What is the intuition behind $R= (\Bbb{Z}/3\Bbb{Z})[X] /\lt X^3 + \bar2 X^2 + X + \bar1 \gt $?

Question

I am studying ring theory and am looking at quotient rings. I have seen this example of a quotient ring: $$R= (\Bbb{Z}/3\Bbb{Z})[X] /\langle X^3 + \bar2 X^2 + X + \bar1 \rangle $$

I believe that $\bar2=2+3\Bbb{Z}$ and $\bar1=1+3\Bbb{Z}$, but I can't get my head around what this ring R means.

Can anyone help with the intuition?

How can I check if this ring is commutative or unital?

Answer 1

Let me show a simpler example:

$$R= (\Bbb{Z}/2\Bbb{Z})[X] /\langle X^2 + \bar1 \rangle $$

Before the quotient, we have all polyonomials with coefficients in $\Bbb Z / 2\Bbb Z$. But the quotient tells us that we can replace $X^2$ with $-\bar1$ (which, because we're working in integers mod 2, is the same as $\bar 1$). So a polynomial like $$ x^4 + x^3 + x $$ can be rewritten as $$ (x^2)^2 + x^2(x) + x = (\bar1)^2 + \bar1 x + x = \bar1 + x + x = \bar1 $$ I can do this rewriting because there's only one relation, and it's nice and simple. But in general, a quotient like this allows you to "simplify away" any terms of power higher than $k$ (where $k$ is the largest exponent in the quotiented polynomial) by replacing them with one or more terms of power lower than $k$. (This is a little bit like "casting out nines", if you learned to do that when you were learning arithmetic; if not, ignore this remark.)

Pro tip: when you want to write angle-brackets, us \langle ... \rangle instead of \lt ... \gt.

Answer 2

$R \cong \mathbb F_3[u]$ where $u^3+2u^2+u+1=0$. In words, polynomials expressions in $u$ with coefficients in the integers mod $3$ and subject to that relation. So, quadratic expressions in $u$.

Compare with $\mathbb C \cong \mathbb{R}[x]/(x^2+1) \cong \mathbb R[i]$ with $i^2+1=0$.