# Projection map for polynomial rings

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?

• 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. – EgoKilla Apr 17 at 18:13
• Cool, thanks for the verification – Sigurd Apr 17 at 18:25

## 1 Answer

It is simple to prove that the restriction map is a morphism of ring so you have that

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

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

• Wait a minute, should the exponents be $X^{i-1}$ and $X^{j-1}$ here? – Sigurd Apr 17 at 18:21
• r(X^k)=0 for each $j\leq k \leq i$ – Federico Fallucca Apr 17 at 18:28
• 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? – Sigurd Apr 17 at 18:31