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Let $k$ be a commutative ring with $\deg(t)=0$, and let $k[t]$ be the ungraded polynomial ring in the variable $t$, centred in degree zero. Let $A$ be an associative $k$-algebra with increasing filtration $F$ given by

$$A_{-1}=0\subseteq A_0\subseteq A_1\subseteq\cdots\subseteq A.$$

What is the induced filtration on the algebra $A[t]$?

Is it $F_n(A[t])=A_n[t]$ or is it $F_n(A[t])=\sum_{i-j=n}A_i\otimes t^jk[t]$,

where we took into account the fact that $k[t]$ has a negative increasing filtration

$$\{0\}\subseteq\cdots\subseteq F_{-2}:=t^2k[t]\subseteq F_{-1}:=t^{-1}k[t]\subseteq F_0:=k[t].$$

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It depends.

If you really consider $t$ to have degree $0$ and therefore to have $k[t]$ unfiltered (or rather with trivial filtration), then the induced filtration on $A[t]$ should be the first one.

If on the other hand you want to consider the filtration on $k[t]$ you give at the end of your question, then the induced filtration on $A[t]$ should reflect that.

Simply said, the filtration on $A[t] = A\otimes k[t]$ depends on your choices of filtration on both $A$ and $k[t]$, so just choose one on $k[t]$ and it will give you one on $A[t]$.

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  • $\begingroup$ Thanks for your answer. If the degree of $t$ is zero, then the ring $k[t]$ is ungraded and concentrated in degree zero. Therefore, the induced filtration on $k[t]$ is trivial. That is what you mean by the trivial filtration on $k[t]$, right? You mean the filtration on $k[t]$ induced by the trivial grading on $k[t]$? $\endgroup$
    – Flavius Aetius
    Mar 19 at 22:48

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