0

Let $F_p[x]$ be a ring of polynomials with coefficients in the field $F_p$. Let $q(x)$ be some irreducible polinomial in the base field $F_p$ with degree $n$. To construct the field extension $F_p^n$, I've seen in literature two different notations/constructions:

1) $F_p^n = F_p[x]/\langle q(x) \rangle$ : quotient ring by the ideal generated by $q(x)$

and

2) $F_p^n = F_p[x]/( q(x) )$ : the residue class by $q(x)$, when operations are done $\mod q(x)$.

Now I ask: These different notations means the same? Why one might be preferable instead the other?

Berk7871
  • 117
  • See this answer for the general relationship between congruences, quotient rings and subalgebras of the square. – Bill Dubuque Sep 17 '17 at 16:56
  • 4
    They are exactly the same. – M. Van Sep 17 '17 at 16:57
  • 3
    quotienting out an ideal $I$ and doing arithmetic with the cosets is exactly what you get when you quotient out by the equivalence relation $a ~ b$ iff $a-b \in I$ and define addition and multiplication on the equivalence classes in the natural way. – M. Van Sep 17 '17 at 16:59
  • @M.Van They are certainly not "exactly the same", but there is a general correspondence. See the link in my prior comment. – Bill Dubuque Sep 17 '17 at 17:12
  • 3
    @BillDubuque How are they not the same sets with the same ring structure on them (not only isomorphic, but the isomorphism is given by the identity -.-)? – M. Van Sep 17 '17 at 17:15
  • @M.Van This is explained in the link I provided (and its links). – Bill Dubuque Sep 17 '17 at 17:16
  • 2
    @Bill Dubuque, is it not true that by definition $R/I = R/!\equiv$ as sets, where $I$ is ideal in $R$ and $\equiv$ is congruence generated by $I$? – Ennar Sep 17 '17 at 21:46
  • @Ennar It depends on how one presents the constructions. I suspect that those who think that they are "exactly the same" have not given enough thought to the general case (e.g. see here for the ideal analog for general algebras - so-called ideal-determined algebras). But it is pointless to argue the matter without a precise definition of "exactly the same". In any case readers may find it insighful to examine these matters from the more general perspective in the links. – Bill Dubuque Sep 17 '17 at 22:38

1 Answers1

4

These are exactly the same (note though that this field is written $\mathbb{F}_{p^n}$, not $\mathbb{F}_p^n$). "Residue classes mod $q(x)$" and "elements of the quotient by the ideal generated by $q(x)$" are two different terms for the same thing: namely, equivalence classes of elements of $\mathbb{F}_p[x]$ under the equivalence relation $f(x)\sim g(x)$ iff $f(x)-g(x)$ is a multiple of $q(x)$. The difference in notation between $(q(x))$ and $\langle q(x)\rangle$ is meaningless--some people use parentheses to denote ideals generated by a list of elements, and other people use angle brackets. You should feel free to use either one, as long as you are consistent.

Eric Wofsey
  • 330,363