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Element $X_1 X_2 \cdots X_n - 1$ is irreducible in $K[X_{1},\ldots,X_{n}]$ for $n\ge 1$, where $K$ is a field. For $n=2,3$ it is easy to see that the element is irreducible but for higher value of $n$ I was trying to use inductive argument but not getting any idea.

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Is $K$ a field? –  Alex Becker Feb 2 '13 at 13:16
    
Sorry forget to mention ,now I have corrected it. –  Shraddha Srivastava Feb 2 '13 at 13:18
    
@Shraddha: how did you prove it for $n=2$ and $3$? –  Weltschmerz Feb 2 '13 at 13:32
    
For $n=0$ the polynomial vanishes, so it's not irreducible ;). –  Martin Brandenburg Feb 2 '13 at 17:30

4 Answers 4

up vote 4 down vote accepted

We know localisation of integral domain is integral domain. So $K[X_{1},...X_{n-1}]$[$1/X_1\cdots X_{n-1}$] is an integral domain hence the ideal ($X_1\cdots X_n - 1$)`is prime and $K[X_{1},\ldots,X_{n}]$ is an UFD so ($X_1\cdots X_n - 1$) has to be irreducible .

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Of course only localizations at non-zero elements. And prime elements are always irreducible, we don't need UFD for that. Best proof by the way. –  Martin Brandenburg Feb 2 '13 at 14:14
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+1 for your proof best proof by far. However you should say: You are using the isomorphism $A_f \cong A[x]/(xf - 1)$ where $A_f$ means the localisation of $A$ at $\{1,f,f^2\dots\}$. –  user38268 Feb 2 '13 at 16:02
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Thanks Martin Brandenburg.. true UFD is not needed ,and BenjaLim for giving the explicit isomorphism . –  Sarjbak Feb 3 '13 at 8:34
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@Sarjbak: thanks for your nice solution. –  Shraddha Srivastava Feb 6 '13 at 3:43

Hint $\, $ It's a special case of: $\rm\:ax\!+\!b\in D[x]\:$ is irreducible for domain $\rm\:D\:$ and $\rm\:0\ne a\:$ and $\rm\,b\,$coprime. Indeed, if reducible then comparing degrees (using $\rm\,D\,$ domain) we infer that one factor would be a constant nonunit, so it would be a nonunit common divisor of $\rm\,a,b,\,$ contra hypothesis. This may fail if $\rm\,D\,$ is not a domain, e.g. $\rm\,(2x\!+\!1)(3x\!-\!1) = x\!-\!1 \in \Bbb Z/6[x].$

Yours is the special case $\rm\:D = K[X_1,\ldots,X_{n-1}],\ a = X_1\cdots X_{n-1},\ x = X_n,\ b = -1.$

Remark $\ $ This "primitivity" criterion for irreducibility generalizes to higher degree polynomials. Namely, if $\rm\,D\,$ is a domain with fraction field $\rm\,K\,$ then, for a nonconstant polynomial $\rm\,f\in D[x]$

$$\rm f\,\ is\ prime\ in\ D[x]\iff f\,\ is\ prime (= irreducible)\ in\ K[x]\ and\,\ f\,\ is\ superprimitive $$

$$\rm where\,\ f\,\ is\ {\bf superprimitive}\ in\ D[x]\,\ :=\,\ d\,|\,cf\, \Rightarrow\, d\,|\,c\,\ \ for\ all\,\ c,d\in D^*$$

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Suppose $f(X_1,\ldots,X_n) | X_1\cdots X_n-1$. Evaluating at $X_i=0$, we see that $f(X_1,\ldots,0,X_{i+1},\ldots,X_n) | -1$ and thus must be an element of $K$, which must be the constant term $f_0$. Thus $f(X_1,\ldots,X_n)-f_0$ is divisible by all $X_i$, so $f=g\cdot X_1\cdots X_n+f_0$ for some $g\in K[X_1,\ldots,X_n]$. Thus for some $h\in K[X_1,\ldots,X_n]$ we have $h\cdot (g\cdot X_1\cdots X_n + f_0)=X_1\ldots X_n-1$. If $g$ is nonzero (and otherwise $f\in K$), both $h$ and $g$ can have degree at most $0$ since otherwise the degree of the LHS would exceed that of the right, so we see that $h_0g_0=1$ and $h_0f_0=1$ so $$f=-f_0\cdot X_1\ldots X_n+f_0=-f_0(X_1\cdots X_n-1)$$ which is an associate of $X_1\cdots X_n-1$. Thus $X_1\cdots X_n-1$ is irreducible.

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Yet another proof!

The degree $1$ polynomial $-X_1 + X_2 \cdots X_n$ is irreducible over $k[X_2,\dotsc,X_n]$, and therefore the same holds for its reverse $X_1 X_2 \dotsc X_n - 1$.

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