I want to prove/disprove that $\mathbb{Z}[\sqrt{2}]/\langle 1 + \sqrt{2} \rangle$ is a field in two ways:

Let $R = \mathbb{Z}[\sqrt{2}]$ and $I = \langle 1 + \sqrt{2} \rangle$.

1) Using the First Ring Isomorphism Theorem, I need to find a ring homomorphism $\phi: R \rightarrow S$ with $I$ as its kernel in order to get $$R/I \cong \text{Im} (\phi)$$ and determine if $\text{Im} (\phi)$ is a field, but I'm unsure of how to construct what $S$ should be and how to construct $\phi$.

2) Using the fact that $R/I$ is a field iff $I$ is a maximal ideal in $R$, I'm thinking that $$R/I = \mathbb{Z}[\sqrt{2}]/\langle 1 + \sqrt{2} \rangle \cong \mathbb{Z}[x]/\langle x^{2}-2,x^{2}-2x-1 \rangle$$ and ideally (heh) I'd have $x^{2}-2x-1 \mid x^{2} - 2$ so I'd be working in with $\mathbb{Z}[x]/ \langle x^{2}-2x-1 \rangle$, in which I'd see if $x^{2}-2x-1$ is irreducible and thus if $R/I$ is a field. However, I feel as if working directly with $\mathbb{Z}[\sqrt{2}]/\langle 1 + \sqrt{2} \rangle$ is what was intended for the problem.

Any advice on how to go about both methods is greatly appreciated.

  • $\begingroup$ Irreducibility of $x^2-2x-1$ would not imply $\mathbb{Z}[x]/(x^2-2x-1)$ is a field, since $\mathbb{Z}$ is not a field. $\endgroup$ – Eric Wofsey Jan 17 '17 at 4:44
  • 3
    $\begingroup$ You're right to try to use the 3rd Isomorphism Theorem. That should be $R/I = \mathbb{Z}[\sqrt{2}]/\langle 1 + \sqrt{2} \rangle \cong \mathbb{Z}[x]/\langle x^{2}-2,1 + x \rangle$, though. $\endgroup$ – Richard D. James Jan 17 '17 at 4:45
  • $\begingroup$ $(x^2-2x-1)$ is not maximal ideal in $\mathbb Z[x]$ because the maximal ideals in $\mathbb Z[x]$ are the forms $(p,f(x))$ where $f(x)$ is irreducible in $\mathbb Z_p$ $\endgroup$ – Mustafa Jan 17 '17 at 14:47

Hint: One particular element of the ideal $I$ is $(1-\sqrt{2})(1+\sqrt{2})$. What does this tell you about $I$?

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