When $\mathbb Z_3[x] / \langle x^2+1\rangle$, why $(2x^2 + 2x + 1 + \langle x^2+1\rangle) = (2x + 2 + \langle x^2+1\rangle)$?

Question

When $\mathbb Z_3[x]/\langle x^2+1\rangle$, why is $(2x^2 + 2x + 1 + \langle x^2+1\rangle) = (2x + 2 + \langle x^2+1\rangle)$?

I know $\langle x^2+1\rangle$ has $2x^2+2$ but I'm confused by the fact that how $(2x^2 + 2x + 1 + \langle x^2+1\rangle) = (2x + 2 + \langle x^2+1\rangle)$?

Answer 1

You're working modulo two things at once: mod 3 and mod $x^2+1$.

Replacing a $2$ with a $-1$ which is allowed because we're working mod 3, we have

$$2x^2+2x+1=-x^2+2x+1.$$

Adding $x^2+1$ which is allowed because we're working mod $x^2+1$, we have

$$-x^2+2x+1=2x+2$$

and we're done.