Strong transcendence over a subring

Let $R\subseteq S$ be an inclusion of commutative rings (Subring: a subset containing $1_R$ and closed under addition, additive inverses and multiplication) and let $x\in S$. The element $x$ is strongly transcendental over $R$ when the implication

$u(a_0+a_{1}x+\cdots +a_k{x}^k)=0⟹u{a}_i=0\text{ for every }i\in \{0,\dots ,k\}$

holds for every integer $k\ge 0$, every multiplier $u\in S$ and all $a_0,a_1,\dots ,a_k\in R$; the polynomial $a_0+a_{1}x+\cdots +a_k{x}^k$ is the evaluation in $S$ of a polynomial over $R$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution), so the condition is a statement about all polynomial relations that hold in $S$ between $x$ and the elements of $R$.

Conventions kept here. (i) The multiplier $u$ is retained on purpose. It records annihilators: when $S$ has zero divisors the vanishing of a product $u\cdot P(x)$ does not force $P(x)$ to vanish, and it is the annihilator form above, not mere linear independence of the monomials, that is used in the minimal-prime argument of this page. (ii) No finiteness or noetherian hypothesis is imposed on $R$, on $S$ or on the polynomial degree; $x$ has trivial annihilator in $S$: applying the condition to $P(X)=X$ gives $ux=0\Rightarrow u=0$. The zero polynomial ($k=0$, $a_0=0$) is included. (iii) The condition is tied to the chosen pair $R\subseteq S$ and is not preserved by an arbitrary quotient: for $R=k$ a field $S=k[z]\times k[y]$ and $x=(z,y)$, the element $x$ is strongly transcendental over $k$ (a multiplier $u=(u_1,u_2)$ with $u\cdot (P(z),P(y))=0$ has $u_{1}P(z)=0$ and $u_{2}P(y)=0$, so if $P=0$ all coefficients vanish, whereas if $P\mathrm{\ne }0$ both $P(z)$ and $P(y)$ are nonzero, so the domain property forces $u_1=u_2=0$; in either case $u{a}_i=0$ for every $i$), yet in the quotient $S\mathrm{/}\mathfrak{q}$ by the prime $\mathfrak{q}=(z-1)k[z]\times k[y]$ the image of $x$ is $1\in k$, which is a root of the monic $X-1$ and therefore not transcendental at all. The descent proved on this page is consequently formulated for minimal primes of reduced rings.

Domains. If $S$ is an integral domain (Zero divisor, and integral domain: a commutative ring with $1\mathrm{\ne }0$ and no zero divisors), then strong transcendence of $x$ over $R$ is the same as saying that $x$ is transcendental over the fraction field $\mathrm{Frac}(R)$, viewed inside $\mathrm{Frac}(S)$ (The field of fractions $\mathrm{Frac}(D)=(D\setminus \{0\}{)}^{-1}D$ of an integral domain). Suppose first that $x$ is strongly transcendental over $R$ and let $P\in \mathrm{Frac}(R)[X]$ be nonzero with $P(x)=0$; clearing denominators gives $u\in R$ nonzero with $uP\in R[X]$ still nonzero, and evaluating gives $u\cdot (uP)(x)=u\cdot 0=0$, so strong transcendence applied to the polynomial $uP$ and the multiplier $u$ forces $u\cdot (u{p}_i)=0$ for every coefficient $p_i$ of $P$; since $S$ is a domain and $u\mathrm{\ne }0$, this gives $p_i=0$ for all $i$, contradicting $P\mathrm{\ne }0$. Conversely, if $x$ is transcendental over $\mathrm{Frac}(R)$ and $u(a_0+\cdots +a_k{x}^k)=0$ with $u\mathrm{\ne }0$, then cancelling the nonzero element $u$ in the domain $S$ exhibits the polynomial $a_0+\cdots +a_k{X}^k\in R[X]$ as vanishing at $x$; if some $a_i$ were nonzero that polynomial would be a nonzero element of $\mathrm{Frac}(R)[X]$ vanishing at $x$, contradicting transcendence, so all $a_i$ vanish and $u{a}_i=0$ holds; for $u=0$ the conclusion is immediate. In this case the annihilator clause is automatic and the condition reduces to the classical notion. Nothing in the argument uses integrality; compare Integral elements over a commutative ring and algebraic integers, where the integral element is the opposite extreme, a root of a monic polynomial.