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Give an element of $ \mathbb{Z}[\sqrt{-17}] $ that is a product of two irreducibles and also a product of three irreducibles.

My thoughts so far:

Using the multiplicative norm $ N(a + b\sqrt{-17}) = a^2 + 17 b^2 $, we see that the units are precisely 1, -1. I can also see that there are no elements of norm $ 2,3,5,6,7,8,10,11,12,13,14,15... $. So if an element has norm 4 or 9 for example, then it is irreducible.

I don't really know where to go from here.

Any help appreciated. Thanks

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$18=2\cdot3\cdot3=(1-\sqrt{-17})(1+\sqrt{-17})$ –  yoyo May 6 '11 at 19:25
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Now that you have the reputation for it, don't forget to vote up answers you find useful and questions you find interesting (as a participant). Also, don't do it just yet (I'll explain why in a second), but don't forget to eventually accept an answer to your questions once you are satisfied. You should accept whatever answer you found most intersting/helpful, etc, by clicking on the checkmark you will see on the left margin, right under the links to vote up and down (you don't have the reputation to vote down yet). (cont...) –  Arturo Magidin May 6 '11 at 19:45
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(cont) You may want to wait before accepting answers to your recent questions because questions without an accepted answer tend to attract a bit more attention, and you may yet receive other answers that prove to be better/more interesting/more informative/ etc. Waiting a day or so is not amiss, but you'll want to eventually accept an answer to each of your questions once you are satisfied. –  Arturo Magidin May 6 '11 at 19:46

1 Answer 1

Hint. How much is $(1+\sqrt{-17})(1-\sqrt{-17})$? Can you express it as a product in a different way? Are all the factors you have in either factorization irreducible?

Added. Why consider this product? If $\mathbb{Z}[\sqrt{-d}]$, with $d$ an odd squarefree integer greater than $1$, is not a UFD, then $1+\sqrt{-d}$ will be part of a witness to this fact. You have $(1+\sqrt{-d})(1-\sqrt{-d}) = d^2+1$ is divisible by $2$, but neither $1+\sqrt{-d}$ nor $1-\sqrt{-d}$ are divisible by $2$ in $\mathbb{Z}[\sqrt{-d}]$. Also, $2$ is irreducible, because $a^2+db^2 = 2$ has no solutions when $d\gt 2$, so that shows that $2$ is an irreducible that is not a prime (since it divides a product but neither of the factors). So $1+\sqrt{-d}$ and $1-\sqrt{-d}$ are usually good sources of examples of things going wrong with factorizations into irreducibles in $\mathbb{Z}[\sqrt{-d}]$, when such things do indeed go wrong.

Coda. Bill Dubuque will no doubt give you a general way to approach this kind of problem once he gets around to it. As I noted in the comments, the above was not meant to be a "method", or an "algorithm", or a "solution", but merely the thought process that led me to consider that product before expending too much effort dissecting this particular problem. Since it immediately gave a solution to the desired problem, that was all she wrote.

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@Bill, Arturo is right - you should explain your way to deal with the problem instead of saying his is bad.. if his actually isn't the best way it will be clear to anyone who understands both. –  quanta May 6 '11 at 20:13
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@quanta: For odd $d\gt 1$, just as I said above: $(1+\sqrt{-d})(1-\sqrt{-d})$ is a multiple of $2$, but neither factor is a multiple of $2$, and $2$ is irreducible because no element can have norm $2$ under these circumstances. So $2$ is an irreducible that is not a prime, hence the ring is not a UFD. If $d$ is even this doesn't work, but you can consider $(2+\sqrt{-d})(2-\sqrt{-d}) = 4+d$, which is divisible by $2$, but neither factor is (and if $d\gt 2$, then no element has norm $2$ either). –  Arturo Magidin May 6 '11 at 20:18
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@quanta: I never claimed I had a "way". It seemed like a good thing to try first and see if it worked, before spending too much time thinking about it. It worked. I was asked why I tried it, I said why. –  Arturo Magidin May 6 '11 at 20:26
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@quanta: In the explanation in the prior comment, I'm assuming $d$ is squarefree for the argument. I'm not including $d=1$ and $d=2$, but in those cases $\mathbb{Z}[\sqrt{-d}]$ is a UFD; and of course, if $d\equiv 3\pmod{4}$, then $\mathbb{Z}[\sqrt{-d}]$ is not the ring of integers of $\mathbb{Q}(\sqrt{-d})$. –  Arturo Magidin May 6 '11 at 20:32
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To answer Bill: I never claimed I had a "method"; I was asked what let me to consider that product, and I said what led me to consider that product. If Bill thinks this is not the right way to think about, I have no objection; if he thinks it's the wrong way to think about it, he may very well be correct; that it worked "by luck" and not by design, I am also quite happy to grant. To claim that the reply "has nothing to do with the problem" suggests that I what I wrote has nothing to do with what led me to try that product, and I find that an astounding statement, whether intended or not. –  Arturo Magidin May 7 '11 at 0:11

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