Let $k\geq 2$ be a fixed integer. If $R$ is a commutative, integral, unital ring, the Waring height of an element $r\in R$ is the smallest number of $k$-ths powers whose sum is $r$ (this height can equal $\infty$, when $r$ cannot be written as a sum of $k$-th powers).
Is a simple example known of a ring containing an element whose Waring height is $\infty$ ?
Edit As noted in Doug Chatham’s comment below, the answer is easy when $k$ is even : take $R={\mathbb Z}$, $r=-1$.
If $k = p$ is prime and $R$ has characteristic $p$ then a sum of $k^{th}$ powers is a $k^{th}$ power, so every element has Waring height either $1$ or $\infty$. To get the latter it suffices to find an element of such a ring which is not a $p^{th}$ power. A simple example is $x \in \mathbb{F}_p[x]$.
Edit: And note that this reasoning lifts to characteristic zero: for any odd prime $p$, if $x$ is any element of a commutative ring $R$ such that $x$ is not a $p^{th}$ power $\bmod p$, then $x$ is not a sum of $p^{th}$ powers. For example, $x \in \mathbb{Z}[x]$ is not a sum of $p^{th}$ powers for any $p$.
(What is an integral ring?)
That worked. the Eisenstein integers and $k=3.$ We have $$ \omega = \frac{-1}{2} + \frac{i \sqrt 3}{2}, $$ so that $\omega^3 = 1$ and $\omega^2 = -1 - \omega.$ The ring is all $$ x + y \, \omega, \; \; \; x,y \in \mathbb Z. $$ I got $$ ( x + y \, \omega)^3 = (x^3 - 3 x y^2 + y^3) + 3 (x^2 y - x y^2) \, \omega $$ which is to say that the $\omega$ coefficient of every cube is divisible by $3.$ The $\omega$ coefficient of every sum of cubes is divisible by $3.$ So $\omega$ itself is not the sum of cubes.
Another example. Let $q$ be a power of a prime, $R=E$ the finite field $E=GF(q^2)$ and $k=q+1$, and $F$ the subfield $GF(q)$. Then we have for all $x\in E$ $$ x^{q+1}=N^E_F(x)\in F. $$ Therefore any sum of $k$th powers in $E$ belongs to the subfield $F$ and thus the elements of $E\setminus F$ all have infinite Waring height.
How about $k=2$ (actually any even number will do), $R = \mathbb{Q}$, and $r=-2$?
