# Basis-free and noncommutative versions of the two-polynomials-over-ring problem (McCoy theorem etc.)

There is a rather canonical bunch of exercises in commutative algebra which tend to come up time and again on math.stackexchange: recently in #948010 and #83121, formerly in #227787 and #413788, and in many other places, such as Messing/Reiner arXiv:1209.6307v2. Probably its most well-known appearance is as Exercise 2 in Chapter 1 of Atiyah/Macdonald's "Introduction to Commutative Algebra". Let me rephrase that exercise:

Let $A$ be a commutative ring, and let $A\left[x\right]$ be the ring of polynomials in one variable $x$ over $A$. Let $f \in A\left[x\right]$.

(a) Show that $f$ is a unit in the ring $A\left[x\right]$ if and only if the constant coefficient of $f$ is a unit and all other coefficients are nilpotent.

(b) Show that $f$ is nilpotent if and only if all coefficients of $f$ are nilpotent.

(c) Show that $f$ is a non-zero-divisor in $A\left[x\right]$ if and only if every $a \in A$ satisfying $af = 0$ satisfies $a = 0$. (We say that an element $u$ of a commutative ring $B$ is a non-zero-divisor if and only if every $v \in B$ satisfying $uv = 0$ satisfies $v = 0$.)

(d) We say that a polynomial in $A\left[x\right]$ is primitive if and only if $1$ is an $A$-linear combination of its coefficients. Show that for any two polynomials $f$ and $g$ in $A\left[x\right]$, the product $fg$ is primitive if and only if $f$ and $g$ are primitive.

Notice that what I call (c) is the contrapositive of Atiyah/Macdonald's Exercise 2 (c), as the notion of a non-zero-divisor is the correct constructive way to formalize statements about zero-divisors.

The exercise is followed by an Exercise 3 which asks for generalizations of all of these results to multivariate polynomial rings $A\left[x_1, x_2, ..., x_n\right]$; I believe these can be done by induction over $n$ (though I have not really checked).

Solutions of the problem given in literature are usually not constructive per se, but can often be rewritten in constructive terms.

$\newcommand\Sym{\operatorname{Sym}}$A question I have posed to myself long ago, but never had the time to seriously think about, is the following: A polynomial ring is a particular case of a symmetric algebra. What happens if we blindly generalize the exercise to symmetric algebras in general?

Let $A$ be a commutative ring, and let $V$ be an $A$-module. Let $\Sym V$ denote the symmetric algebra of $V$ over $A$. Let $f \in \Sym V$.

(Sa) Prove or disprove that $f$ is a unit in the ring $\Sym V$ if and only if the $0$-th homogeneous component of $f$ is a unit and all other homogeneous components are nilpotent.

(Sb) Prove that $f$ is nilpotent if and only if all homogeneous components of $f$ are nilpotent. [This one is actually true by an easy induction argument.]

(Sc) Prove or disprove that $f$ is a non-zero-divisor in $\Sym V$ if and only if every $a \in A$ satisfying $af = 0$ satisfies $a = 0$.

(Sd) We say that an element $h$ of $\Sym V$ is primitive if and only if $F\left(h\right) = 1$ for some $F \in \left(\Sym V\right)^\ast$ (linear dual). Prove or disprove that for any two elements $f$ and $g$ of $\Sym V$, the product $fg$ is primitive if and only if $f$ and $g$ are primitive.

These generalizations of the polynomial-ring exercise are by no means the only ones, the right ones or the canonical ones. I had to generalize the notion of a coefficient differently for (Sa) and for (Sd) to not get something obviously stupid, and I would not be totally surprised if the result is still wrong.

Another direction to generalize things in is that of noncommutative polynomials. There is no difference between commutative polynomial rings $A\left[x\right]$ and noncommutative polynomial rings $A\left<x\right>$ in one variable, so let us state the question in multiple variables:

Let $A$ be a commutative ring, and let $n \in \mathbb{N}$. Let $A\left<x_1, x_2, ..., x_n\right>$ be the ring of noncommutative polynomials in the variables $x_1, x_2, ..., x_n$ over $A$. (This is the monoid ring of the free monoid generated by $x_1, x_2, ..., x_n$.) Let $f \in A\left<x_1, x_2, ..., x_n\right>$.

(Na) Prove or disprove that $f$ is a unit in the ring $A\left<x_1, x_2, ..., x_n\right>$ if and only if the constant coefficient of $f$ is a unit in $A$ and all other coefficients are nilpotent.

(Nb) Prove or disprove that $f$ is nilpotent if and only if all coefficients of $f$ are nilpotent.

(Nc) Prove or disprove that $f$ is a left non-zero-divisor in $A\left<x_1, x_2, ..., x_n\right>$ if and only if every $a \in A$ satisfying $af = 0$ satisfies $a = 0$. (We say that an element $u$ of a ring $B$ is a left non-zero-divisor if and only if every $v \in B$ satisfying $uv = 0$ satisfies $v = 0$.)

(Nd) We say that a noncommutative polynomial in $A\left<x_1, x_2, ..., x_n\right>$ is primitive if and only if $1$ is an $A$-linear combination of its coefficients. Prove or disprove that for any two noncommutative polynomials $f$ and $g$ in $A\left<x_1, x_2, ..., x_n\right>$, the product $fg$ is primitive if and only if $f$ and $g$ are primitive.

Finally, the two generalizations can be combined into one that concerns the tensor algebra. (This is not to say that it will entail the symmetric-algebra version as a corollary.) I'll leave stating the conjectures to the reader, as this post is long enough.

• General remark: Some things reduce to polynomial rings, since $V$ is wlog finitely generated and thus a quotient of some $A^n$. – Martin Brandenburg Sep 27 '14 at 15:36

(Sa) and (Sb) may be proven in an even more general setting: arbitrary commutative $\mathbb{N}$-graded rings.

(Sa) In any commutative ring, elements of the form "unit + nilpotent" are units. Now assume that $A$ is a commutative $\mathbb{N}$-graded ring and let $a \in A$ be a unit. Write $a=\sum_{i \geq 0} a_i$ with $a_i \in A_i$, and also write $a^{-1}=\sum_{i \geq 0} b_i$ with $b_i \in A_i$. Assume $n$ is maximal with $a_n \neq 0$ and $m$ is maximal with $b_m \neq 0$. Then we get $a_n b_m = 0$ by looking at degree $n+m$, and $a_{n-1} b_m + a_n b_{m-1}=0$ by looking at degree $n+m-1$. This implies $a_n^2 b_{m-1}=0$. Next, by looking at degree $n+m-2$, we find $a_n^3 b_{m-2}=0$, etc. Finally we get $a_n^{m+1} b_0=0$. But since $a_0 b_0=1$ by looking at degree $0$, we get $a_n^{m+1}=0$, i.e. $a_n$ is nilpotent. Then $a-a_n$ is a unit, too, and we may repeat the argument and see that $a_1,\dotsc,a_n$ are nilpotent.

(Sb) Let $A$ be a commutative $\mathbb{N}$-graded ring and let $a \in A$ be nilpotent. Write $a=a_s+a_{s+1}+\dotsc$ with $a_i \in A_i$. Choose some $k \geq 0$ with $a^k=0$. Then $a_s^k=0$ since this is the lowest homogeneous component of $a^k$. Since nilpotent elements form a subgroup ($*$), it follows that $a_{s+1}+\dotsc$ is nilpotent. By induction, we see that all $a_i$ are nilpotent. The converse is also true, again using ($*$).

As for (Sc), I have proven here that every zero divisor of a graded ring $A$ (as above) is killed by some homogeneous element $\neq 0$. Now if $A=\mathrm{Sym}(V)$, this element has the form $v_1 \cdot \dotsc \cdot v_n$ for some $v_i \in V$. But we cannot do better than that. There are non-trivial modules $V$ with $V \otimes V = 0$ and therefore $\mathrm{Sym}^2(V)=0$. Then any $v \in V$ is a zero divisor in $\mathrm{Sym}(V)$, but it doesn't have to be killed by some $0 \neq a \in A$.

• "This means that $f$ is nilpotent" -- why? – darij grinberg Sep 27 '14 at 16:02
• Since it lies in the direct sum, the infinite sum has to stabilize somewhere, i.e. adding some $f^n$ doesn't change the partial sum anymore. – Martin Brandenburg Sep 27 '14 at 16:04
• I don't think so. The sum converges in the completion's topology, not in the discrete topology, so convergence $\neq$ stabilization. – darij grinberg Sep 27 '14 at 16:05
• Yes, but the argument is that if $f^n \neq 0$ for all $n$, then the homogeneous components "filled" by the partial sums get bigger and bigger. – Martin Brandenburg Sep 27 '14 at 16:08
• Really? What if each $f^n$ "removes" the noise made by the previous $f^{n-1}$ and adds some noise of its own in higher degrees? – darij grinberg Sep 27 '14 at 16:08