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Consider the polynomials $xw-yz\in A[x,y,z,w]$ and $x^n+y^n+z^n\in A[x,y,z]$, where $A$ is a commutative ring. I am curious to know what conditions on $A$ (factorial ring, algebraically closed field, characteristic 0, etc..) make these polynomials irreducible? I know that the first should be irreducible for $A=\mathbf{C}$ but don't know how this is proven.

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  • $\begingroup$ Consider $xw - yz$ as a first-order polynomial in $x$ with coefficients in $A[y,z,w]$. How might it factor? $\endgroup$ – hardmath Jul 18 '14 at 21:51
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    $\begingroup$ well I guess it would be $fg$ with $f$ degree $1$ in $x$ and $g$ with degree $0$ in $x$. but does $g$ have to be a unit? $\endgroup$ – user165145 Jul 18 '14 at 21:53
  • $\begingroup$ Certainly over a domain $A[y,z,w]$ that would be easy to argue. $\endgroup$ – hardmath Jul 18 '14 at 21:54
  • $\begingroup$ oh well you are right. but just to know: it is true over ANY commutative ring? this would surprise me. for the second one can we somehow use eisensteins criterion? does $y^n+z^n$ have prime factors not occuring twice? here the problem for me is just the generality in which it might hold. perhaps i should restrict myself to $\mathbf{C}$ at first? $\endgroup$ – user165145 Jul 18 '14 at 21:55
  • $\begingroup$ Yes it is true in any commutative ring. It's not too hard to argue by contradiction. Factor it as either (ax+by)(xw+dz) or (ax+bz)(cw+dy), then expand and start making deductions. $\endgroup$ – blue Jul 18 '14 at 21:58
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Let's assume that $A$ is a domain (otherwise irreducibility is a weird notion).

Eisenstein's criterion states in a more general form: If $D$ is a domain, $\mathfrak{p}$ is a prime ideal of $D$, and $f=\sum_{i=0}^{n} a_i t^i \in D[t]$ is a polynomial such that $a_0,\dotsc,a_{n-1} \in \mathfrak{p}$, $a_n \notin \mathfrak{p}$, $a_0 \notin \mathfrak{p}^2$, then $f$ is not a product of two non-constant polynomials. Hence, if $f$ is also primitive (i.e. $\mathrm{gcd}(a_0,\dotsc,a_n)=1$), then $f$ is irreducible in $D[t]$.

Apply this to $D=A[x,y,z]$ and $f = xw - yz \in D[w]$, i.e. $n=1$ and $a_1=w$, $a_0 = -yz$. Take the prime ideal $\mathfrak{p}=(y)$ of $D$. Then Eisenstein's criterion applies and shows that $f$ is irreducible.

Notice that $x^n+y^n+z^n$ is reducible when $n \geq 1$ has a prime factor $p$ such that $pA=0$, because then $n=pm$ for some $m$ and $x^n+y^n+z^n= (x^m+y^m+z^m)^p$. However, if $K$ is a field whose characteristic $p \neq 2$ doesn't divide $n$ (in particular when $p=0$), then $x^n+y^n+z^n \in K[x,y,z]$ is irreducible: Since this polynomial has no factors in $\overline{K} \setminus K$, it suffices to prove that $x^n+y^n+z^n \in \overline{K}[x,y,z]$ is irreducible. In $\overline{K}$ we have a primitive $2n$th root of unity $\zeta$ (since $p \nmid 2n$). Then we may write $y^n+z^n = y^n-(\zeta z)^n$, which has a simple factor $y-\zeta z$. This is a prime element in $\overline{K}[y,z]$. Hence, Eisenstein's criterion shows that $x^n+(y^n+z^n) \in \overline{K}[y,z][x]$ is irreducible.

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  • $\begingroup$ PS: I am pretty sure that the last paragraph can be generalized to (certain) integral domains. $\endgroup$ – Martin Brandenburg Jul 18 '14 at 23:02
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    $\begingroup$ To use the Eisenstein's criterion for proving that a degree one polynomial is irreducible is definitely an overkill! $\endgroup$ – user26857 Jul 19 '14 at 7:43
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    $\begingroup$ A more general case of the second polynomial, that is, with different exponents, is already solved here. (This site tends to become repetitive.) $\endgroup$ – user26857 Jul 19 '14 at 7:49
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    $\begingroup$ @user26857: Sure. But what about $x^n w^m - y^m z^n$? It is better to learn methods which apply in other situations, too. $\endgroup$ – Martin Brandenburg Jul 19 '14 at 9:06
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    $\begingroup$ Your example is confusing. I can't see why and when it is irreducible (thought I can see many cases when it's not). $\endgroup$ – user26857 Jul 19 '14 at 9:23

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