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Proposition 7.13 of Neukirch's ANT states that for a primitive $p^{m}$-th root of unity $\zeta$ ($p$ prime) the extension $\mathbb{Q}_{p}(\zeta)/\mathbb{Q}_{p}$ is totally ramified of degree $(p-1)p^{m-1}$. In the proof he shows that the polynomial $\phi(X)=X^{(p-1)p^{m-1}}+X^{(p-2)p^{m-1}}+\cdots+1$ is the minimal polynomial of $\zeta$ over $\mathbb{Q}_p$ .

I can't see how to obtain the congruence $\phi(X)=(X^{p^m}-1)/(X^{p^{m-1}}-1)\equiv(X-1)^{p^{m-1}(p-1)}$ mod $p$ as claimed in the book. Here, I take it that mod $p$ means mod $p\mathbb{Z}$ and not mod $p\mathbb{Z}_{p}$ ?

Also, does Eisenstein's irreducibility criterion extend to polynomials in $\mathbb{Q}_{p}[X]$ ?

Any help would be very much appreciated.

user0
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Josh F.
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    The congruence is just a consequence of the so called Freshman's dream: $(a+b)^p=a^p+b^p\pmod p$ that holds in any commutative ring. And, yes, Eisenstein's criterion works in the p-adic domain as well. – Jyrki Lahtonen Jun 23 '13 at 18:43
  • I'm a bit confused because the above relation only holds if the ring is of characteristic $p$ and I thought $\mathbb{Z}_{p}$ is of characteristic 0 as it contains $\mathbb{Z}$? – Josh F. Jun 23 '13 at 19:09
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    You are right in that this characteristic zero. That's why I added $\pmod p$ there. In general this is meant to mean: modulo the principal ideal generated by $p$. So in your case it is modulo $p\mathbb{Z}_p$, not modulo $p\mathbb{Z}$. – Jyrki Lahtonen Jun 23 '13 at 19:11
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    Dear @user68418, Keep in mind that the congruence $(a+b)^p\equiv a^p+b^p\pmod{p}$ takes place in any commutative ring, but the actual equality $(a+b)^p=a^p+b^p$ (no $\pmod{p}$) holds in a ring of characteristic $p$. – Keenan Kidwell Jun 23 '13 at 19:14
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    One way of thinking about it is to observe that in the binomial theorem the terms ${p\choose i}a^ib^{p-i}$ are clearly in the ideal generated by $p$. BTW, this time the congruencee really is modulo the ideal $p\mathbb{Z}_p[X]$ of the ring $\mathbb{Z}_p[X]$. I was a bit hasty above. – Jyrki Lahtonen Jun 23 '13 at 19:14
  • @Jyrki Lahtonen many thanks for your replies – Josh F. Jun 23 '13 at 19:44
  • For the case $m=1$, you may use the identity $\sum_{j=k}^{n}\binom{j}{k}=\binom{n+1}{k+1}$ from this post to show that $f(X)=\phi(X+1)$ is Eisenstein at $p$. – 19021605 Nov 19 '25 at 10:13

1 Answers1

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In $\mathbf{F}_p[X]$ (you are looking at the polynomial $p\mathbf{Z}$, so as an element of $\mathbf{F}_p[X]$), because you're in characteristic $p$, $X^{p^m}-1=(X-1)^{p^m}$, and similarly $X^{p^{m-1}}-1=(X-1)^{p^{m-1}}$, so the quotient, modulo $p$, is $(X-1)^{p^m-p^{m-1}}=(X-1)^{p^{m-1}(p-1)}$. This is the congruence

$\phi(X)=(X^{p^m}-1)/(X^{p^{m-1}}-1)\equiv (X-1)^{p^m(p-1)}\pmod{p}$.

Eisenstein's criterion holds for any UFD $R$ and its field of fractions. In particular it holds for $\mathbf{Z}_p$ and $\mathbf{Q}_p$ (using the prime ideal $(p)$ of $\mathbf{Z}_p$).

EDIT: To answer the question posed in the coments, in $\mathbf{F}_p[X]$, $X^{p^m}-1=(X-1)^{p^m}$ and $X^{p^{m-1}}-1=(X-1)^{p^{m-1}}$. Since the latter clearly divides the former (they are both powers of the same polynomial and the former is a larger power) the quotient is an element of $\mathbf{F}_p[X]$, which can be checked directly to be the image (i.e. reduction) of $\phi(X)$ in $\mathbf{F}_p[X]$, simply because, in $\mathbf{Z}[X]$, $\phi(X)=(X^{p^m}-1)/(X^{p^{m-1}}-1)$ (this equality is written in the OP's question so I assume it is understood).

Keenan Kidwell
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  • I thought the field $\mathbb{Q}_{p}$ of $p$-adic numbers is of characteristic 0 since it contains $\mathbb{Q}$. – Josh F. Jun 23 '13 at 18:52
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    It is. Your polynomial $\phi(X)$ is in $\mathbf{Z}[X]$ so you can reduce its coefficients modulo $p$ to obtain an element of $\mathbf{F}_p[X]$. Since $\mathbf{Z}\subseteq\mathbf{Z}_p$, you can also regard the polynomial as an element of $\mathbf{Z}_p[X]$ and reduce its coefficients modulo $p\mathbf{Z}_p$, again to get an element of $\mathbf{F}_p[X]$ (since $\mathbf{Z}/p\mathbf{Z}\rightarrow\mathbf{Z}_p/p\mathbf{Z}_p$ is an isomorphism), which is the same element you got before. The equation $X^{p^m}-1=(X-1)^{p^m}$ takes place in $\mathbf{F}_p[X]$. – Keenan Kidwell Jun 23 '13 at 19:12
  • I am not sure how you can take the quotient of the two powers of $X-1$ in the ring $\mathbb{F}_{p}[X]$. Could you explain this? Thanks. – Josh F. Jun 23 '13 at 19:20
  • Another silly question- how do you know that $\mathbb{Z}_{p}$ is a UFD? – Josh F. Jun 23 '13 at 19:26
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    Dear @user68418, There is nothing special about $\mathbf{F}_p[X]$. If you have any commutative ring $R$ and any polynomial $f\in R[X]$, then $f^{p^{m-1}}$ divides $f^{p^m}$ with quotient $f$ in the sense that $f^{p^m}=f(f^{p^{m-1}})$. The fact that $\mathbf{Z}_p$ is a principal ideal domain (in particular a UFD) is proved in most algebraic number theory texts that describe the process of completion. You can also prove directly using the $p$-adic valuation that its only ideals are those of the form $(p^n)$ for $n\geq 0$, together with $(0)$. – Keenan Kidwell Jun 23 '13 at 19:27
  • Many thanks. So $\phi(X)=(X^{p^m}-1)/(X^{p^{m-1}}-1)$ is equivalent to $(X^{p^{m-1}}-1)\phi(X)=(X^{p^m}-1)$ in $\mathbb{Z}[X]$ (or $\mathbb{Z}_{p}[X]$) ? – Josh F. Jun 23 '13 at 19:41
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    That's right. When you have a commutative ring $R[X]$ and some polynomials $f,g\in R[X]$ such that $g$ is not a zero divisor in $R[X]$, and $g$ divides $f$ in $R[X]$, meaning there is some $q\in R[X]$ such that $f=qg$, then the notation $f/g$ means (by definition) $q\in R[X]$. This is okay to do because the assumption that $g$ is not a zero divisor means that the quotient $q$ is unique, so the notation $f/g$ is well-defined. If $R$ is an integral domain, then $R[X]$ is as well, and then you can view $f/g$ – Keenan Kidwell Jun 23 '13 at 19:46
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    (for any $f\in R[X]$ and $g\in R[X]\setminus{0}$) as an element of the field of fractions of $R[X]$, and if $g$ divides $f$ in $R[X]$, $f/g$ will be an element of $R[X]$, the unique one such that $f=(f/g)g$. – Keenan Kidwell Jun 23 '13 at 19:47
  • I see. So the unique $q$ exists by the division algorithm for polynomial rings? Also I take it that the equality $f=(f/g)g$ should be viewed as an equality in the quotient field of $R[X]$ (for $R$ an integral domain)? – Josh F. Jun 23 '13 at 20:08
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    The existence of $q$ is the definition of the statement "$g$ divides $f$," and its uniqueness when $g$ is a non-zero divisor is immediate, because if there is another such $q^\prime$, then $gq=f=gq^\prime$ Implies $g(q-q^\prime)=0$, so $q-q^\prime=0$ because $g$ is not a zero divisor. You don't need the division algorithm for any of this. The equality $f=(f/g)g$ is a matter of definition; $f/g$ is the element defined by this relation. When $g$ divides $f$, $f/g$ is an element of $R[X]$. When $R$ is a domain $f/g$ can also be viewed as the element $fg^{-1}$ of the fraction field of $R[X]$. – Keenan Kidwell Jun 23 '13 at 20:19
  • I see now- thanks very much. – Josh F. Jun 23 '13 at 20:28