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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?

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  • $\begingroup$ Your polynomial in the quotient ring becomes $1-2X$. Doesn't it? Because $X^2=47$ in your quotient ring. $\endgroup$
    – stressed out
    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 $(\Bbb F_2[X]/(X^2))/(1) \cong (0)$ $\endgroup$
    – Edward Evans
    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. $\endgroup$
    – Dietrich Burde
    Dec 4, 2017 at 12:35

2 Answers 2

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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

$$\mathbb{Z}\left[\sqrt{47}\right]/\left<2,1+\sqrt{47}\right>=R/\left(I+J\right)\cong\left(R/J\right)/\left(\left(I+J\right)/J\right)$$

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

$$R/J=\mathbb{Z}\left[\sqrt{47}\right]/\left<1+\sqrt{47}\right>\cong\mathbb{Z}_{46}$$

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

$$\left(a+b\sqrt{47}\right)\left(1+\sqrt{47}\right)=\left(a+47b\right)+\left(a+b\right)\sqrt{47}$$

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

$$\mathbb{Z}\left[\sqrt{47}\right]/\left<2,1+\sqrt{47}\right>\cong\mathbb{Z}_{46}/\left<2\right>=\left(\mathbb{Z}/46\mathbb{Z}\right)/\left(2\mathbb{Z}/46\mathbb{Z}\right)\cong\mathbb{Z}/2\mathbb{Z}=\mathbb{Z}_{2}$$

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

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  • $\begingroup$ 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. $\endgroup$
    – Viktor Vaughn
    Dec 4, 2017 at 15:31
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    $\begingroup$ @Quasicoherent Thanks! Your are of course correct. I have edited my answer. $\endgroup$
    – eranreches
    Dec 4, 2017 at 15:46
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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 $\cong \mathbb{Z}[\sqrt{47}]/(2,(1+\sqrt{47})^2)=\mathbb{Z}[\sqrt{47}]/(2) \cong \mathbb{Z}_2[X]/(1+X)^2$

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  • $\begingroup$ 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. $\endgroup$
    – stressed out
    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. $\endgroup$
    – stressed out
    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

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