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 anon-zero-divisorif and only if every $v \in B$ satisfying $uv = 0$ satisfies $v = 0$.)

(d)We say that a polynomial in $A\left[x\right]$ isprimitiveif 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 $\Sym V$ isprimitiveif 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 aleft non-zero-divisorif and only if every $v \in B$ satisfying $uv = 0$ satisfies $v = 0$.)

(Nd)We say that a noncommutative polynomial in $A\left<x_1, x_2, ..., x_n\right>$ isprimitiveif 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.