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I am teaching an elementary number theory class, mainly for non-technical majors. Today, I mentioned $\mathbb Z[\sqrt{-5}]$ to show that unique prime factorization is not "obvious."

Is there some application (e.g. finding solutions to some Diophantine equation) they could understand which shows why mathematicians might care about the failure of unique prime factorization in $\mathbb Z[\sqrt{-5}]$?

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  • $\begingroup$ The most obvious thing would be to study equations of the form $x^2+5y^2=k$. If $k$ is composite, say $k=ab$, then you could obtain such solutions by solving $x^2+5y^2=a$ and $x^2+5y^2=b$, and then combining such solutions in a suitable way. It seems the failure of UPF might alter what that algorithm looks like, as compared to solving the otherwise-similar $x^2+2y^2=k$. $\endgroup$
    – G Tony Jacobs
    Commented Nov 3, 2016 at 16:06
  • $\begingroup$ Ah yes! I remember that solutions to $p = x^2 + y^2$ (for $p$ prime) could be characterized using UPF of $\mathbb Z[i]$. Is there something that could be said for $p = x^2 + 5y^2$? $\endgroup$
    – Alan C
    Commented Nov 3, 2016 at 16:48
  • $\begingroup$ It relies on this: The norm of a number in $\mathbb{Z}[\sqrt{-5}]$ is given by the formula: $N(a+b\sqrt{-5}) = a^2 + 5b^2$. Thus, solving $a^2+5b^2=k$ is equivalent to solving $N(\alpha)=k$ for $\alpha$. Then you use the fact that the norm is a multiplicative function: $N(xy)=N(x)N(y)$. So, if $k=mn$, and we want to solve $N(\alpha)=k$, we can find $\beta$ and $\gamma$ so that $N(\beta)=m$ and $N(\gamma)=n$. That's how factorizations come into it. $\endgroup$
    – G Tony Jacobs
    Commented Nov 3, 2016 at 17:00

2 Answers 2

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Actually, concerning Diophantine equations we have that Fermat's equation $$ x^p+y^p=z^p $$ has no non-trivial solution for $p>2$, with an easy proof if the ring of integers $\mathbb{Z}[\zeta_p]$ of the cyclotomic field $\mathbb{Q}[\zeta_p]$ is UFD. In this context it is very enlightening that the rings $\mathbb{Z}[\zeta_p]$ have class number one, i.e., are PIDs and hence factorial if and only the prime satisfies $p\le 19$. Unfortunately, $\mathbb{Z}[\sqrt{-5}]$ is not the ring of integers of a cyclotomic field. But Fermat shows that mathematicians do care for "similar" rings being PIDs or UFDs. For example, the ring of integers $\mathbb{Z}[\zeta_3]$ for exponent $p=3$, see the introduction here.

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One manifestation involves primitive solutiins for $x^2+5y^2$, where $x$ and $y$ are whole numbers. "Primitive" means $x$ amd $y$ are relatively prime.

We cannot in general factor such numbers into prime factors having the form above. For instance, we have

$1189=41×29\implies8^2+(5×15^2)=[6^2+(5×1^2)][3^2+(5×2^2)]$

but

$6=3×2\implies1^2+(5×1^2)=\text{no }x^2+5y^2\text{ factors}$

This shows up as $1189$ actually having a unique factorization in $\mathbb{Z}[\sqrt{-5}]$ but $6$ famously does not. The nonunique factorization of $6$ goes along with it having prime factors that fail to have the form $x^2+5y^2$.

In class field theiry we learn that $\mathbb{Z}[\sqrt{-5}]$ has class number $2$. With domains corresponding to positive definite quadratic forms this implies that full factorability of numbers having the given quadratic form ($x^2+5y^2$) can be gained by introducing a second quadratic form. In the case of $x^2+5y^2$ we may choose $2x^2+2xy+3y^2$ as the second form. Then we can factor such numbers as $6$:

$6=3×2\implies1^2+(5×1^2)=[(2×0^2)+(2×0×1)+(3×1^2)][(2×1^2)+(2×1×0)+(3×0^2)]$

Introducing the second quadratic form to repair the prime factorization of $x^2+5y^2$ numbers corresponds to adding another set of algebraic integers to $\mathbb{Z}[\sqrt{-5}]$ such that the second set has norms with this alternate form. One such set is given by $\frac12(a\sqrt2+b\sqrt{-10})$. With this augmentation $\mathbb{Z}[\sqrt{-5}]$ is converted to a UFD.

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