# Remainders Modulo for Polynomials

I'm very confused about the remainders modulo of polynomials. For instance $$\mathbb{R}[t]/(t^3).$$ Does this simply mean that you keep dividing any polynomial $P\in\mathbb{R}[t]$ by $t^3$ until you can't anymore? Hence in this algebra $t^4\equiv t$ and $t^5+t^3\equiv t^2+1$ and so on?

How does this work for things like $$\mathbb{R}[X,Y]/(X^2+Y^2-1)?$$ What polynomials would belong to such an algebra?

The point of quotient rings is to “kill” things. So, in $\mathbb{R}[t]/(t^3)$, we have $t^3=0$. More precisely, $\mathbb{R}[t]/(t^3)=\mathbb{R}[u]$, with $u^3=0$. So, $\mathbb{R}[u]$ is the set of polynomial expressions in $u$ subject to $u^3=0$. This leaves just quadratic expressions in $u$.
Likewise, $\mathbb{R}[X,Y]/(X^2+Y^2-1)= \mathbb{R}[u,v]$ with $u^2+v^2=1$.
• Oh I think I see. But does that also mean there are no polynomials of degree higher than 2 in $\mathbb{R}[t]/(t^3)$? Because, for instance $t^4+1$ isn't divisible by $t^3$. Oh and also, how is the unit $1\in\mathbb{R}[t]/(t^3)$? It's an algebra, so it has to be in there. That would just be 1 in this case, right? – Joffysloffy Jan 21 '14 at 10:39
• @Joffysloffy, $u^4+1=1$ in $\mathbb R[u]$ because $u^4= u^3u=0$. – lhf Jan 21 '14 at 10:46