This looks extremely easy, but then again it's late at night...


Let $k$ be a commutative ring with unity. An element $a$ of a $k$-algebra $A$ is said to be *transcendental* over $k$ if and only if every polynomial $P\in k\left[X\right]$ (with $X$ being an indeterminate) such that $P\left(a\right)=0$ must satisfy $P=0$.


Let $n$ be a positive integer. Let $A$ be a $k$-algebra, and $t$ be an element of $A$ such that $t^n$ is transcendental over $k$. Does this yield that $t$ is transcendental over $k$ ?


There is a rather standard approach to a problem like this which works if $k$ is reduced (namely, assume that $t$ is not transcendental, take a nonzero polynomial $P$ annihilated by $t$, and consider the product $\prod\limits\_\omega P\left(\omega X^{1/n}\right)$, where $\omega$ runs over a full multiset of $n$-th roots of unity adjoined to $k$; this product can be seen to lie in $k\left[x\right]$ and annihilate $t^n$; this all requires a lot more work to put on a solid footing). There are even easier proofs around when $k$ is an integral domain or a field (indeed, in this case, if $t$ is not transcendental over $k$, then $t$ is algebraic over $k$, so that, by a known theorem, $t^n$ is algebraic over $k$ as well, hence not transcendental). I am wondering if there is a counterexample in the general case or I am just blind...



Are roots of transcendental elements transcendental?
$\def\F{{\mathbb F}}$
This is just a proof of Qiaochu's example.


Let $k=\F[Y,Z]/(Y^2,YZ,Z^2)$, $a=\overline{Y}$, $b=\overline{Z}$. In the ring $k[X]$ we have $I=(aX^3-b)=\{aP(X)X^3-bP(X): P(X)\in \F[X]\}$, since $a(aX^3-b)=b(aX^3-b)=0$. So each element of this ideal contains both even and odd powers of $X$. Thus in $k[X]/I$ the element $t=\overline{X}$ is algebraic, while $t^2$ is not.


