I'm trying to understand the quotient $\Bbb Z[\sqrt{47}]/(2, 1 +\sqrt{47})$, in order to find out whether or not $(2, 1 + \sqrt{47})$ is a prime ideal in $\Bbb Z[\sqrt{47}]$. I think it is but my calculations seem to be giving me something that doesn't agree with this, so either it isn't a prime ideal or I'm doing something very wrong.

Since $\Bbb Z[\sqrt{47}] \cong \Bbb Z[X]/(X^2 - 47)$, I'm writing

$$ (\Bbb Z[X]/(X^2 - 47))/(2, X^2 - 2X - 46) $$

where $X^2 - 2X - 46$ is a monic irreducible polynomial with $1 + \sqrt{47}$ as a root. Is this step correct? If so, then I think it follows that

$$ (\Bbb F_2[X]/(X^2))/(\overline{X^2 - 2X - 46}) $$

where $\overline{.}$ denotes the reduction map. The problem is, this is the zero-ideal in $\Bbb F_2[X]/(X^2)$, and this finite ring is not an integral domain, so the conclusion is that $(2, 1 + \sqrt{47})$ is not a prime ideal.

Which steps here (if any) are correct? Is any body able to show me how they might do it if it is not correct?

  • $\begingroup$ Your polynomial in the quotient ring becomes $1-2X$. Doesn't it? Because $X^2=47$ in your quotient ring. $\endgroup$ Dec 4, 2017 at 12:26
  • $\begingroup$ @stressed-out I'm not sure, if I reduce first then I have $(\Bbb Z[X]/(X^2 - 47))/(2, 1 - 2X)$, but after that I end up with just $(\Bbb F_2[X]/(X^2))/(1) \cong (0)$ $\endgroup$ Dec 4, 2017 at 12:28
  • $\begingroup$ Instead of taking the quotient you can directly look at the ideal $(2,1+\sqrt{47})$, exactly like here. $\endgroup$ Dec 4, 2017 at 12:35

2 Answers 2


Lets denote $R=\mathbb{Z}\left[\sqrt{47}\right]$, $I=\left<2\right>$ and $J=\left<1+\sqrt{47}\right>$. By the third isomorphism theorem


As @Quasicoherent pointed out, I have made a mistake saying that $\mathbb{Z}\left[\sqrt{47}\right]/\left<1+\sqrt{47}\right>\cong\mathbb{Z}$. In fact


since $\left<1+\sqrt{47}\right>\ni-\left(1-\sqrt{47}\right)\left(1+\sqrt{47}\right)=-\left(1-47\right)=46$ and


is an integer iff $a=-b$ iff it is $46b$ for some $b$. Thus


and $I+J=\left<2,1+\sqrt{47}\right>$ is prime.

  • $\begingroup$ It's not true that $\mathbb{Z}[\sqrt{47}]/\langle 1 + \sqrt{47} \rangle \cong \mathbb{Z}$. Note that $-46 = 1 - \sqrt{47}^2$ maps to $0$ under your proposed $\varphi$, so the map is not injective. Another way to see that your proposed isomorphism is impossible: $\mathbb{Z}[\sqrt{47}]$ has Krull dimension $1$, so the chain of primes $(0) \subsetneq (2, 1 + \sqrt{47})$ is already maximal, hence $(1 + \sqrt{47})$ can't be prime. $\endgroup$ Dec 4, 2017 at 15:31
  • 1
    $\begingroup$ @Quasicoherent Thanks! Your are of course correct. I have edited my answer. $\endgroup$
    – eranreches
    Dec 4, 2017 at 15:46

The ring is $$\mathbb{Z}[\sqrt{47}]/(2,1+\sqrt{47}) \cong(\mathbb{Z}[X]/(X^2-47))/(2,1+X) \cong\mathbb{Z}[X]/(X^2-47,2,1+X)\\ \cong \mathbb{Z}_2[X]/(X^2-47,1+X)= \mathbb{Z}_2[X]/(X^2-1,1+X) =\mathbb{Z}_2[X]/(1+X)\cong \mathbb{Z}_2$$ where $\mathbb{Z}_2 = \mathbb{Z}/(2)$.

How did you come to $(\Bbb Z[X]/(X^2 - 47))/(2, X^2 - 2X - 46)$ ?

It is $\cong \mathbb{Z}[\sqrt{47}]/(2,(1+\sqrt{47})^2)=\mathbb{Z}[\sqrt{47}]/(2) \cong \mathbb{Z}_2[X]/(1+X)^2$

  • $\begingroup$ Would you please, in layman terms, demonstrate why $(\mathbb{Z}[X]/(X^2-47))/(2,1+X) \cong\mathbb{Z}[X]/(X^2-47,2,1+X)$ is true? I'm new to all this. Thanks. $\endgroup$ Dec 4, 2017 at 13:56
  • $\begingroup$ @stressed-out $(R/I)/J \cong R/(I,J)$ is really obvious. $(I,J) = \{ a+b \in R, a \in I, b \in J\}$. Remember an element of $R/I$ is a subset of $R$, and an element of $(R/I)/J$ too. $\endgroup$
    – reuns
    Dec 4, 2017 at 14:01
  • $\begingroup$ @stressed-out That's just the third isomorphism theorem. Let $R=\mathbb{Z}\left[x\right]$, $I=\left<x^{2}-47\right>$, $J=\left<2\right>$ and $K=\left<1+x\right>$. Then the above statement is $$\left(R/I\right)/\left(\left(J+K+I\right)/I\right)\cong R/\left(J+K+I\right)$$ $\endgroup$
    – eranreches
    Dec 4, 2017 at 14:01
  • $\begingroup$ Thanks. That's handy. The abuse of notation was kind of confusing at first. $\endgroup$ Dec 4, 2017 at 14:04
  • $\begingroup$ If $J$ is an ideal of $R/I$ then it is a subset of $R/I$, thus a subset of a set of subsets of $R$, thus a subset of $R$, thus $JR= \{ab, a \in J, b\in R\}$ is an ideal of $R$, thus no abuse of notation here. @stressed-out $\endgroup$
    – reuns
    Dec 4, 2017 at 14:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.