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$$5 = (-1) \left( \frac{3 - \sqrt{29}}{2} \right) \left( \frac{3 + \sqrt{29}}{2} \right)$$

or

$$5 = \left( \frac{7 - \sqrt{29}}{2} \right) \left( \frac{7 + \sqrt{29}}{2} \right)?$$

$\mathcal{O}_{\mathbb{Q}(\sqrt{29})}$ is supposed to be a unique factorization domain, so the two above factorizations are not distinct. But I can divide factors in both of them by units to obtain yet more seemingly different factorizations.

The presence of the unit $-1$ in the first factorization does not trouble me, since for example on this page http://userpages.umbc.edu/~rcampbel/Math413Spr05/Notes/QuadNumbFlds.html the factorization of $3$ in $\mathcal{O}_{\mathbb{Q}(\sqrt{13})}$ is given as $(-1)(7 - 2 \sqrt{13})(7 + 2 \sqrt{13})$.

I honestly find rings of complex numbers far easier to understand!

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    $\begingroup$ I don't understand what you mean by "more valid." $\endgroup$ – Qiaochu Yuan Mar 30 '15 at 22:07
  • $\begingroup$ I was tempted to answer the one with the $-1$, but it's easy to accomplish the sign change with a judicious choice of units to multiply by. And heck, given that the fundamental unit in this ring is a so-called "half-integer," you could even argue that a factorization like $(11 - 2\sqrt{29})(11 + 2\sqrt{29})$ is also valid. $\endgroup$ – Robert Soupe Mar 31 '15 at 0:30
  • $\begingroup$ @Robert: Yes, because $\frac{3+\sqrt{29}}2\cdot\frac{5+\sqrt{29}}2=11+2\sqrt{29}$, so that's still the same prime. $\endgroup$ – hmakholm left over Monica Mar 31 '15 at 16:28
  • $\begingroup$ @Henning: Alright, what do you make of this factorization of 4 in $\mathcal{O}_{\mathbb{Q}(\sqrt{17})}$? $(-1)(8 - 2\sqrt{17})(8 + 2\sqrt{17})$ Isn't that less "fundamental" than this other one? $$\left(\frac{5}{2} - \frac{\sqrt{17}}{2}\right)^2 \left(\frac{5}{2} + \frac{\sqrt{17}}{2}\right)^2$$ Bob, you too feel free to chime in on this one. $\endgroup$ – Robert Soupe Apr 1 '15 at 0:43
  • $\begingroup$ @RobertSoupe: Yes, because $8\pm2\sqrt{17}$ is not irreducible. $\endgroup$ – hmakholm left over Monica Apr 1 '15 at 2:14
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In $\mathbb{Z}^+$, if $n > 1$ and $n = ab$, we expect that $n > a$ or $n > b$, maybe both. Our choices for $a$ and $b$ are considerably limited if $n$ is prime. With $n$ composite, we can choose $a$ and $b$ such that both $n > a$ and $n > b$ are true. When we broaden our view to all of $\mathbb{Z}$, we can make similar statements: if $|n| > 1$ and $n = ab$, then $|n| > |a|$ or $|n| > |b|$, maybe both.

In a ring like $\mathcal{O}_{\mathbb{Q}(\sqrt{29})}$, we can make $a$ or $b$ arbitrarily large. One of the comments to another answer alludes to the formula $$\left(\frac{5}{2} + \frac{\sqrt{29}}{2}\right)^k \left(\frac{3}{2} + \frac{\sqrt{29}}{2}\right),$$ with $k \in \mathbb{Z}$. But is it possible to choose $ab = 5$ such that both $|a| < 5$ and $|b| < 5$ hold true? In fact, ignoring multiplication by $-1$, there is only one possible choice: $$\left(-\frac{3}{2} + \frac{\sqrt{29}}{2}\right) \left(\frac{3}{2} + \frac{\sqrt{29}}{2}\right).$$ Any other choice can make $a$ very small, but $b$ concomitantly much larger than $5$. From this it is much easier to arrive at $$5 = (-1)\left(\frac{3}{2} - \frac{\sqrt{29}}{2}\right) \left(\frac{3}{2} + \frac{\sqrt{29}}{2}\right)$$ than it is to get to the factorization involving $\frac{7}{2}$. For this reason, I think the factorization with $\frac{3}{2}$ is more valid.

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  • $\begingroup$ This might not necessarily work in a ring like $\mathbb{Z}[\sqrt{6}]$, but it is the best effort I have seen among these answers in carrying over from $\mathbb{Z}$ what we feel is right about a factorization. $\endgroup$ – Bob Happ Apr 11 '15 at 18:55
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Note that $$ \frac{7-\sqrt{29}}2 \cdot \frac{5+\sqrt{29}}2 = \frac{3+\sqrt{29}}2 $$ and $$ \frac{5+\sqrt{29}}2 \cdot \frac{-5+\sqrt{29}}2 = 1 $$

So the two factorizations are equally good, just related by unit factors. (Remember that even in an UFD, factorizations are only unique up to associatedness).

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    $\begingroup$ Good. It’s really hard doing arithmetic in a real quadratic number field, ’cause the ominipresence of units thoroughly obscures the situation. $\endgroup$ – Lubin Mar 30 '15 at 23:06
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    $\begingroup$ These two factorizations are related by multiplication by a fundamental unit. Though I'm not sure I yet fully understand what makes a fundamental unit fundamental. But I could also be wrong about $\frac{5 + \sqrt{29}}{2}$ being a fundamental unit. $\endgroup$ – user153918 Apr 1 '15 at 14:30
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    $\begingroup$ @AlonsodelArte: I'm not sure it matters that the unit is fundamental. Unique factorizations are unique up to multiplication with any unit, not restricted to fundamental ones. (MathWorld agrees with you that this particular unit is fundamental, though). $\endgroup$ – hmakholm left over Monica Apr 1 '15 at 14:42
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    $\begingroup$ @AlonsodelArte: If someone's trying to single out one factorization as more valid or canonical than another (among ones related by unit factors), they're doing it wrong. $\endgroup$ – hmakholm left over Monica Apr 1 '15 at 17:49
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    $\begingroup$ They could be doing it wrong. Let's suppose one of the factorizations in the OP's question is the canonical one. Let's also suppose someone has somehow obtained a ridiculous but technically good factorization with $2373704925213782176 \pm 440785938820106883 \sqrt{29}$. Multiply that by the wrong unit and they'd get themselves further away from the canonical factorization. Fortunately there's only two directions to go here. Assuming there is a factorization that can be called more canonical, what would be the more efficient way of getting there? $\endgroup$ – David R. Apr 1 '15 at 21:13
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This question is a real humdinger. In $\frac{a + b \sqrt{29}}{2}$ its worthwhile to make the absolute values of it's $a$ and $b$ as small as possible, which makes the $3/2$ factorization look much better. But I feel the unit in that factorization is misleading because $5$ is a positive number, not a negative number. I think if a unit is included in a factorization, it should give you an idea as to what side of $0$ (or in what quadrant, in the case of a imaginary ring) the composite number is.

So I'm going to have to say the $7/2$ factorization is more valid, in my opinion.

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Fundamental units provide an important clue to this vexing question, but they also provide a red herring.

If $u$ is the fundamental unit in $\mathbb{Z}\left[\frac{1 + \sqrt{29}}{2}\right]$ and $p$ is a prime such that $|N(p)| = 5$, then it is $\pm u^n p$ that will lead us to all the other factorizations of $5$, not $\pm up^n$ (this is the red herring).

Then we must look to see if there is another property of $u$ that we can try to carry over to $p$. There is. If $u$ is the fundamental unit, then it is the smallest unit greater than $1$. Verify this numerically: $$u = \frac{5 + \sqrt{29}}{2} \approx 5.192582403567252.$$

See then that $$p = \frac{3 + \sqrt{29}}{2} \approx 4.192582403567252$$ but $$p = \frac{7 + \sqrt{29}}{2} \approx 6.192582403567252.$$ The first $p$ is closer to $1$ than the second $p$. Therefore the more valid factorization is $$(-1) \left( \frac{3 - \sqrt{29}}{2} \right) \left( \frac{3 + \sqrt{29}}{2} \right) = 5.$$

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  • $\begingroup$ I don't think proximity to $1$ is that important. But still a thumbs up from me. $\endgroup$ – Bob Happ Apr 11 '15 at 18:57
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I hesitated to answer because I thought I'd have to invent a scale with which to weigh factorizations by validity, and you might disagree with my priorities for the scale. Plus I have to define a bunch of terms, like "unnecessary," and you might not agree with my definitions. But then I thought the worst that could happen is that you and a few others downvote my answer, so here goes.

Given multiple factorizations that differ by multiplication by units, weigh them according to the following priorities:

  1. In the highest priority, the factorization contains no unnecessary multiplication by units. For example, $-6 = (-1)^{756478254} \times 2 \times 3$ has unnecessary units, but $-6 = (-1) \times 2 \times 3$ does not. Also, no unnecessary divisions that get canceled out by unnecessary factors. For example, do $(\sqrt{29})^2$ but not $$\left(\frac{0}{2} + \frac{2\sqrt{29}}{2}\right)^2.$$
  2. If it can be done while also satisfying the highest priority, the next priority is that the factorization conveys as much information as possible about the forms algebraic integers may take in the ring at hand, including, but not limited to, the presence of integers with negative norms and the presence of integers of the form $$\sum_{i = 1}^n \frac{a_i \theta^{i - 1}}{n},$$ where $n$ is the algebraic degree of $\theta$, and each $a_i \in \mathbb{Z}$.
  3. With the above priorities satisfied, the absolute value of each $a_i$ is the smallest possible.

Maybe you think I have my priorities backwards. Maybe you completely disagree with my concept and you hope an answer much better than the ones you've gotten so far comes along.

But if you agree with these priorities, then $$(-1)\left(\frac{3}{2} - \frac{\sqrt{29}}{2}\right)\left(\frac{3}{2} + \frac{\sqrt{29}}{2}\right) = 5$$ is the more valid factorization, by a nose.

  1. Both of the factorizations you presented satisfy the first priority.
  2. Only the first factorization you presented conveys the fact that numbers in $\mathcal{O}_{\mathbb{Q}(\sqrt{29})}$ can have negative norms, though both convey the fact that numbers in this domain can be of the form $$\frac{a_1}{2} + \frac{a_2\sqrt{29}}{2}.$$
  3. In both factorizations $|a_2| = 1$, but $|a_1|$ in the first factorization is smaller.
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  • $\begingroup$ First, wouldn't $(-2)\times 3$ be a better factorization than $(-1)\times 2\times 3$? A better example would be factoring $-4$ in the usual integers, where arguably $(-1)\times 2^2$ is better than $(-2)\times 2$ because the latter contains different factors related by units. $\endgroup$ – hmakholm left over Monica Apr 2 '15 at 14:02
  • $\begingroup$ Second, $2^2\left(\frac{\sqrt{29}}2\right)^2$ is not a valid factorization at all, because $\frac{\sqrt{29}}2$ is not an algebraic integer. $\frac{a+b\sqrt{29}}2$ is an algebraic integer if and only if $a$ and $b$ are both odd (or both even, in which case the division is redundant). $\endgroup$ – hmakholm left over Monica Apr 2 '15 at 14:04
  • $\begingroup$ Third, it is not clear what you mean by "conveying information" in your second criterion. Are you trying to say that a factorization is better if the factors look pedagogically instructive? That would be somewhat hard to make precise, I think. $\endgroup$ – hmakholm left over Monica Apr 2 '15 at 14:11
  • $\begingroup$ @HenningMakholm I think its a good answer but not worthy of the bounty. He doesn't make it clear enough that the invalid factorizations he presents are meant to be invalid. My answer won't be worth the bounty ether and I look forward to you ripping it apart. $\endgroup$ – user155234 Apr 2 '15 at 21:53
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    $\begingroup$ Yes, it does seem rather unnecessary, kind of like raising $-1$ to an odd exponent, or 1 to any exponent. $\endgroup$ – Robert Soupe Apr 3 '15 at 19:51

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