I would like to discuss a fine observation that I made which is not discussed in the literarure. Maybe because it is easy. Nevertheless, I think that it is quite important.
Let $k$ be a commutative ring. Let $k[t]$ be the ring of polynomials in one variable $t$ with $\deg t=0$ and coefficients in $k$. Then, the ring $k[t]$ is not graded and it is concentrated in degree zero. The filtration associated with this trivial grading is the trivial filtration
$$k[t]\supseteq k[t]\supseteq k[t]\supseteq\cdots.$$
Correspondingly, the associated graded of the trivially filtered polynomial ring $k[t]$ yields the isomorphism of ungraded rings
$$\mathrm{Gr}^{\bullet}(k[t])\cong k[t].$$
Nevertheless, the ungraded ring $k[t]$ also has a nontrivial decreasing filtration $F$, which is given by the sequence
$$F_0=k[t]\supseteq F_1:=tk[t]\supseteq\cdots\supseteq F_p:=t^pk[t]\supseteq\cdots\supseteq\{0\}.$$
The associated graded ring $\mathrm{Gr}_F^{\bullet}(k[t])$ associated with the nontrivial decreasing filtration $F$ yields an isomorphism of graded rings
$$\mathrm{Gr}_F^{\bullet}(k[t])\cong k[t].$$
The fact that the isomorphism is graded implies necessarily however that $\deg t=1$, which means the the associated graded of the ungraded ring $k[t]$ is not isomorphic as a graded ring to the original ungraded ring $k[t]$ but it is isomorphic to a different polynomial ring - namely the graded polynomial ring $k[t]$ with $\deg(t)=1$. Is this observation correct?
Moreover, is there an ungraded isomorphism between the associated graded $\mathrm{Gr}_F^{\bullet}(k[t])$ and the ungraded polynomial ring $k[t]$? Is it the same as the graded isomorphism forgetting the grading of $\mathrm{Gr}_F^{\bullet}(k[t])$?