0

I want to prove that any element $a+b{\omega}$ in ${\bf Z}[{\omega}]$ is congruent to either $0$, $1$, or $-1$ in the quotient ring ${\bf Z}[{\omega}]/(1-{\omega})$. Here $(1-{\omega})$ is the ideal generated by $1-{\omega}$. We recall that ${\bf Z}[{\omega}]$ is a Euclidean ring and is hence a principal ideal domain.

My attempt is as follows. I am not entirely sure about my reasoning! In the quotient ring, since $1={\omega}$, $1={\omega}^2$, and so $1+{\omega}+{\omega}^2$ must be $3$. But $1+{\omega}+{\omega}^2=0$ in ${\bf Z}[{\omega}]$ (remember $\omega$ is a complex cube root of unity). Since $3=0$ in this quotient ring, and $\omega = 1$, any element $a+b{\omega}$ must be congruent to $a+b$ modulo the ideal $(1-{\omega}).$ Therefore dividing $a+b$ by $3$ yields one of the remainders $0$, $1$, 0r $-1$. Where am I making a mistake? (Perhaps irrationally), I feel uncomfortable about concluding that $3=0$ in the quotient ring.

student
  • 1,324
  • 2
    What makes you think you’re making a mistake? – J. W. Tanner Jan 09 '24 at 17:23
  • 1
    For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Jan 09 '24 at 17:28
  • Sorry. Will edit my question underscoring the precise point which troubles me! – student Jan 09 '24 at 17:35
  • 1
    You essentially used $w\equiv 1\Rightarrow f(w)\equiv f(1),$ so $, 0 \equiv w^2+w+1\equiv 1^2+1+1,,$ which is rigorous - being a special case of the Polynomial Congruence Rule, see the linked dupe. $\ \ $ – Bill Dubuque Jan 09 '24 at 18:39
  • 1
    i.e. we're working modulo the ideal $,I = (x!-!1,\color{#c00}{f(x)}) = (x!-!1,\color{#c00}{f(1)}) = (x!-!1,3),,$ by $,\color{#c00}{f(x)\equiv f(1)}\pmod{x!-!1}.,$ Finally $,\Bbb Z[x]/I :!\cong, \Bbb Z/3,$ as here. $\ \ $ – Bill Dubuque Jan 09 '24 at 19:02

1 Answers1

1

$(1-\omega)(2+\omega)=2-\omega-\omega^2=3-(1+\omega+\omega^2)=3$, so $3 \in (1-\omega)$.

Robert Shore
  • 23,332
  • Cleared up doubts!! Should I withdraw the question? – student Jan 09 '24 at 17:53
  • 1
    No reason that I can see to do so. – Robert Shore Jan 09 '24 at 17:53
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Jan 09 '24 at 18:42
  • 1
    @stu This answer can be viewed more conceptually as computing the norm of $,1!-!w,,$ i.e. $,N(1!-!w) = (1!-!w)(1!-!\bar w) = (1-w)(2+w) = 3,,$ since the conjugate $\bar w$ satisfies $,\bar w + w = -\color{#c00}1,$ by $,w^2+ \color{#c00}1 w + 1 = 0,,$ i.e. sum of roots $= -\color{#c00}1,$ by Vieta. But it is wasteful to compute the "Bezout" coefs $,g_i,$ that verify the ideal membership $,3 = (x!-!1) g_1 + f g_2\in (x!-!1,f)$ since we need only the "remainder" - not the "quotients", so congruence (or modular) arithmetic is simpler (much simpler for larger problems) – Bill Dubuque Jan 09 '24 at 19:20
  • Bill, should I withdraw this question? I genuinely had this doubt, but can see how it could be misinterpreted. – student Jan 09 '24 at 19:27
  • 1
    @BillDubuque (a) When I answered the question (roughly an hour after it had been asked), no one had identified it as a dupe. $$$$ (b) Based on the suggested dupe, I still don't think it's a dupe. Getting from the suggested dupe to an answer to the specific question asked requires some reasoning. I see the mathematical equivalence, of course, but that doesn't make it a dupe to the question that's been asked. $$$$ (c) I respectfully suggest that questions tagged as "abstract algebra" are closed as dupes somewhat too frequently, based on concerns similar to those I've expressed here. – Robert Shore Jan 09 '24 at 23:34
  • @RobertShore The answer follows immediately by applying the polynomial remainder theorem - as explained in the closing comment on the question. Said timing does not change the dupe status. – Bill Dubuque Jan 12 '24 at 15:53