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I have a little doubt.
How do I find representants for classes of the quotient $$\big( \mathbb{F}_{11}[X]/(X^3-4x^2+23) \big)^{\ast} /\Big( \big( \mathbb{F}_{11}[X]/(X^3-4x^2+23) \big)^{\ast}\Big)^7 $$ I think that such quotient should be isomorphic to $\mathbb{Z}/7\mathbb{Z}$ using the fact that $\mathbb{F^\ast_{11^3}} \cong \mathbb{Z}/1330\mathbb{Z}$ and raising to the 7-th power in the LHS is equivalent to multiplication by 7 in the RHS. But this is not particularly helpful in order to find explicitly representative in the 1st quotient

FBruzzesi
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  • $\Bbb{F}_{11^3}^{*}$ is a cyclic group. You're correct that it's isomorphic to $\Bbb{Z}/1330\Bbb{Z}$, but think carefully about what happens to $\Bbb{Z}/1330\Bbb{Z}$ under multiplication by 7. Is $\mathbb{Z}/7\mathbb{Z}$ a quotient or a kernel? – sharding4 May 17 '17 at 18:20
  • If you know of a generator $\gamma$ (aka a primitive element) of the multiplicative group $\Bbb{F}{11^3}^*$, then $1,\gamma,\gamma^2,\cdots,\gamma^{7-1}$ is an obvious set of representatives. Because $1330=7\cdot190$ and $\gcd(7,190)=1$ the theory of finite abelian groups also tells us that $$C{1330}\simeq C_7\times C_{190}.$$ Furthermore, here $(C_{1330})^7$ becomes identified with ${1}\times C_{190}$. This implies that another natural set of representatives of those cosets consists of the seventh roots of unity in your field. In many a setting that may be a more attractive choice. – Jyrki Lahtonen May 17 '17 at 18:45
  • I totally agree with the 1st part: calling $\alpha$ the real root of $X^3-4X^2+23$, shouldn't that be a primitive element? Moreover, is there any way to find roots of unity in function of $\alpha$ ? – FBruzzesi May 17 '17 at 20:03
  • Luke, a few problems with that idea: 1) there are no real numbers in this field, 2) there is no reason why a zero of that cubic (we can call it $\alpha$ all right, but it is the coset $\alpha=x+\langle x^3-4x^2+23\rangle$) should necessarily be a generator of the multiplicative group of $\Bbb{F}_{11^3}$. We do know that it generates that as a field, but not necessarily as a group. – Jyrki Lahtonen May 18 '17 at 06:02
  • It is actually easy to prove that $\alpha$ is not a generator of the multiplicative group. By Galois theory of finite fields $$(x-\alpha)(x-\alpha^{11})(x-\alpha^{11^2})=x^3-4x^2+23=x^3-4x^2+1.$$ Expanding and comparing the constant terms shows that $\alpha^{1+11+11^2}=-1$. Therefore $\alpha^{266}=1$, so its order is a proper factor of $1330$. – Jyrki Lahtonen May 18 '17 at 06:07
  • Anyway, $266/7=38$, and $\alpha^{38}\neq1$, so we have found a primitive seventh root of unity $$\beta=\alpha^{38}=7\alpha^2+7\alpha+6.$$ I cheater and used Mathematica to tell me that $$x^{38}\equiv 7x^2+7x+6\pmod{\langle 11, x^3-4x^2+1\rangle}.$$ Anyway $1,\beta,\beta^2,\ldots,\beta^6$ is then a set of representatives of the cosets of the subgroup of seventh powers. – Jyrki Lahtonen May 18 '17 at 06:13
  • Thank you for your time sir, this fully answer my question. I guess i should start to use some program myself to do these computations – FBruzzesi May 19 '17 at 08:14

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