Let $k$ be a commutative ring. Let $M$ be a $k$-module. Let $M$ be a graded module, concentrated in degree one. Then, the shifted module $M[1]$ is concentrated in degree zero. Hence, $M[1]$ is trivially graded or equivalently ungraded. The symmetric algebra of the ungraded module $M[1]$ is defined as
$$\mathrm{Sym}^{\bullet}(M[1]):=T^{\bullet}(M[1])/I,$$
where $I:=\langle x\otimes y- y\otimes x~|~x, y\in M[1]\rangle$.
What is the degree of $\mathrm{Sym}^{n}(M[1])$? Is it zero or is it $n$? If it is zero, doesn't this contradict Bourbaki's Algebra $I$, according to which the symmetric algebra of a graded module is graded, too?