# Induced filtration on polynomial ring with coefficients in a filtered associative algebra

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

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

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 $$A[t]$$.
• 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 $k[t]$? Mar 19 at 22:48