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Let $R$ be the subring of the real numbers such that $R=\left\{a+b\sqrt{2}:a,b \in \Bbb Z\right\}$ Let $M$ be the ideal in $R$ given by $M=\left\{a+b\sqrt{2}:\text{ a and b are divisible by 5}\right\}$ Prove that $M$is a maximal ideal of $R$.

I was thinking that suppose $M \subseteq P\subseteq R$ and prove that $1\in P$ but I don't know how to prove that. Hope somebody can help me with this

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Assume that $M$ is not maximal. So $M\subset P\subset R$. Let $a\in P$, with $a$ an integer such that $5$ does not divide $a$. We have that $5\in M\subset P$. By the ideal property this means that $5+a\in P$. But we also have that $\gcd(5,a)=1$. So we know there exists an integer combination of $5$ and $a$ summing to 1. Say

$$5n+am=1 $$

$5n,am\in P$ Therefore, since an ideal is closed under addition we have that $1\in P$ and therefore, $P=R$. Therefore, no proper ideal of $R$ contains $M$. Also, $M$ is not the whole ring because it doesn't contain $1$. Thus $M$ is maximal. I feel like there may be a hole to be found, so I'm open to suggestions.

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  • $\begingroup$ In fact I really don't think my answer works, because $5$ is not in $M$. $\endgroup$ – monroej Jan 21 '14 at 5:12
  • $\begingroup$ yes it is: $\;5=5+0\sqrt2\;,\;\;5\mid 5\;,\;5\mid 0\;$ $\endgroup$ – DonAntonio Jan 21 '14 at 5:13
  • $\begingroup$ Ok. I don't know why I was thinking $5$ didn't divide zero...Thanks. $\endgroup$ – monroej Jan 21 '14 at 5:13
  • $\begingroup$ I don't think you can assume such an $a$ exists. $\endgroup$ – Dustan Levenstein Jan 21 '14 at 5:44
  • $\begingroup$ If no such $a$ existed, then $M$ would be maximal from the start. $\endgroup$ – monroej Jan 21 '14 at 5:45
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Try the following :

  1. Prove that $(x^2-2) \in \mathbb{Z}_5[x]$ is irreducible, and let $F = \mathbb{Z}_5[x]/(x^2-2)$ be the corresponding quotient, which is a field.

  2. Define a surjective homomorphism $\varphi : R\to F$ such that $\ker\varphi = M$.

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