1
$\begingroup$
  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!

$\endgroup$
5
  • 1
    $\begingroup$ $\Bbb Z[X]/(X^4+1)$ 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 $4$. $\endgroup$
    – Sassatelli Giulio
    May 18, 2022 at 17:02
  • $\begingroup$ 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? $\endgroup$
    – fuo55631
    May 18, 2022 at 17:08
  • 1
    $\begingroup$ No, a quotient $R[x]/(f(x))$ is not a subset of $R[x]$ if $f$ is nontrivial. $\endgroup$
    – Dietrich Burde
    May 18, 2022 at 17:10
  • 1
    $\begingroup$ @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. $\endgroup$
    – Sassatelli Giulio
    May 18, 2022 at 17:13
  • $\begingroup$ Got it. I need some time to understand it further, Thanks again! $\endgroup$
    – fuo55631
    May 18, 2022 at 17:18

1 Answer 1

Reset to default
2
$\begingroup$

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 $\le 3$.

$\endgroup$
6
  • $\begingroup$ Does it means that the ring contains only constant terms (degrees of < 1) because it can be factorized into (x+15)(x+9)(x+8)(x+2), Thanks! $\endgroup$
    – fuo55631
    May 18, 2022 at 16:59
  • 1
    $\begingroup$ 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. $\endgroup$
    – Dietrich Burde
    May 18, 2022 at 17:00
  • $\begingroup$ I see. But I got confused with the following answer math.stackexchange.com/questions/3200710/…. In that question it says that because x^3 - x can be factored into x(x + 1)(x - 1), so the ring will consists only constant terms, what makes the differences between this two questions, Thanks! $\endgroup$
    – fuo55631
    May 18, 2022 at 17:05
  • 1
    $\begingroup$ 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. $\endgroup$
    – Dietrich Burde
    May 18, 2022 at 17:06
  • $\begingroup$ I somewhat got the idea. The main idea I was originally think was this, where it choose an irreducible polynomial phi(x), and a prime number q, I originally thought that it chooses phi(x) because it is irreducible, but to my surprise if it is reducible under Z_q, then what are the advantages of choosing an irreducible polynomial phi(x) at the first place, Thanks! $\endgroup$
    – fuo55631
    May 18, 2022 at 17:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .