Show that $2,3, 1-\sqrt{-5}, 1+\sqrt{-5}$ are irreducible over $\mathbb{Z}[\sqrt{-5}]$, but not prime and that 1 and -1 are the only units.

Let $N$ be the norm map into $\mathbb{Z}$ and let u denote the unit, then because it is a homomorphism it follows that :

$N(xy)=N(x)N(y)$ and thus with $N(u)N(u^{-1})=N(uu^{-1})=N(1)$ it follows that the norm of u is a divider of 1 and thus also a unit of $\mathbb{Z}$. Therefore we have in $\mathbb{Z}[\sqrt{-5}]$: $\forall a,b \in \mathbb{Z}: N(a+b\sqrt{-5})=(a+b\sqrt{-5})(a-b\sqrt{-5})=a^{2}+5b^{2}=1 \ \ \Rightarrow u= \pm 1$

Now it is to show that $2,3,1-\sqrt{-5}, 1+\sqrt{-5}$ are irreducible over $\mathbb{Z}[\sqrt{-5}]: $

Assume $1-\sqrt{-5}$ is reducible, then there must exist $a,b \in \mathbb{Z}[\sqrt{-5}]$ so that $N(1-\sqrt{-5})=N(a)N(b) \Rightarrow N(a)=N(b)= \pm (1-\sqrt{-5})$ But since $1-\sqrt{-5}$ is not a quadratic remainder of $5$, there doesn't exist a solution for the equations $a^{2}+5b^{2}= \pm (1-\sqrt{-5})$ And the exactly same argument also works for $2,3$ and $1+\sqrt{-5}$.

Thus we have shown that $2,3,1\pm \sqrt{-5}$ are not reducible over $\mathbb{Z}[\sqrt{-5}]$.

Now we show that they are not prime:

Assume that 2 is a prime in $\mathbb{Z}[\sqrt{-5}]$, then because of $2\cdot 3 = (1-\sqrt{-5})(1+\sqrt{-5})=6$ it must hold that $2|(1-\sqrt{-5})$ or $2|(1+\sqrt{-5})$. But with $a,b \in \mathbb{Z}[\sqrt{-5}]$ it immediately follows that for :

$(1\pm\sqrt{-5})= 2(a+b\sqrt{-5})$ $2b = \pm 1$. So 2 is not a prime in $\mathbb{Z}[\sqrt{-5}]$.

Assume that 3 is a prime in $\mathbb{Z}[\sqrt{-5}$, then: $(1\pm \sqrt{-5}) = 3(a+b\sqrt{-5}) \Rightarrow 3b= \pm 1$ it follows that 3 is not a prime in $\mathbb{Z}[\sqrt{-5}]$.

Assume that $1\pm \sqrt{-5}$ is a prime in $\mathbb{Z}[\sqrt{-5}]$, then: $3 = (1\pm \sqrt{-5})(a+b\sqrt{-5}) \Rightarrow (1\pm \sqrt{-5} )a = 3$ which is not solvable in $\mathbb{Z}$. And also $2=(1\pm \sqrt{-5})(a+b\sqrt{-5})=(1\pm \sqrt{-5})a$ which is also not solvable in $\mathbb{Z}$ and thus $(1\pm \sqrt{-5})$ can not be a prime in $\mathbb{Z}[\sqrt{-5}]$.

Tell me if this proof is correct. Please.

  • 2
    $\begingroup$ There are a number of errors. The norm is an integer. And the usual quadratic residues/non-residues are ordinary integers. For example to show $1+\sqrt{-5}$ is irreducible, calculate the norm. The only non-trivial factors are $2$ and $3$, and it is obvious we cannot have $x^2+5y^2=2$ (or $3$), no need to appeal to fancier stuff. $\endgroup$ – André Nicolas Oct 29 '11 at 21:10
  • $\begingroup$ The proof that $1-\sqrt{-5}$ is not prime is not clear. What you should do is show say that $1-\sqrt{-5}$ divides $6$ (easy), but divides neither $2$ nor $3$. If it divided $2$, its norm $6$ would divide the norm of $2$, which is $4$, in the ordinary integer sense. But $6$ does not divide $4$. Neither does it divide $9$, the norm of $3$. $\endgroup$ – André Nicolas Oct 29 '11 at 21:21
  • $\begingroup$ the proof about units is clear .? $\endgroup$ – VVV Oct 29 '11 at 21:26
  • $\begingroup$ It could be clearer, but there are no mistakes. You should say something like this. Clearly $1$ and $-1$ are units. We show there are no others. If $u$ is a unit, then $uv=1$ for some $v$. Then $N(uv)=N(u)N(v)=1$ and therefore $N(u)=1$. But $N(a+b\sqrt{-5})=a^2+5b^2$, and this can be $1$ only for $a=\pm 1$, $b=0$. $\endgroup$ – André Nicolas Oct 29 '11 at 21:35
  • $\begingroup$ @VVV: There is no need to "sign" your posts: every post you make has your user name on the bottom right corner, which acts like a "signature" $\endgroup$ – Arturo Magidin Oct 29 '11 at 22:26

The proof that $1-\sqrt{-5}$ is irreducible is incorrect. In addition to a minor (and alas, common) mistake in writing that makes what you write not what you intended to write, there is an assertion which is just plain false.

Explicilty, you write:

Assume $1-\sqrt{-5}$ is reducible, then there must exist $a,b \in \mathbb{Z}[\sqrt{-5}]$ so that $N(1-\sqrt{-5})=N(a)N(b) \Rightarrow N(a)=N(b)= \pm (1-\sqrt{-5})$

First: the use of $\Rightarrow$ is incorrect. What you have written is that there exist $a$ and $b$ in $\mathbb{Z}[\sqrt{-5}]$ for which the following statement holds:

If $N(1-\sqrt{-5}) = N(a)N(b)$, then $N(a)=N(b) = \pm (1-\sqrt{-5})$.

What you actually wanted to write was that

If there exist $a,b\in\mathbb{Z}[\sqrt{-5}]$ such that $N(1-\sqrt{-5}) = N(a)N(b)$, then $N(a)=N(b)=\pm(1-\sqrt{-5})$.

What's the difference? The first statement will be true if you can find an $a$ and a $b$ for which $N(1-\sqrt{-5})$ is not equal to $N(a)N(b)$! It will be true that the implication holds, because the antecedent will be false. So, exhibiting $a=7$ and $b=10578432$ makes the statement you wrote true. However, they are irrelevant towards the second statement (and towards establishing what you want to establish, namely, that no such $a$ and $b$ exist).

Second: this is incorrect. What you want to assume is that there exist $a$ and $b$ such that $1-\sqrt{-5} = ab$, and neither $a$ nor $b$ are units; you do not simply want to assume that the product of the norms of $a$ and $b$ equals the norm of $1-\sqrt{-5}$.

Third: Even so, you conclusion is nonsense. The norm of any element of $\mathbb{Z}[\sqrt{-5}]$ must be an integer. What you want to conclude is that $N(a)N(b) = N(1-\sqrt{-5}) = 6$, and then get a contradiction. The norm cannot equal $\pm(1-\sqrt{-5})$.

It's also nonsense to say "since $1-\sqrt{-5}$ is not a quadratic remainder modulo $5$". It's not even a remainder modulo $5$, because it's not an integer!

The argument about $2$ not being prime is likewise incorrect in its use of the norm, which quickly reduces to nonsense symbols being strewn around:

But with $a,b \in \mathbb{Z}[\sqrt{-5}]$ it immediately follows that for : $(1\pm\sqrt{-5})= 2(a+b\sqrt{-5})$ $2b = \pm 1$. So 2 is not a prime in $\mathbb{Z}[\sqrt{-5}]$.

This is nonsensical as written. I suspect you wanted to say something like

But there cannot exist $a,b\in\mathbb{Z}$ (not in $\mathbb{Z}[\sqrt{-5}]$) such that $2(a+b\sqrt{-5}) = 2a+2b\sqrt{-5} = 1-\sqrt{-5}$.

Same issues with the rest of the arguments. They are either terse to the point of nonsense, or contain incorrect or incoherent claims. I strongly urge you to write complete sentences, use words, and don't over rely on symbols. And to read your own arguments with a critical eye after you are done.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.