# If $X^n+Y^n+1$ is reducible, is the degree divisible by the characteristic?

I was playing around with the Frobenius map, and made a small observation.

Suppose $F$ is a field, and $F[X,Y]$ is the corresponding polynomial ring in two indeterminates. If $\text{char}(F)=p$ divides some integer $n$, then $n=pq$ for some $q$. Consider the polynomial $X^n+Y^n+1$. Then $$X^n+Y^n+1=X^{pq}+Y^{pq}+1^{pq}=(X^q+Y^q+1^q)^p$$ so $X^n+Y^n+1$ is reducible. I hope this observation is correct.

Does the converse also hold? That is, if $X^n+Y^n+1$ is reducible, can you conclude that $\text{char}(F)$ divides $n$? I tried finding some factorization, and I think it should look something like $(X^r+\text{stuff}+1)$ and $(X^s+\text{stuff}+1)$ where $r+s=n$, but couldn't actually find an explicit factorization.

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I will show that if the characteristic, say $p$, of the field does not divide $n$ then $X^n + Y^n + 1$ is irreducible in $F[X,Y]$. In $F[Y]$ the polynomial $Y^n + 1$ is separable and of course not constant. Let $\pi(Y)$ be an irreducible factor of $Y^n+1$ in $F[Y]$. Then in the ring $F[X,Y] = F[Y][X]$, we consider the polynomial $X^n + Y^n + 1$ as a polynomial in $X$ with coefficients in $F[Y]$ and see that it is Eisenstein with respect to $\pi(Y)$. Therefore by the Eisenstein irreducibility criterion (applied to the PID $F[Y]$, not ${\mathbf Z}$) our polynomial in irreducible in $F[Y][X] = F[X,Y]$.

I like to use this kind of example in some courses for $X^n + Y^n - 1$ instead, since then $Y-1$ is a visible linear factor of $Y^n - 1$ and one can say directly (without the abstract choice of $\pi(Y)$) that $X^n + Y^n - 1$ is Eisenstein with respect to $Y-1$ when $p$ doesn't divide $n$ so it is irreducible in $F[Y][X] = F[X,Y]$. This makes (some) students get a new appreciation for the scope of the Eisenstein irreducibility criterion.

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Dear Keith, Very nice! Best wishes, –  Matt E Nov 17 '11 at 4:39
Waldott: that is correct. And the same argument goes through with $X^n + Y^n + c$ for any nonzero $c$ in $F$. I happen to like $c = -1$ instead of $c = 1$, for reasons indicated in the answer. –  KCd Nov 17 '11 at 5:48
We can push the argument into three variables, with a surprise twist. To show $X^n + Y^n + Z^n + 1$ is irreducible in $F[X,Y,Z]$ when $p$ doesn't divide $n$, view the ring as $F[Y,Z][X]$. Since $F[Y,Z]$ is a UFD and $Y^n + Z^n + 1$ is irreducible by the case of two variables, $X^n + Y^n + Z^n + 1$ is Eisenstein with respect to the prime $Y^n + Z^n + 1$ (no need to talk about an unknown irred. factor of this "constant term", as it is irred. itself). Similarly, $X_1^n + \cdots + X_k^n + 1$ is irred. in $F[X_1,\dots,X_k]$ for all $k \geq 2$ when $p$ doesn't divide $n$. –  KCd Nov 17 '11 at 5:53
Great!, that's a very interesting generalization, thanks for this. –  Waldott Nov 17 '11 at 5:55
Waldott: you made a simple typo in your comment, as $\pi(X)$ should of course be $\pi(Y)$. –  KCd Nov 17 '11 at 5:55

If $n$ is coprime to the characteristic (say $p$) then you can check that the projective curve in $\mathbb P^2$ cut out by $X^n + Y^n + Z^n = 0$ is smooth, and so in particular irreducible. Thus $X^n + Y^n + Z^n$ is irreducible, and hence so is $X^n + Y^n + 1$.

So for reducibility, you do need $p$ to divide $n$.

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Thank you Matt E, this is a nice quick argument. –  Waldott Nov 17 '11 at 5:16