Difference between 2 Quotient Rings Consider the rings, $Z_2[x]/(1 + x^2)$ and $ Z_2[x]/(1 + x + x^2)$, despite having different polynomial as divisor, I have been told that -
$$Z_2[x]/(1 + x^2) = \{0, 1, x, 1 + x\}$$
and
$$Z_2[x]/(1 + x + x^2)= \{0, 1, x, 1 + x\}$$
Then what is the difference between $Z_2[x]/(1 + x^2)$ and $ Z_2[x]/(1 + x + x^2)$?
 A: Well, both quotient rings are vector spaces over $Z_2$ of dimension 2, both with bases $\{1,x\}$.
In general, if $K$ is a finite field and $f(x)$ is a polynomial over $K$ of degree $n$, then $K[x]/\langle f(x)\rangle$ is a vector space over $K$ of dimension $n$.
The difference above is that $x^2+x+1$ is irreducible over $Z_2$, while $x^2+1=(x+1)^2$ is not. So $Z_2[x]/\langle x^2+x+1\rangle$ is a field, while $Z_2[x]/\langle x^2+1\rangle$ is not (it has zero divisors).
A: This is a notation issue. When you write $\mathbb{Z}_2[x]/(1 + x^2) = \{0, 1, x, 1 + x\}$, that's just an equality of equivalence classes. In addition, on the RHS of the equality, you're missing some relations that hold in $\mathbb{Z}_2[x]/(1 + x^2)$ as the notation $\{0, 1, x, 1 + x\}$ alone doesn't tell anything about relations.
The same with $\mathbb{Z}_2[x]/(1 + x^2) = \{0, 1, x, 1 + x\}$. Moreover the equivalence classes in the latter are not the same as the one in the former, because you're modding out be different ideals, so even as sets  $\{0, 1, x, 1 + x\}$ and $\{0, 1, x, 1 + x\}$ are are basically different. Hence one should use different notation to avoid confusion.
