- Consider the below polynomial ring
$$\mathbb Z[X]/(X^4 + 1)$$
I think the above is a quotient ring,
and because $X^4 + 1$ can't be further factorized under $Z$,
the above ring consists of all the polynomials of degree $<4$,
whose coefficients are integers.
Is my reasoning for the above statement right?
- Consider the below polynomial ring
$$\mathbb Z_{17}[X]/(X^4 + 1)$$
Although $X^4 + 1$ can't be further factorized under $Z$,
I think it doesn't hold true
considering we are under $Z_{17}$ now.
Consider the following,
$$\mathbb (X^2 + 4)(X^2 – 4) = X^4 – 16 = X^4 + 1$$
because of $Z_{17}$. So that means that $X^4 + 1$ can be factorized under $Z_{17}$, right?
- Continuing the above polynomial ring
$$\mathbb Z_{17}[X]/(X^4 + 1)$$
Since $X^4 + 1$ can be factorized into $(X^2 + 4)(X^2 – 4)$,
what elements are in the ring now?
Is it still polynomials of degree $<4$ according to Q1?
Or is it polynomials of degree $<2$ because of $(X^2 + 4)$ and $(X^2 – 4)$?
Thanks!
Best Answer
Hint: Over the finite field $\Bbb F_{17}$ we have $$ x^4+1=(x + 15)(x + 9)(x + 8)(x + 2). $$ Therefore the quotient has zero divisors and looks different from what you have over $\Bbb Z$, where it is a field. Still, the classes consist of polynomials of degree $\le 3$.