Let $k$ be a field and let $R$ be an $\mathbb{N}$-graded $k$-algebra such that $R_0=k$. Let $\theta$ be a not necessarily homogeneous element of $R$ that is transcendental over $k$ and is a non-zerodivisor in $R$. My question is,
Can $R$ have torsion as a $k[\theta]$-module? (If yes, example? If no, proof?)
I can prove that if $\theta$ is homogeneous, then $R$ is torsion-free as a $k[\theta]$-module. (Basically, as all powers of $\theta$ are linearly independent and homogeneous of distinct degrees, a polynomial in $\theta$ cannot annihilate an element $x$ of $R$ unless its top-degree component annihilates $x$'s top degree component; but $\theta$ is a non-zerodivisor and $k$ is a field, so the top degree component of a polynomial in $\theta$ is a non-zerodivisor.)
Also, if I drop the assumption that $\theta$ is transcendental over $k$, I have an example of torsion. Take $\theta = x+1$ in $R=k[x,y]/(x,y)^2$. $\theta$ is a non-zerodivisor since it is a unit, but $\theta-1=x$ is clearly a zerodivisor.
Or, if I retain the assumption that $\theta$ is transcendental, but I drop the assumption that $R$ is graded, then again I have an example of torsion. Take $V$ some finite-dimensional vector space over $k$, let $\theta$ be a nonsingular linear transformation of $V$, so that $V$ becomes a $k[\theta]$-module, and then take $R=k[\theta]\oplus V$, giving it an algebra structure by having the multiplication be identically zero on $V\otimes V$. Then $\theta$ is a non-zerodivisor, but its characteristic polynomial annihilates $V$.
Let $R=k[x,y]/(xy)$ with $|x|=|y|=1$, and let $\theta=x+1$. It is easy to see $\theta$ is transcendental over $k$ and a nonzerodivisor. But $(\theta-1)y=0$, so $R$ has torsion as a $k[\theta]$-module.