# Polynomial Ring Explanation

1. 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?

1. 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?

1. 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!

• $\Bbb Z[X]/(X^4+1)$ cannot be embedded as a ring in $\Bbb Z[X]$, so it's a bit unclear what you mean by it consisting of polynomials with of degree $<4$ with integer coefficients. $\Bbb Z[X]/(X^4+1)$ is a free $\Bbb Z$-module with basis $1,X, X^2, X^3$, but that would be true even if you were quotienting by a reducible polynomial, as long as it's monic of degree $4$. May 18, 2022 at 17:02
• I was looking at en.wikipedia.org/wiki/Quotient_ring, so to my understanding, $\mathbb R[X]/(X^4 + 1)$ is a ring in $R[X]$ but not for $Z$, right? May 18, 2022 at 17:08
• No, a quotient $R[x]/(f(x))$ is not a subset of $R[x]$ if $f$ is nontrivial. May 18, 2022 at 17:10
• @fuo55631 If you replace every occurrence of $\Bbb Z$ with $\Bbb R$ and "integer" with "real" in my previous statement you obtain an equally true statement. The instances when a quotient ring $A/I$ embeds as a subring in $A$ are very rare and often uninteresting. May 18, 2022 at 17:13
• Got it. I need some time to understand it further, Thanks again! May 18, 2022 at 17:18

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 $$\le 3$$.
• Have a look at this answer by A. Caranti, which is very similar to your case, namely $\Bbb F_2[x]/(x^4+1)$. The (classes of) polynomials are still of degree $\le 3$ in the quotient, but a nontrivial product can be zero. May 18, 2022 at 17:00
• No, this is no difference. You just write the quotient as a direct product of quotients using the CRT. This doesn't change what $K[x]/f(x))$ is by definition. It only shows that it is isomorphic to a direct product of fields. May 18, 2022 at 17:06