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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.

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see… Its the famous proof by Gauss. – Paramanand Singh Oct 20 '13 at 7:52
It is hard to see... May I carry that in this website as an answer? – NH.Jeong Oct 20 '13 at 7:58
Do you want me to copy from blog the entire proof and paste it here as an answer? – Paramanand Singh Oct 20 '13 at 8:02
Yeah... Because the post is hard to see.. Would you give me the permission? – NH.Jeong Oct 20 '13 at 8:06
I am posting that as an answer. Dont worry – Paramanand Singh Oct 20 '13 at 8:10
up vote 12 down vote accepted

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 modulo 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.

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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 $x^{10}+x^5+1$? – NH.Jeong Oct 20 '13 at 8:26
Yes $\Phi_{15}(x)$ must divide $x^{10} + x^{5} + 1$ – Paramanand Singh Oct 20 '13 at 8:28
I see... Thank you very much:) – NH.Jeong Oct 20 '13 at 8:32
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. – Paramanand Singh Oct 20 '13 at 8:41
@ParamanandSingh: If it is not too much trouble, I would like to hear what you think of my attempted version of a proof:… – String Jan 20 '14 at 14:08

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.

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Ok, I see my title is historically misleading in the following question: $$\quad$$… $$\quad$$ But I would like to ask what you make of my attempt to prove it differently... – String Jan 20 '14 at 14:07

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.

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Thank you for your advice and document:) – NH.Jeong Oct 20 '13 at 8:35
Onlt the special case of this result where $n$ is prime is due to Gauss (see the note at the end of my answer). – Marc van Leeuwen Jan 19 '15 at 5:09

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.

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What is the relationship between irreducibility and these "spans" containing no fractions? – Slade Oct 20 '13 at 9:37
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 $Q$. – wendy.krieger Oct 20 '13 at 9:50
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. – Slade Oct 20 '13 at 9:59
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. – wendy.krieger Oct 20 '13 at 10:29
I could do without the superior attitude, especially as I am mostly struggling with the grammar, not the mathematics. But thank you for updating with some details (though it is beginning to look like constructing the entire approach myself from scratch will be less effort than understanding what you have written there!). – Slade Oct 20 '13 at 10:50

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