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I was worried that my proof isn't right so I want to know if there are any mistakes in this and if this way can work? Thank you very much.

We want to show every maximal ideal $\mathfrak m$ of $\mathbb Z[\varphi]$ is principal. My idea is that the quotient $\mathbb Z[\varphi]/\mathfrak m$ must be a finite field, so we can use the classification of finite fields to produce a principal ideal equal to $\mathfrak m$.


Some lemmas which are used.

Lemma $\mathbb Z[\varphi]/\mathfrak m$ is isomorphic to either $\mathbb F_p$ (case 1) or $\mathbb F_{p^2}$ (case 2) (some $p$). proof: The quotient is a field since the ideal is maximal and the field must be finite (and have degree $\le 2$) because it's the quotient of a finite dimensional space ($\mathbb Z[\varphi]$ has dimension $2$) and doesn't contain $\mathbb Q$.

Edit, the next lemma is wrong but I don't use it anymore.

Lemma Suppose $\mathbb Z[\varphi]/\mathfrak m \simeq \mathbb Z[\varphi]/\mathfrak m'$ is an isomorphism then $\mathfrak m = \mathfrak m'$. proof: In general $R/I \simeq R/J$ implies $J$ is an automorphism of $R$ applied to $J$ but $\mathbb Z[\varphi]$ has no automorphisms.

Lemma $p$ is of the form $x^2 + xy - y^2$ iff $X^2 - X - 1$ has a solution mod $p$. proof: I don't actually prove this but I assume it could be done like this.

Note, if $p$ is of that form it can be written that way for infinitely many $(x,y)$ but $x+\varphi y$ will be an associate of $x'+\varphi y'$ or $x'+\bar\varphi y'$ if so.


The main theorem:

Lemma In case 1, $X^2 - X - 1$ has a solution $\mod p$. proof: The quotient map is a ring homomorphism.

Lemma If $X^2 - X - 1$ has a solution $\mod p$ then case 2 is impossible. proof: The basis $\{1,\varphi\}$ collapses as the image of $\varphi$ is expressible as an integer multiple of $1$, so $M$ is one dimensional.

Theorem $\mathfrak m$ is principal. proof:

In case 2, $\mathbb Z[\varphi]/(p)$ has size $p^2$ and it is the only ideal which gives a finite field of this size so $\mathfrak m = (p)$.

In case 1, Let $x$,$y$ such that $p = x^2 + xy - y^2 = (x + \varphi y)(x + \bar \varphi y)$ and both quotients like $\mathbb Z[\varphi]/(x + \varphi y)$ have size $p$ and there's no other way to quotient to get a field of the right size so either $\mathfrak m = (x + \varphi y)$ or $\mathfrak m = (x + \bar \varphi y)$ for any such associate pair $x$,$y$ (it doesn't matter which you choose).

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Do you know anything about the ideal class group? Your general approach of using the fact that the residue fields are all finite is not going to be a good strategy in general for rings like this, because such quadratic rings always have finite residue fields but many of them are not PIDs. –  KCd Nov 20 '12 at 1:39
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Dear sperners lemma, Your second lemma is not true. Your third lemma is the key point, and note that it uses more than just the finiteness of the residue fields (and thus gets around @KCd's objection above) --- the estimates needed to prove it will be sensitive to the particular value of the norm of $\varphi$, and related facts. Regards, –  Matt E Nov 20 '12 at 1:46
    
@KCd, I have studied unique factorization of ideals and the class group being size 1 iff the integers have unique factorization. –  sperners lemma Nov 20 '12 at 13:30
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@spernerslemma: If you know about the ideal class group, and have ever done calculations of it, then I think a better way to approach your problem is to show the class number is 1 (thus showing all ideals are principal) rather than to focus only on the maximal ideals. –  KCd Nov 20 '12 at 16:06
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Nowhere have you said what $\phi$ means. If it means $(1+\sqrt5)/2$, then just the other day I posted an approach to proving your ring is Euclidean (a fortiori, a PID). See math.stackexchange.com/questions/240700/… –  Gerry Myerson Nov 21 '12 at 5:14
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