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Let $K$ be a field and Consider the projection map $\pi_{i,j} : K[X]/(X^i) \to K[X]/(X^j)$, for $j \leq i$. This is well-defined since $(X^i) \subseteq (X^j)$. I'm wondering what it looks like, is it just restriction in the sense: $$ a_0 + a_1 X + \ldots + a_{i-1}X^{i-1} \mapsto a_0 + a_1 X + \ldots + a_{j-1}X^{j-1} $$ Can someone verify that?

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  • $\begingroup$ This looks right to me, you are essentially just setting $X^j$ to $0$ in the image. So any term with $X^k$ for $k > j$ should die. $\endgroup$ – EgoKilla Apr 17 at 18:13
  • $\begingroup$ Cool, thanks for the verification $\endgroup$ – Sigurd Apr 17 at 18:25
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It is simple to prove that the restriction map is a morphism of ring so you have that

$r(a_0+\dots+a_{i-1}X^i)=$

$=a_0+\dots+a_{i-1}r(X^i)=a_0+\dots+a_{j-1}X^j$

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  • $\begingroup$ Wait a minute, should the exponents be $X^{i-1}$ and $X^{j-1}$ here? $\endgroup$ – Sigurd Apr 17 at 18:21
  • $\begingroup$ r(X^k)=0 for each $j\leq k \leq i$ $\endgroup$ – Federico Fallucca Apr 17 at 18:28
  • $\begingroup$ Right. But there is the term $a_{i-1} X^i$ in the ring $K[X]/(X^i)$, but $X^i = 0$ in this ring right? $\endgroup$ – Sigurd Apr 17 at 18:31

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