TITLE: Is $x_1^d + x_2^d + x_3^d + x_4^d + x_5^d= 0$ a geometrically integral hypersurface in $\mathbb{P}^4$?
QUESTION [0 upvotes]: Let $d>2. $Let $X$ be a surface defined by $x_1^d + x_2^d + x_3^d + x_4^d + x_5^d= 0$ in $\mathbb{P}^4_{\mathbb{Q}}$? I am interested in finding out if this is a geometrically integral hypersurface or not. 
I was wondering if someone could tell me what are simple ways to verify this fact (or that is is not) if there are any? I would appreciate any reference as well. Thank you very much.

REPLY [2 votes]: Here's an elementary but somewhat cumbersome proof that $x_1^d + \ldots + x_n^d$ is irreducible for $n \geq 3$. (In characteristic $p > 0$ we also need that $p \nmid d$; e.g. note that $(x_1 + \ldots + x_n)^p = x_1^p + \ldots + x_n^p$.) We will prove this by induction, but first recall some facts.
Fact. (Kummer theory.) Let $K$ be a field containing a primitive $d$-th root of unity $\zeta$, with $d$ not divisible by $\operatorname{char} K$. Let $a \in K$, and let $\alpha \in \bar K$ be an element such that $\alpha^d = a$. Then the extension $K \subseteq K(\alpha)$ is Galois, and the following numbers are equal:

The degree of the field extension $K \subseteq K(\alpha)$;
the least natural number $m$ such that $\alpha^m \in K$;
the least natural number $n$ such that $a^n$ is a $d$-th power in $K$.

Remark. The same is false if we drop the assumption on the roots of unity in $K$. Indeed, consider $K = \mathbb Q$ and $d = 3$, with $a = 1$ and $\alpha = \zeta_3$; then the field extension $\mathbb Q \subseteq \mathbb Q(\zeta_3)$ has degree $2$. On the other hand, we get $m = 3$ and $n = 1$. In this case, using the notation $\sqrt[3]{1}$ is misleading to say the least; hence my use of $\alpha$ in stead of $\sqrt[d]{a}$.
There might be a small improvement of the fact above (minus the Galois statement) to the case where $d$ is possibly divisible by $\operatorname{char} K$, but we would run into trouble later anyway.
Fact. (Gauß's lemma.) Let $R$ be a UFD with fraction field $K$, and let $f \in R[x]$ be a polynomial. Then $f$ is irreducible in $R[x]$ if and only if $f$ is irreducible in $K[x]$ and $f$ is primitive, i.e. the coefficients of $f$ share no common factor.
Corollary. Let $R$ be a UFD containing a primitive $d$-th root of unity, with $d$ not divisible by $\operatorname{char} K$. Let $a \in R$ be an element that contains a factor $g$ with multiplicity one. Then $x^d - a$ is irreducible in $R[x]$.
Proof. Indeed, the least integer $m$ such that $a^m$ is a $d$-th power in $R$ is $d$, since $a^m$ contains the factor $g$ with multiplicity $m$. Since a UFD is normal, this is also the least integer $m$ such that $a^m$ is a $d$-th power in $K$. Thus, the extension $K \subseteq K(\sqrt[d]{a})$ has degree $d$, hence the polynomials $x^d - a$ is irreducible in $K[x]$. By Gauß's lemma, it is irreducible in $R[x]$. $\square$
Remark. More generally, the result is true if the $\gcd$ of the multiplicities of the irreducible factors of $a$ is $1$.
Example. If $a$ is itself irreducible, then $x^d - a$ is irreducible.
Proposition. Let $k$ be a field, and let $d \in \mathbb Z_{>0}$ not divisible by $\operatorname{char} k$. If $n \geq 3$, then the polynomial $x_1^d + \ldots + x_n^d$ is irreducible in $k[x_1, \ldots, x_n]$.
Proof. We may replace $k$ by $\bar k$; that proves a stronger statement. As promised, we proceed by induction. The case $n = 3$ follows from the corollary since $ - (x_1^d + x_2^d)$ contains the factor $x_1 - \zeta x_2$ with multiplicity one (where $\zeta$ is a primitive $2d$-th root of unity).
Once we have the base case, the induction step is easy: apply the corollary to the case where $a = - g$ with $g = x_1^d + \ldots + x_{n-1}^d$ irreducible by the induction hypothesis. $\square$
Remark. On the other hand, for $n \leq 2$ it's reducible: $x_1^d = (x_1)^d$, and
$$x_1^d + x_2^d = \prod_{i \text{ odd}} (x_1 - \zeta^i x_2),$$
where $\zeta$ is a primitive $2d$-th root of unity. (Multiplying also with the factors $x_1 - \zeta^i x_2$ for $i$ even gives $x_1^{2d} - x_2^{2d}$.)
Corollary For $d$ invertible in $K$, and for $n \geq 3$, the ideal $(x_1^d + \ldots + x_n^d) \subseteq K[x_1, \ldots, x_n]$ is prime.
Proof. In a UFD, the ideal generated by an irreducible element is prime (this is not true in general! See e.g. the remarks/examples at the end of this post). $\square$
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