Relationship between a polynomial and its constant term (ring)

Question

There is an example illustrating ring isomorphism that surprises me so much. It essentially talks about the relationship between a polynomial and the constant term:

Consider the ring $\mathbb{Q}[t]$ with the ideal $<t>$. For $p \in \mathbb{Q}[t]$,

$$ \bar{p} = p + I = p_0 + I$$ where $p_0$ is the constant term of the polynomial $p$. Then $\mathbb{Q}[t]/I \to \mathbb{Q}$ is an isomorphism.

Problem

How can $p+I = p_0 +I$ hold at all? Considering $p$ is a polynomial, and $p_0$ is a constant in $\mathbb{Q}$.

Answer 1

Hint: $p(t)= p_0 + p_1 t + p_2 t^2 + \cdots + p_n t^n = p_0 +t \,q(t)$.

For a slightly more sophisticated approach, consider the map $\mathbb{Q}[t] \to \mathbb{Q}$ given by $p(t) \mapsto p(0)=p_0$. Then this map is a surjective ring homomorphism with kernel $\langle t \rangle$.

Answer 2

That is because $\;p(t)=p_0+p_1t+p_2t^2+\dots+p_nt^n$ for some $n$, and that $$p(t)-p_0=t(\underbrace{p_1+p_2t+\dots=p_,t^{n-1}}_{q(t)})\in\langle \,t \,\rangle$$

Answer 3

Let $p=\sum_{j=0}^n p_jt^j$ be a polynomial in ${\Bbb Q}[t]$. Then $p\equiv p_0\mod t$, since $p-p_0$ is divisible by $t$. Thus $p+I = p_0+I$ as claimed.