}+x^5+1$? $\endgroup$
    – NNNN
    Commented Oct 20, 2013 at 8:26
  • $\begingroup$ Yes $\Phi_{15}(x)$ must divide $x^{10} + x^{5} + 1$ $\endgroup$
    – Paramanand Singh
    Commented Oct 20, 2013 at 8:28
  • $\begingroup$ I see... Thank you very much:) $\endgroup$
    – NNNN
    Commented Oct 20, 2013 at 8:32
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    $\begingroup$ In fact $\Phi_{15}(x) = x^{8} - x^{7} + x^{5} - x^{4} + x^{3} - x + 1$ and $x^{10} + x^{5} + 1 = (x^{2} + x + 1)\Phi_{15}(x)$ which can be checked by hand calculation. $\endgroup$
    – Paramanand Singh
    Commented Oct 20, 2013 at 8:41
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    $\begingroup$ Awesome proof! The theorem on the end is actually very easy to prove: just use that $\textbf{Q}(\zeta_n, \zeta_m)=\textbf{Q}(\zeta_{nm})$ and use degrees of field extensions (using the irreducibility of $\Phi_n$). $\endgroup$
    – Mar
    Commented Jan 26, 2018 at 20:58

I will try to rephrase the essence of answer by Paramanand Singh so as to somewhat better isolate the arguments used (though probably less faithful to what Gauss wrote*), for the sake of transparency.

The starting point is that $\def\Z{\Bbb Z}\Phi_n\in\Z[X]$ are inductively defined for $n>0$ by the recurrence relation $\prod_{k\mid n}\Phi_k=X^n-1$ (just like the numbers $\phi(n)$, which ar their degrees, are defined by $\sum_{k\mid n}\phi_k=n$); one can compute $\Phi_n$ in $\def\Q{\Bbb Q}\Q[X]$ by polynomial exact division starting from $X^n-1$, and since (by induction) all divisions are by monic polynomials with integer coefficients, $\Phi_n$ is monic and has integer coefficients.

If $\Phi_n$ were reducible over$~\Q$, this would partition the primitive $n$-th roots of unity (in $\def\C{\Bbb C}\C$) into more than one subset, each characterised as the roots of a different rational polynomial. To show that this is impossible, one argues that for any irreducible factor $F$ of $\Phi_n$ in $\Q[X]$, and for any prime$~p$ not dividing$~n$, the set of roots of $F$ is closed under the operation $\zeta\mapsto\zeta^p$. Since (by consideration of the cyclic group of $n$-th roots of unity) this operation maps primitive $n$-th roots to primitive $n$-th roots, and compositions of such operations for different primes$~p$ allow going from any one of the $\phi(n)$ primitive $n$-th roots to any other, this will show that $F$ has $\phi(n)$ distinct roots in$~\C$, and therefore equals $\Phi_n$.

Now fix such $F$ and $p$. The operation $\zeta\mapsto\zeta^p$ in $\C$ on roots gives rise to an operation on irreducible polynomials in $\Q[X]$: the minimal polynomial$~G$ over$~\Q$ of the image of $X^p$ in the field $\Q[X]/(F)$ is a monic irreducible polynomial in $\Q[X]$ determined by$~F$, and by construction $F$ divides $G[X^p]$, the result of substituting $X^p$ for $X$ in $G$. Whenever $\zeta\in\C$ is a root of$~F$ it is clear that $\zeta^p$ is a root of$~G$, and since $\deg G\leq[\Q[X]/(F):\Q]=\deg F$ one gets all roots of $G$ this way. One must show that $F=G$.

A key point is that $F$ and $G$, both of which divide $\Phi_n$ in $\Q[X]$ since their (distinct) roots in$~\C$ are among the roots of the latter, have integer coefficients. This is because of Gauss's lemma stating that the product of primitive polynomials in $\Z[X]$ (i.e., those not divsible by any prime number) is again primitive. (A detailed argument goes: $F\in\Q[X]$ is a monic divisor of the monic $\Phi_n\in\Z[X]$, so the quotient $Q\in\Q[X]$ with $FQ=\Phi_n$ is monic; then some positive integer multiples $aF,bQ\in\Z[X]$ are primitive (namely the minimal such multiples with all integer coefficients), and by the lemma $aFbQ=ab\Phi_n$ is primitive, but since $\Phi_n\in\Z[X]$ this forces $a=b=1$.)

So one can apply modular reduction $\def\Fp{\Bbb F_p}\Z[X]\to\Fp[X]$ to our polynomials, which I shall write $P\mapsto\overline P$. Being a morphism of rings, it preserves divisibility of polynomials. An important point is that $\overline{X^n-1}$ is still square-free over$~\Fp$ (it has no multiple roots in an extension field), since it is coprime with its derivative $\overline{nX^{n-1}}$ (as $\overline n\neq0$ in$~\Fp$ by hypothesis). Now if $F$ and $G$, which are both irreducible factors of$~X^n-1$, were distinct, then $\def\oF{\overline F}\oF$ and $\def\oG{\overline G}\oG$ would (also) be coprime in$~\Fp[X]$. We will show this to be false.

Since in any ring of characteristic$~p$ the map $\eta:x\mapsto x^p$ is a morphism of rings (called the Frobenius endomorphism), and $\eta$ fixes all elements of$~\Fp$, one has in $\Fp[X]$ that $\oG^p=\eta(\oG)=\oG[X^p]$, which is also clearly equal to $\overline{G[X^p]}$ and therefore divisible by $\oF$. This contradicts that $\oF$ and $\oG$ are coprime, and this contradiction finishes the proof.

*Googling around a bit I found evidence that the result is not due to Gauss at all for general $n$, but only for prime$~n$ (where $\Phi_n=1+X+\ldots,X^{n-1}$). See this note and this one. Apparently the general case was first proved by Dedekind, and the simplification leading to above proof is due to van der Waerden.

  • $\begingroup$ Ok, I see my title is historically misleading in the following question: $$\quad$$ math.stackexchange.com/questions/644899/… $$\quad$$ But I would like to ask what you make of my attempt to prove it differently... $\endgroup$
    – String
    Commented Jan 20, 2014 at 14:07
  • 1
    $\begingroup$ Interesting to dispel the myth that the result was due to Gauss, etc! $\endgroup$ Commented Apr 12, 2017 at 21:14
  • $\begingroup$ According to Tignol's book (see end of Paramanand Singh's answer), the result for general $n$ is due to Gauss in 1808 after all ? $\endgroup$
    – PatrickR
    Commented Feb 17, 2021 at 3:20

According to the Book Algebra of Serge Lang, the "fact that $\Phi_n$ is an irreducible polynomial of degree $\varphi(n)$ in the ring $\mathbb{Z}[x]$ is a nontrivial result due to Gauss". So there is no short answer to your question.

Anyway, just open your favorite book on abstract algebra, and find the proof there. (If not, maybe it's time to change your "favorite" book...)

Also, google brought up this document where several different proofs are given.

  • $\begingroup$ Thank you for your advice and document:) $\endgroup$
    – NNNN
    Commented Oct 20, 2013 at 8:35
  • $\begingroup$ Onlt the special case of this result where $n$ is prime is due to Gauss (see the note at the end of my answer). $\endgroup$ Commented Jan 19, 2015 at 5:09

There is also a non-elementary proof that perhaps is more explanatory than the "elementary" argument, using some (but not too much) algebraic number theory, and primes in arithmetic progressions. To prove that $\mathbb Q(\zeta_{p^kN})$ has the expected degree over $\mathbb Q$ (where $p$ does not divide $N$), it suffices to do an induction, namely, that $\mathbb Q(\zeta_{p^kN})$ has the expected degree over $\mathbb Q(\zeta_N)$. (Good so far!) For this, it suffices to find a prime $q$ so that $\mathbb Q_q$ already contains a primitive $N$th root of unity, so that $\mathbb Q_q(\zeta_N)$ just collapses to $\mathbb Q_q$, but at the same time so that $\mathbb Q_q(\zeta_{p^kN})=\mathbb Q_q(\zeta_{p^k})$ has the expected degree over $\mathbb Q_q(\zeta_N)=\mathbb Q_q$. By Dirichlet's theorem on primes in an arithmetic progress, there is a prime $q$ such that $q=1\mod N$, while $q$ is congruent to a primitive root mod $p^k$. Then the finite field $\mathbb F_q$ contains a primitive $N$th root of unity, which lifts to a primitive $N$th root of unity in $\mathbb Q_q$. Meanwhile, since the degree of an extension is at least the residue class field extension degree) $\mathbb Q_q(\zeta_{p^k})$ is of the expected (maximal possible) degree, namely, $(p-1)p^{k-1}$. (Unless I've maintained old typos or created new ones... Hopefully the verbal description makes things more robust...)

(EDIT: Thanks to @M.A.Sarkar for gentle prodding to make corrections above.)


Here is my version of the argument that does not use complex numbers (explicitly). It is based on the proofs presented by Paramanand Singh and by Marc van Leeuwen.

Let it be taken for granted that the cyclotomic polynomials $\Phi_n\in\mathbf{Z}[X]$ can be defined recursively using the formulae $$ \prod_{d:\,d|n}\Phi_d = X^n - 1 $$ (where $n$ and $d$ are assumed to be positive integers). It follows in particular that each $\Phi_n$ is monic.

Since the Euler's function $\phi$ can be recursively defined using the formulae $$ \sum_{d:\,d|n}\phi(d) = n, $$ it follows by induction on $n$ that the degree of $\Phi_n$ is $\phi(n)$.

If $\mathbf{k}$ is a field, two polynomials $P$ and $Q$ in $\mathbf{k}[X]$ shall be called coprime if and only if $(P) + (Q) = (1) = \mathbf{k}[X]$ or, equivalently,1 if and only if they have no common non-unit (i.e., in this case, non-constant) factors in $\mathbf{k}[X]$.

1 These two properties are not equivalent in general in an arbitrary commutative ring.

Lemma 1. $X^n - 1$ is square-free in $\mathbf{Q}[X]$, and also in $(\mathbf{Z}/(p))[X]$ for every prime $p$ that does not divide $n$.

Proof idea. The derivative of $X^n - 1$ is $nX^{n-1}$; it is coprime with $X^n - 1$ in $\mathbf{Q}[X]$, and also in $(\mathbf{Z}/(p))[X]$ for every prime $p$ that does not divide $n$. $\square$

Lemma 2. $\Phi_m$ and $\Phi_n$ are coprime in $\mathbf{Q}[X]$ for all positive integers $m$ and $n$ such that $m\ne n$.

Proof. Let $k$ be a positive integer divisible by both $m$ and $n$ (for example: $k = mn$). Then $\Phi_m\Phi_n$ divides $X^k - 1$ in $\mathbf{Q}[X]$. Since $X^k - 1$ is square-free in $\mathbf{Q}[X]$, $\Phi_m$ and $\Phi_n$ are coprime in $\mathbf{Q}[X]$. Q.E.D. $\square$

Lemma 2'. $\Phi_m$ and $X^n - 1$ are coprime in $\mathbf{Q}[X]$ for all positive integers $m$ and $n$ such that $m$ does not divide $n$.

Proof idea. This is an easy corollary of lemma 2. It can also be proved analogously to lemma 2. $\square$

Lemma 3. Let $F$ be a non-unit (i.e., in this case, non-constant) factor of $\Phi_n$ in $\mathbf{Q}[X]$. If $s$ and $t$ are non-negative integers such that $X^s = X^t$ in $\mathbf{Q}[X]/(F)$, then $s = t$ in $\mathbf{Z}/(n)$.

Proof. Let $s$ and $t$ be non-negative integers such that $X^s = X^t$ in $\mathbf{Q}[X]/(F)$. Without loss of generality, assume that $s\ge t$. Then $F$ divides $(X^{s-t} - 1)X^t$ in $\mathbf{Q}[X]$.

$F$ is coprime with $X^t$ in $\mathbf{Q}[X]$ since so is $X^n - 1$. Therefore, $F$ divides $X^{s-t} - 1$ in $\mathbf{Q}[X]$. Therefore, $X^{s-t} - 1$ and $\Phi_n$ are not coprime in $\mathbf{Q}[X]$. By lemma 2', this implies that $n$ divides $s - t$ (though the case $s - t = 0$ needs to be treated separately).

Thus, $s = t$ in $\mathbf{Z}/(n)$. Q.E.D. $\square$

The following lemma seems to be the central part of the argument.

Lemma 4. Suppose $FG = X^n - 1$ in $\mathbf{Z}[X]$. Let $p$ be a positive prime that does not divide $n$. Then $F$ has no common non-unit factors with $G(X^p)$ and divides $F(X^p)$ in $\mathbf{Z}[X]$.

Proof. Clearly, either $F$ and $G$ are monic, or $-F$ and $-G$ are monic. Without loss of generality, assume that $F$ and $G$ are monic.

First of all, observe that $FG = X^n - 1$ divides $F(X^p)G(X^p) = X^{pn} - 1$ in $\mathbf{Z}[X]$. Since $\mathbf{Z}[X]$ is a unique factorisation domain, in order to prove that $F$ divides $F(X^p)$ in $\mathbf{Z}[X]$, it shall be enough to show that $F$ has no common non-unit factors with $G(X^p)$ in $\mathbf{Z}[X]$.

Recall that the reduction of coefficients modulo $p$ defines an epimorphism $\mathbf{Z}[X]\to(\mathbf{Z}/(p))[X]$.

Recall that the application $(\mathbf{Z}/(p))[X]\to(\mathbf{Z}/(p))[X]$ that maps $P\mapsto P^p$ is the so-called Frobenius endomorphism of $(\mathbf{Z}/(p))[X]$, and that it fixes all constant polynomials. Therefore, for every $P\in\mathbf{Z}[X]$, $P(X^p) = P^p$ in $(\mathbf{Z}/(p))[X]$. In particular, $F$ divides $F(X^p) = F^p$ in $(\mathbf{Z}/(p))[X]$.

Since $p$ does not divide $n$, $X^n - 1$ is square-free in $(\mathbf{Z}/(p))[X]$ (by lemma 1). Therefore, $F$ is coprime with $G$ in $(\mathbf{Z}/(p))[X]$. It follows that for every polynomial $S\in(\mathbf{Z}/(p))[X]$, $F(S)$ is coprime with $G(S)$ in $(\mathbf{Z}/(p))[X]$, and in particular $F^p = F(X^p)$ is coprime with $G(X^p)$ in $(\mathbf{Z}/(p))[X]$. Hence, $F$ is coprime with $G(X^p)$ in $(\mathbf{Z}/(p))[X]$.

Let $D$ be an arbitrary common divisor of $F$ and $G(X^p)$ in $\mathbf{Z}[X]$. Since $F$ and $G(X^p)$ are both monic, either $D$ or $-D$ is monic. Since $F$ and $G(X^p)$ are coprime in $(\mathbf{Z}/(p))[X]$, $D$ is constant in $(\mathbf{Z}/(p))[X]$. Since $D$ or $-D$ is monic, it follows easily that $D =\pm1$ in $\mathbf{Z}[X]$.

Thus $F$ has no non-unit common divisors with $G(X^p)$ in $\mathbf{Z}[X]$, and hence $F$ divides $F(X^p)$ in $\mathbf{Z}[X]$. Q.E.D. $\square$

Lemma 5. Suppose $\mathbf{r}$ is a commutative ring, $P,S_1,S_2\in\mathbf{r}[X]$, and $P$ divides both $P(S_1)$ and $P(S_2)$ in $\mathbf{r}[X]$. Then $P$ divides $P(S_1(S_2))$ in $\mathbf{r}[X]$.

(In particular, if $P$ divides both $P(X^m)$ and $P(X^n)$, then $P$ divides $P(X^{mn})$.)

Proof. Let $P(S_1) = PQ_1$ and $P(S_2) = PQ_2$ in $\mathbf{r}[X]$. Then $P(S_1(S_2)) = (P(S_1))(S_2) = (PQ_1)(S_2) = P(S_2)Q_1(S_2) = PQ_2Q_1(S_2)$ in $\mathbf{r}[X]$. $\square$

Lemma 6. Let $F$ be a factor of $X^n - 1$ in $\mathbf{Q}[X]$ and $m$ be a positive integer coprime with $n$. Then $F$ divides $F(X^m)$ in $\mathbf{Q}[X]$.

Proof idea. Using Gauss's lemma about polynomial content, it can be easily shown that $F$ is a rational multiple of a monic polynomial with integer coefficients. Without loss of generality, assume that $F$ is monic with integer coefficients itself. Then $F$ divides $X^n - 1$ in $\mathbf{Z}[X]$.

Let $m = p_1p_2\dotsb p_k$, where $p_1,\dotsc,p_k$ are positive primes (not pairwise distinct in general). By lemma 4, $F$ is divides each of the polynomials $F(X^{p_1}),\dotsc,F(X^{p_k})$ in $\mathbf{Z}[X]$. Now lemma 5 can be used to show that $F$ divides $F(X^m)$ in $\mathbf{Z}[X]$. It follows that $F$ divides $F(X^m)$ in $\mathbf{Q}[X]$. Q.E.D. $\square$

Lemma 7. Let $F$ be a non-unit factor of $\Phi_n$ in $\mathbf{Q}[X]$. Then $F$ has at least $\phi(n)$ distinct roots in $\mathbf{Q}[X]/(F)$. ($F$ can be viewed as an element of $(\mathbf{Q}[X]/(F))[X]$.)

Proof idea. If $m$ is an integer coprime with $n$ and satisfying $0 < m\le n$, then, by lemma 6, $F$ divides $F(X^m)$ in $\mathbf{Q}[X]$, and, therefore, $X^m$ is a root of $F$ in $\mathbf{Q}[X]/(F)$. ($F$ is viewed here as an element of $(\mathbf{Q}[X]/(F))[X]$ and $X^m$ -- as an element of $\mathbf{Q}[X]/(F)$.)

If $m_1$ and $m_2$ are distinct positive integers $\le n$ coprime with $n$, then $X^{m_1}$ and $X^{m_2}$ are distinct elements of $\mathbf{Q}[X]/(F)$ by lemma 3.

There are $\phi(n)$ distinct positive integers $m$ that are coprime with $n$ and satisfy $0 < m\le n$. $\square$

Theorem. $\Phi_n$ is irreducible in $\mathbf{Q}[X]$.

Proof. Let $F$ be an irreducible factor of $\Phi_n$ in $\mathbf{Q}[X]$. Then $\mathbf{Q}[X]/(F)$ is a field extension of $\mathbf{Q}$.

By lemma 7, $F$ has at least $\phi(n)$ distinct roots in $\mathbf{Q}[X]/(F)$. Therefore, $\deg F\ge\phi(n) =\deg\Phi_n$ in $(\mathbf{Q}[X]/(F))[X]$, as well as in $\mathbf{Q}[X]$, and hence $F$ is a rational multiple of $\Phi_n$ in $\mathbf{Q}[X]$. Thus $\Phi_n$ is irreducible itself in $\mathbf{Q}[X]$. Q.E.D. $\square$


I think it's worthy to mention the proof in Washington's cyclotomic fields book, Proposition 2.4.

This proof is really easy to remember when the basic ramification theory is established, but there is also something nontrivial. That is, if $K$ is a number field, then $|d(K)|>1$ if $K\neq \mathbb{Q}$, where $d(K)$ is the discriminant of $K$. This is due to Minkowski.

Since we know the irreducibility for prime power cases. Then the irreducibility is equivalent to $K:=\mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\zeta_m)=\mathbb{Q}$ when $(n,m)=1$. By consider ramify condition, we know $K$ is unramify for every prime numbers, then by the above theorem, $K=\mathbb{Q}$.

  • 1
    $\begingroup$ Indeed, this is the approach that I prefer, though it’s hardly elementary. $\endgroup$
    – Lubin
    Commented Jan 24, 2020 at 23:21

In order to emphasize the fundamental motivations that guide it, I’ve laid the proof below out “backwards” relative to the usual lemma-theorem-corollary mode of exposition, so that its overarching structure is that of a series of reductions of the desideratum (i.e., the irreducibility of the cyclotomic polynomials) to increasingly simple auxiliary claims. Hopefully this makes it more legible, but who knows.

The argument itself is the byproduct of my attempts to make sense of the wonderful proof attributed to Landau—which deserves a post of its own here—as presented by Antoine Chambert-Loir on his blog. Although I haven’t encountered it (i.e., the argument below) anywhere else, it’s neither too complicated nor particularly deep, and as such I’m sure that it does indeed appear elsewhere.

Let $\text{n}$ be a positive natural, $\Phi_{\text{n}}\left(x\right)\in\mathbb{Z}\left[x\right]$ be the (monic) $\text{n}^{\text{th}}$ cyclotomic polynomial, $\zeta_{\text{n}}$ be your favorite primitive $\text{n}^{\text{th}}$ root of unity, and $\mathbb{Z}\left[\zeta_{\text{n}}\right]$ be the algebra that it generates over $\mathbb{Z}$. The latter has in particular underlying Abelian group free on the basis $$\left(\zeta_{\text{n}}^{\text{k}}\right)_{\text{k}\in\left\{0,\ \dots,\ d_{\zeta_{\text{n}}}-1\right\}}$$ with $1\leq d_{\zeta_{\text{n}}}\leq\varphi\left(n\right)$ the degree of the minimal polynomial of $\zeta_{\text{n}}$.

The expository convention is, as your question reflects, to prove the irreducibility of $\Phi_{\text{n}}$ as a step in the computation of $\text{Gal}\left(\mathbb{Q}\left(\zeta_{\text{n}}\right)/\mathbb{Q}\right)$; we will instead compute $\text{Gal}\left(\mathbb{Q}\left(\zeta_{\text{n}}\right)/\mathbb{Q}\right)$ as a step in the proof of the irreducibility of $\Phi_{\text{n}}$.

Specifically, consider the group $\text{Aut}\left(\mathbb{Z}\left[\zeta_{\text{n}}\right]\right)$ of ($\mathbb{Z}$-algebra) automorphisms of $\mathbb{Z}\left[\zeta_{\text{n}}\right]$. As every such automorphism descends (via the fraction field construction) to an element of $\text{Gal}\left(\mathbb{Q}\left(\zeta_{\text{n}}\right)/\mathbb{Q}\right)$, and as every element of $\text{Gal}\left(\mathbb{Q}\left(\zeta_{\text{n}}\right)/\mathbb{Q}\right)$ is determined by its action on $\zeta_{\text{n}}$, which it must send to $\zeta_{\text{n}}^{\alpha}$ for a unique residue $\alpha\in\left(\mathbb{Z}/\text{n}\right)^{\times}$, there results a sequence of injective group homomorphisms $$\text{Aut}\left(\mathbb{Z}\left[\zeta_{\text{n}}\right]\right) \overset{\iota_{\zeta_{\text{n}}}}{\longrightarrow} \text{Gal}\left(\mathbb{Q}\left(\zeta_{\text{n}}\right)/\mathbb{Q}\right) \overset{\chi_{\zeta_{\text{n}}}}{\longrightarrow} \left(\mathbb{Z}/\text{n}\right)^{\times}.$$ It suffices to show that $\chi_{\zeta_{\text{n}}}\circ\iota_{\zeta_{\text{n}}}$ is surjective, as then every primitive $\text{n}^{\text{th}}$ root of unity will be a conjugate of $\zeta_{\text{n}}$. (And by the same token, the injective homomorphisms in the sequence will collapse to isomorphisms, so that $\text{Gal}\left(\mathbb{Q}\left(\zeta_{\text{n}}\right)/\mathbb{Q}\right)\simeq \left(\mathbb{Z}/\text{n}\right)^{\times}$ will follow simultaneously.)

On the other hand, if for $\alpha\in\left(\mathbb{Z}/\text{n}\right)^{\times}$ there exists $\psi\in \text{Aut}\left(\mathbb{Z}\left[\zeta_{\text{n}}\right]\right)$ with $\chi_{\zeta_{\text{n}}}\circ\iota_{\zeta_{\text{n}}}\left(\psi\right)=\alpha$, then the underlying $\mathbb{Z}$-linear map of $\psi$ must be (the unique—using that in every case $\left(\zeta_{\text{n}}^{\text{k}}\right)^{\text{n}}$$\mathbb{Z}$-linear map determined by) $$\left(\psi\colon \zeta_{\text{n}}^{\text{k}}\mapsto \left(\zeta_{\text{n}}^{\text{k}}\right)^{\alpha}\right)_{\text{k}\in\left\{0,\ \dots, d_{\zeta_{\text{n}}}-1\right\}}.$$ Call, given such $\alpha$, this $\mathbb{Z}$-linear map $\psi_{\alpha}$. Our goal amounts to showing that for all $\alpha\in\left(\mathbb{Z}/\text{n}\right)^{\times}$, $\psi_{\alpha}$ is the underlying $\mathbb{Z}$-linear map of a $\mathbb{Z}$-algebra automorphism of $\mathbb{Z}\left[\zeta_{\text{n}}\right]$.

Actually, it suffices to merely show that this is true for $\alpha$ in some (multiplicatively) generating subset of $\left(\mathbb{Z}/\text{n}\right)^{\times}$, as the homomorphic image of $\chi_{\zeta_{\text{n}}}\circ\iota_{\zeta_{\text{n}}}$ will then generate its codomain, so will equal it. Moreover, it suffices merely to show that a generating subset of such $\psi_{\alpha}$ are endomorphisms; indeed, if $\psi_{\alpha}$ is an endomorphism then $\psi_{\alpha}^{\varphi\left(\text{n}\right)}$ is the identity of $\mathbb{Z}\left[\zeta_{\text{n}}\right]$ (as $\psi_{\alpha}^{\varphi\left(\text{n}\right)}\left(\zeta_{\text{n}}\right)=\zeta_{\text{n}}^{\alpha^{\varphi\left(\text{n}\right)}}=\zeta_{\text{n}}$), so $\psi_{\alpha}$ is an automorphism.

To that end, $\psi_{\alpha}$ clearly preserves the multiplicative identity of $\mathbb{Z}\left[\zeta_{\text{n}}\right]$, so we need only check that $\psi_{\alpha}$ commutes with binary multiplication, i.e., that if we denote the ($\mathbb{Z}$-(bi)linear) multiplication map of $\mathbb{Z}\left[\zeta_{\text{n}}\right]$ as $$\mu_{\zeta_{\text{n}}}\ \colon\ \mathbb{Z}\left[\zeta_{\text{n}}\right]\otimes_{\mathbb{Z}}\mathbb{Z}\left[\zeta_{\text{n}}\right]\to \mathbb{Z}\left[\zeta_{\text{n}}\right],$$ then the square $$\require{AMScd} \begin{CD} \mathbb{Z}\left[\zeta_{\text{n}}\right]\otimes_{\mathbb{Z}}\mathbb{Z}\left[\zeta_{\text{n}}\right] @>\psi_{\alpha}\otimes_{\mathbb{Z}}\psi_{\alpha}>> \mathbb{Z}\left[\zeta_{\text{n}}\right]\otimes_{\mathbb{Z}}\mathbb{Z}\left[\zeta_{\text{n}}\right]\\ @V\mu_{\zeta_{\text{n}}}VV @VV\mu_{\zeta_{\text{n}}}V \\ \mathbb{Z}\left[\zeta_{\text{n}}\right] @>\psi_{\alpha}>> \mathbb{Z}\left[\zeta_{\text{n}}\right]; \end{CD}$$ commutes; equivalently, that the $\mathbb{Z}$-linear map $$\Delta_{\alpha}\ :=\ \psi_{\alpha}\circ\mu_{\zeta_{\text{n}}} - \mu_{\zeta_{\text{n}}}\circ\left(\psi_{\alpha}\otimes_{\mathbb{Z}}\psi_{\alpha}\right)\ \colon\ \mathbb{Z}\left[\zeta_{\text{n}}\right]\otimes_{\mathbb{Z}}\mathbb{Z}\left[\zeta_{\text{n}}\right]\to \mathbb{Z}\left[\zeta_{\text{n}}\right]$$ vanishes. (We might think of $\Delta_{\alpha}$ as the obstruction to $\psi_{\alpha}$’s endomorphicity.)

Consider the subset $X_{\text{n}}\subseteq \left(\mathbb{Z}/\text{n}\right)^{\times}$ entailing those residue classes represented by infinitely many primes. We claim that $X_{\text{n}}$ generates $\left(\mathbb{Z}/\text{n}\right)^{\times}$ under multiplication. At the very least, the set of primes not in $X_{\text{n}}$ is (almost tautologically) finite, so for any residue class $\alpha\in \left(\mathbb{Z}/\text{n}\right)^{\times}$, Sun Tzu’s theorem ensures that the compatible system of congruences $$\begin{cases}a=\alpha\pmod{\text{n}} \\ a\neq 0\pmod{p}\text{ if }p\text{ is a prime not in }X_{\text{n}}\end{cases}$$ has a positive natural solution $a$. As the prime divisors of such $a$ are necessarily all in $X_{\text{n}}$, $X_{\text{n}}$ indeed generates $\left(\mathbb{Z}/\text{n}\right)^{\times}$.

Now, for any $\alpha\in \left(\mathbb{Z}/\text{n}\right)^{\times}$ (or even in $\mathbb{Z}/\text{n}$) and prime $p=\alpha\pmod{\text{n}}$, $\psi_{\alpha}\otimes_{\mathbb{Z}} \mathbb{Z}/p$ manifestly agrees (as a $\mathbb{Z}/p$-linear map) with the Frobenius endomorphism of $\mathbb{Z}\left[\zeta_{\text{n}}\right]\otimes_{\mathbb{Z}} \mathbb{Z}/p$. As this is, well, an endomorphism (in fact, by the same argument as before, an automorphism), it follows that $$\Delta_{\alpha}\otimes_{\mathbb{Z}} \mathbb{Z}/p\ =\ 0.$$ In particular, for $\alpha\in X_{\text{n}}$, $\Delta_{\alpha}$ vanishes modulo infinitely many primes. But $\Delta_{\alpha}$ can be expressed as some $d_{\zeta_{\text{n}}}\times d_{\zeta_{\text{n}}}^{2}$ matrix of integers, so for $\alpha\in X_{\text{n}}$, $\Delta_{\alpha}$ itself vanishes.

By the above discussion, this establishes the result. $\blacksquare$


The simple proof, is to show that a span $\mathbb{S}$ of a finite set closed to multiplication, when intersected with $\mathbb{Q}$ gives $\mathbb{Z}$. Since cyclotomic numbers are of this form, one can show that the span of such a set contains no fractions.

The proof is based on an attempt to construct a set $S$ whose span includes a fraction. Since one can show that some $\frac 1{z^n}$ can not belong to $S$, then the intersection between $S$ and $Q$ is the same as between $S$ and $Z$.

Since the cyclotomic numbers form a span closed to multiplication, and the base set is finite, the intersection of any cyclotomic set and the rationals is the same as that of the integers.

Since also, one can show that the chords of a polygon $\{p\}$, which can be constructed from the cyclotomic numbers, are governed by this rule, and since this contains the chords of every rational angle, ditto.

If one takes the product of $a-\operatorname{cis}(x/n)$ for x=1 to n, it gives an algebraic equation of the form $a^n-1$, which has a unique factor for every $m \mid n$. If the GCD of $x, n$, is greater than 1, then the particular root occurs at a lesser n. It only gives the unique factor for $n$, if the GCD is 1. Since the Euler totient gives the number of co-primes to n between $0$ and $n$, this is the order of the equation.

The second stage of the proof is to show that the equation is irreducable. This is done by a process of automorphism: that the set $\operatorname{cis}(x/n)$ (where the gcd of x, n is 1), is identical to the set $\operatorname{cis}(xy,n)$. This particular process has the effect that if a function $F(x) = F(xy)$, where F is some element in the span of $\operatorname{cis}(x,n)$, for all y, then F(x) must belong in the set $Z$.

The span is then a reduction of a sparse array in $\phi(n)$D onto $2$D, which means that the equations are irreducable.

  • $\begingroup$ What is the relationship between irreducibility and these "spans" containing no fractions? $\endgroup$ Commented Oct 20, 2013 at 9:37
  • $\begingroup$ The implication is explained in the answer: there is no intersection between any number constructed over the span (ie $\sum z_i x_i$, where $z \element \mathbb Z$), and $Q$. $\endgroup$ Commented Oct 20, 2013 at 9:50
  • 4
    $\begingroup$ I'm not so sure that you can say it is "explained"—after all, you haven't mentioned polynomials or irreducibility. There is also some ambiguity in your language... one cannot "intersect" numbers, "contains no fractions" is not a transparent phrasing, "ditto" refers to something but I'm not certain of what... so there is really quite a lot that is implicit here. My best guess is that what you are implying with your last comment is that if you do not have all the primitive roots, you cannot get only integers with their symmetric polynomials, but I can hardly connect the dots into an argument. $\endgroup$ Commented Oct 20, 2013 at 9:59
  • $\begingroup$ This is because people use different terms to those i use of the cyclotomic numbers. One does not need to look at the history to find these things: back in the 1970s books of that nature were scarse and brains were cheap. So it's not impossible to implement cyclotomic numbers and hyperbolic geometry without recourse to any text to it. $\endgroup$ Commented Oct 20, 2013 at 10:29
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    $\begingroup$ Finite set of what? Span as as additive group span? This answer is a mess. $\endgroup$ Commented Dec 21, 2015 at 19:36

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