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...

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    $\begingroup$ Perhaps this is wrong as you say it is late: if t where not transcendental it would be algebraic so k[t] is finite dimensional over k, but k[t] contains k[t^n] so t^n is algebraic, contradiction. $\endgroup$ – user9072 May 29 '13 at 2:05
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    $\begingroup$ @quid: $k$ is a commutative ring, not a field. You can't conclude that $k[t]$ is finitely generated as a $k$-module either (since we may have for example $k = \mathbb{Z}, t = \frac{1}{2}$). $\endgroup$ – Qiaochu Yuan May 29 '13 at 2:37
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    $\begingroup$ @darij: Let $k$ contain two elements $a, b$ such that $a^2 = ab = b^2 = 0$ and let $A = k[t]/(at^3 - b)$. By construction, $t$ is algebraic. It looks like $t^2$ might be transcendental, although I'm not sure how to prove it. $\endgroup$ – Qiaochu Yuan May 29 '13 at 3:03
  • $\begingroup$ @Qiaochu Yuan: but it is denoted $k$, it must be a field! :-) Thanks for the correction, somehow I thought I was missing something obvious, but that $k$ is not a field I just did not notice (while it is clearly stated). $\endgroup$ – user9072 May 29 '13 at 10:25

$\def\F{{\mathbb F}}$ 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.

  • $\begingroup$ I believe the term is "integral" for which one requires the equation of integrally to be monic. It might be easy, but I actually don't see why $t$ is "algebraic" over $k$. Can you tell me a polynomial $P$ in $k[x]/I$ such that $P(t) = 0$? If $\mathbb{F}$ is a field, I thought that $k[X] \cong \mathbb{F}[X,Y,Z]/(Y^2,YZ,Z^2, YX^3-Z)$. Isn't this isomorphic to $\mathbb{F}[Y,X]/ (Y^2)$? $\endgroup$ – Youngsu May 29 '13 at 8:53
  • $\begingroup$ Yes, I was wrong about the term, sorry. The polynomial is simply $at^3-b$... $\endgroup$ – Ilya Bogdanov May 29 '13 at 9:17
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    $\begingroup$ And yes, we may say that $k[t]={\mathbb F}[X,Y]/(Y^2)$, $t=\overline{X}$, and $k$ is a subring generated by $a=\overline{Y}$ and $b=\overline{X}^3\overline{Y}$. Then $at^3-b=0$, but $t^2$ is transcendental. $\endgroup$ – Ilya Bogdanov May 29 '13 at 9:22

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