An exercise asks me to show that $3(1+\sqrt{-5})$ and $3(1-\sqrt{-5})$ have no greatest common divisor in $\mathbb Z[\sqrt{-5}]$.

I think I have to find two maximal common divisors which are not associated. I know 3 is one, however, I cannot find another one. Could anyone give me a hint?

  • 2
    $\begingroup$ Hint: ignore the 3s. $\endgroup$ Feb 9, 2012 at 2:24
  • $\begingroup$ Think about norms; any common divisor must have a norm dividing $54$; cannot be $54$, since the two elements are not associates; no elements have norm $2$ or $3$; so common divisors have norm either $6$ or $9$. $\endgroup$ Feb 9, 2012 at 2:34
  • $\begingroup$ It's a bad idea to put two different things in the title and the body of the question. As far as I can tell, Thomas' hint helps you with what the title seems to suggest the question is about, but not with what the body says you want to do. For that, I'd suggest labelling the innermost grid points with their squared magnitudes and looking for two for which the product of the squared magnitudes is that of, say, $3(1+\sqrt{-5})$. There aren't many options. $\endgroup$
    – joriki
    Feb 9, 2012 at 2:34
  • $\begingroup$ @joriki, I thought that this was a good title. I'll think of a new one. $\endgroup$
    – sxd
    Feb 9, 2012 at 2:35
  • 1
    $\begingroup$ @Dimitri Surinx: Persist!. You have very few candidates left, just $1\pm\sqrt{-5}$ and $2\pm\sqrt{-5}$. Let's look at $1+\sqrt{-5}$. It divides the first guy. What about the second? Do the division. $\endgroup$ Feb 9, 2012 at 3:15

2 Answers 2


Now, let me show off some of the nice number theory.

From the point of view of Kummer and his "ideal numbers", it is not hard to verify that $2$ behaves like "the square of a prime" in the sense that it satisfies the following two properties in $\mathbb{Z}[\sqrt{-5}]$:

  1. If $x|a^2b^2$, then $x|a^2$ or $x|b^2$, for all $a,b\in\mathbb{Z}[\sqrt{-5}]$;
  2. There exists $a\in\mathbb{Z}[\sqrt{-5}]$ such that $x|a^2$ but $x$ does not divide $a$.

In the integers, these two properties together characterize the integers that are squares of primes; there is no such prime in $\mathbb{Z}[\sqrt{-5}]$ whose square is $2$, but let's ignore that annoying fact and "invent" an "ideal" prime number ('ideal' as in "imaginary") $\alpha$ such that $\alpha^2=2$.

Similarly, each of $3$, $1+\sqrt{-5}$, and $1-\sqrt{-5}$ behaves like the "product of two distinct primes", in the sense that they satisfy the following three properties:

  1. If $x|abc$, then $x$ divides at least one of $ab$, $ac$, and $bc$, for all $a,b,c\in\mathbb{Z}[\sqrt{-5}]$.
  2. If $x|a^2$, then $x|a$ for all $a\in\mathbb{Z}[\sqrt{-5}]$.
  3. There exist $a,b\in\mathbb{Z}[\sqrt{-5}]$ such that $x$ divides $ab$, but does not divide $a$ and does not divide $b$.

Again, these three conditions would characterize integers that are products of two distinct primes in $\mathbb{Z}$, but no such elements exist in $\mathbb{Z}[\sqrt{-5}]$. We again "invent" such primes. Since we are hoping that unique factorization might hold in terms of these "ideal" primes, and since $$2\times 3 = (1+\sqrt{-5}) \times (1-\sqrt{-5})$$ this suggests that there are two of these "ideal primes", $\beta$ and $\gamma$, with $3=\beta\gamma$, $1+\sqrt{-5}=\alpha\beta$, and $1-\sqrt{-5}=\alpha\gamma$. Though it is not easy, one can verify that, at least as far as divisibility is concerned, everything works out nicely, so long as we don't actually go looking for these "ideal primes". Some products of these ideal primes will yield actual numbers, some products will yield "ideal" numbers. (Doing lots of experimentation may even reveal that any product of an even number of "ideal primes" is an actual number, whereas products of an odd number of "ideal primes" is never an actual number).

If this is the case, then $3(1+\sqrt{-5}) = \alpha\beta^2\gamma$, and $3(1-\sqrt{-5})=\alpha\beta\gamma^2$. So the greatest common divisor in terms of these ideal primes will be $\alpha\beta\gamma$. That means that each of $\alpha\beta=1+\sqrt{-5}$, $\alpha\gamma=1-\sqrt{-5}$, and $\beta\gamma=3$ will be common divisors of $3(1+\sqrt{-5})$ and $3(1-\sqrt{-5})$. And indeed, $3$ clearly divides both; $(1+\sqrt{-5})$ divides the first, and $$(1+\sqrt{-5})(-2-\sqrt{-5})=3-3\sqrt{-5} = 3(1-\sqrt{-5});$$ and similarly, $1-\sqrt{-5}$ divides the second and $$(1-\sqrt{-5})(-2+\sqrt{-5}) = 3+\sqrt{-5} = 3(1+\sqrt{-5}).$$ So the common divisors in $\mathbb{Z}[\sqrt{-5}]$ are $\pm 1$, $\pm 3$, $\pm(1+\sqrt{-5})$, and $\pm(1-\sqrt{-5})$. (The product of three ideal primes is not an actual number, so we can find no actual counterpart to $\alpha\beta\gamma$ in $\mathbb{Z}[\sqrt{-5}]$). None of them is a multiple of all the others.

In terms of the ideal class group, from the point of view of Dedekind and Kronecker, the ring $\mathbb{Z}[\sqrt{-5}]$ is a Dedekind domain; every ideal can be written as a product of prime ideals in a unique way. It is not hard to show that: $$\begin{align*} (2) &= (2,1+\sqrt{-5})^2,& (3)&=(3,1+\sqrt{-5})(3,1-\sqrt{-5}),\\ (1+\sqrt{-5}) &= (2,1+\sqrt{-5})(3,1+\sqrt{-5}) & (1-\sqrt{-5})&=(2,1-\sqrt{-5})(3,1-\sqrt{-5}) \end{align*}$$ Thus, $$\begin{align*} (3)(1+\sqrt{-5}) &= (2,1+\sqrt{-5})(3,1+\sqrt{-5})^2(3,1-\sqrt{-5})\\ (3)(1-\sqrt{-5}) &= (2,1+\sqrt{-5})(3,1+\sqrt{-5})(3,1-\sqrt{-5})^2. \end{align*}$$ So the gcd, in terms of ideals, is the (nonprincipal) ideal $$(2,1+\sqrt{-5})(3,1+\sqrt{-5})(3,1-\sqrt{-5}).$$ Since the ideal class group of $\mathbb{Z}[\sqrt{-5}]$ is cyclic of order $2$, any product of two nonprincipal ideals is principal; the three possible products of two ideal factors of the gcd above yield the three non-unit common divisors mentioned above.


There are slicker ways to proceed, but let us compute. Each of $3(1+\sqrt{-5})$ and $3(1-\sqrt{-5})$ has norm $54$. Make a list of all the elements of $\mathbb{Z}[\sqrt{-5}]$ whose norm divides $54$. Since we are dealing with divisibility, of any two associates you need only mention one. The list will be short.

Among the numbers in your list, which ones divide both $3(1+\sqrt{-5})$ and $3(1-\sqrt{-5})$? Settle this question maybe the hard way, by actually dividing. Recall the trick of multiplying the top and bottom by the conjugate of the bottom. Be careful, there may be a mild surprise.

Show by calculation that for every one of the common divisors $x$ of $3(1+\sqrt{-5})$ and $3(1-\sqrt{-5})$ in your list, there is a common divisor $y$ in your list such that $y$ does not divide $x$. Your list of common divisors will be short, so this task will not take long.


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