When do algebraic elements form a subalgebra?

Question

If $R$ is a commutative ring and $A$ is a commutative $R$-algebra, we say that an element $x\in A$ is algebraic over $R$ if $x$ is a root of a nonzero polynomial $f \in R[X]$, or equivalently, if the "evaluation at $x$" $R$-algebra homomorphism $R[X]\to A$ is not injective. If we require $f$ to be monic, we get instead the notion of integral elements, and it is well-known that the elements of $A$ that are integral over $R$ form an $R$-subalgebra. It is therefore natural to ask whether the algebraic elements also form a subalgebra.

If $R$ is a field, then algebraic elements coincide with integral elements and therefore form a subalgebra; if $R$ is only an integral domain, we may argue as follows: it suffices to show that for any two algebraic elements $x,y\in A$, the subalgebra $S$ generated by $x$ and $y$ consists solely of algebraic elements, i.e. there is no injective $R$-algebra homomorphism from $R[X]$ to $S$. Consider $K=\text{Frac}(A)$ and the $K$-algebra $K\otimes_R S$ generated by $1\otimes x$ and $1\otimes y$, which are still algebraic because $R\to K$ is injective. Any injective $R[X]\to S$ would induce injective $K[X]\to K\otimes_R S$ due to flatness of $K/R$, which is impossible by the field case.

We may consider a simplified version of this, namely whether an algebra generated by a single algebraic element $x$ consists solely of algebraic elements. For this simplified question, the argument above allows us to reduce to local rings $R$: if $x$ is a root of a nonzero polynomial in $R[X]$, the polynomial remains nonzero in $R_{\mathfrak{m}}[X]$ for some maximal ideal $\mathfrak{m}\subset R$. (But given two elements and two nonzero polynomials they are roots of, there may not exist a single $\mathfrak{m}$ such that both polynomials remain nonzero in $R_{\mathfrak{m}}[X]$.)

There are in fact ways to produce polynomials having $x+y$, $xy$ and $f(x)$ (for $f\in R[X]$) as roots via resultants, given polynomials having $x$ and $y$ as roots, but the problem is that we may get the zero polynomial. For example, zero is what we get if we try to produce a polynomial having $1+X+X^2\in \mathbb{Z}/4\mathbb{Z}[X]/\langle2X^3\rangle$ as a root by computing $\text{res}_X(2X^3,Z-(1+X+X^2))$. However, the following Singular code nonetheless shows that $1+X+X^2$ is a root of $2(X^3+X^2+X+1)$:

ring r=(integer, 4), (x), lp;
map f=r,1+x+x2;
ideal i=2x3;
preimage(r,f,i);

and I've been unable to find a counterexample after experimentations along this line.

I'm therefore asking: for which classes of commutative rings $R$ do we have

(1) every element in an $R$-algebra generated by an algebraic element is algebraic;

(2) algebraic elements form a subalgebra;

(3) idempotency of "algebraic closure" / transitivity of algebraicity: if $x \in A$ is a root of a nonzero polynomial in $A[X]$ with $R$-algebraic coefficients, then $x$ is itself algebraic.

YCor gave a counterexample for (2) (and therefore (3)) in the comments, and darij grinberg pointed out that this question contains a counterexample to (1).

Answer 1

Partial answer: for the case of reduced $R$, one has (2) $\Leftrightarrow$ $R$ is a domain.

For, let $R$ be any ring, reduced or not, containing non-zero elements $a,b$ such that $aR$ $\cap$ $bR=0$. (E.g.: $R$ is a reduced ring but not a domain; then if $ab=0$ and $x\in aR\cap bR$, say $x$ $=$ $au$ $=$ $bv$, one has $x^2$ $=$ $abuv$ $=0$; hence $x=0$; thus indeed $aR$ $\cap$ $bR=0$.) We produce an $R$-algebra $A$ for which (2) fails.

Take $A=R[X,Y]/(aX,bY)$. Then $\overline{X},\overline{Y}\in A$ are algebraic over $R$, but $z$ $:=$ $\overline{X}$ $+$ $\overline{Y}$ is not. For suppose $H(z)$ $=$ $0$ in $A$, for some $H$ $=$ $\sum_{\,0\le i\le n}c_iZ^i$ $\in R[Z]$. Then $\sum_{\,0\le i\le n}c_i(X+Y)^i$ $=$ $aXF$ $+$$ bYG$ in $B$ $:=$$R[X,Y]$, for suitable $F,G$ $\in$ $B$. Clearly, $c_0$ $=$ $0$. And for $i$ $>$ $0$, the coefficient of $X^i$ on the left-hand side is $c_i$, while on the right it is $a$ times the coefficient in $F$ of the monomial $X^{i-1}$. So $c_i$ $\in$ $aR$. Likewise, comparing coefficients for $Y^i$ on both sides, $c_i$ $\in$ $bR$. Since $aR$ $\cap$ $bR=0$, we obtain $c_i$ $=$ $0$, and therefore $H$ $=$ $0$. Thus (2) does not hold for $R$. (Indeed, the algebraic elements do not even form an additive subgroup of $A$.)

Added 2024/12/03:

On the positive side, (2) is true for every finite local principal ideal ring $(R,\mathfrak{m})$, such as $\mathbb{Z}/4\mathbb{Z}$. One has $\mathfrak{m}$ $=$ $\pi R$, with $\pi^n$ $=$ $0$ for some $n$, minimal. And every nonzero element of $R$ is of the form $\varepsilon\pi^i$ with $\varepsilon$ a unit and $0$ $\le$ $i$ $<$ $n$. Let $A$ be a $R$-algebra. Then $\pi^m$ $=$ $0$ in $A$, for a minimal $m$ $\le$ $n$. If $m$ $<$ $n$, all elements of $A$ are algebraic over $R$: they are zeros of $\pi^mZ$ $\in$ $R[Z]-\{0\}$. So assume $m$ $=$ $n$, and let $x,y$ $\in$ $A$ be algebraic over $R$. Say $\sum_{0\le i\le t}$ $c_ix^i$ $=$ $0$ in $A$, with the $c_i$ $\in$ $R$ not all zero. Put $\lambda$ $:=$ $\pi^{n-1}$. There exists a unique $0$ $\le$ $j$ $<$ $n$ such that, for all $0$ $\le$ $i$ $\le$ $t$, $d_i:$ $=$ $\pi^jc_i$ is either zero or an associate of $\lambda$ in $R$, and not every $d_i$ vanishes. Then $\sum_{0\le i\le s}$ $d_ix^i$ $=$ $0$ for some $s$ $\le$ $t$ with $d_s$ $\ne$ $0$. As $\lambda$ $\ne$ $0$ in $A$, necessarily $s$ $>$ $0$. So $\lambda x^s$ $\in$ $\sum_{0\le i<s}$ $\lambda Rx^i$. It follows that, for every $k$ $>$ $s$, $\lambda x^k$ is also in this $R$-submodule of $A$. Hence $\lambda R[x]$ $=$ $\sum_{0\le i<s}$ $\lambda Rx^i$. In the same vein, $\lambda R[y]$ $=$ $\sum_{0\le j<t}$ $\lambda Ry^j$ for some $0$ $<$ $t$ $\in$ $\mathbb{N}$. Therefore, $\lambda R[x,y]$ $=$ $\sum_{0\le i<s,\,0\le j<t}$ $\lambda Rx^iy^j$. But this is a finite set. Since it contains $\lambda(x+y)^k$ for every $k$ $>$ $0$, we must have $\lambda(x+y)^{k_1}$ $=$ $\lambda(x+y)^{k_2}$ for certain positive integers $k_1$ $>$ $k_2$, so that $x+y$ is a zero of $\lambda Z^{k_1}$ $-$ $\lambda Z^{k_2}$ $\in$ $R[Z]$. Similarly, $xy$ is a zero of a nontrivial polynomial of the same type.