4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ General remark: Some things reduce to polynomial rings, since $V$ is wlog finitely generated and thus a quotient of some $A^n$. $\endgroup$ – Martin Brandenburg Sep 27 '14 at 15:36
2
$\begingroup$

(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 $0 \neq a \in A$.

$\endgroup$
  • $\begingroup$ "This means that $f$ is nilpotent" -- why? $\endgroup$ – darij grinberg Sep 27 '14 at 16:02
  • $\begingroup$ 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. $\endgroup$ – Martin Brandenburg Sep 27 '14 at 16:04
  • $\begingroup$ I don't think so. The sum converges in the completion's topology, not in the discrete topology, so convergence $\neq$ stabilization. $\endgroup$ – darij grinberg Sep 27 '14 at 16:05
  • $\begingroup$ 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. $\endgroup$ – Martin Brandenburg Sep 27 '14 at 16:08
  • $\begingroup$ 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? $\endgroup$ – darij grinberg Sep 27 '14 at 16:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.