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Prove $4+\sqrt{5}$ is prime in $\mathbb{Z}[\sqrt{5}]$. I already proved it's irreducible but have no idea what to do next since I dom't know much about $\mathbb{Z}[\sqrt{5}]$, if it is a UFD or not. Anyway, help me with this problem

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    $\begingroup$ It's not a UFD, because $2\cdot 2 = (1+\sqrt5)(-1+\sqrt5)$. $\endgroup$ – Arthur Dec 11 '16 at 14:55
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    $\begingroup$ It’s not UFD, because it’s not the ring of integers of $\Bbb Q(\sqrt5\,)$. $\endgroup$ – Lubin Dec 11 '16 at 17:30
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Extended hint: Show that $\mathfrak{p}=(4+\sqrt5)$ is a prime ideal by showing that $$ \Bbb{Z}[\sqrt5]/\mathfrak{p}\simeq\Bbb{Z}_{11}, $$ which is an integral domain. This is accomplished (do you see why?), if you prove that

  1. $11\in\mathfrak{p}$, and
  2. every coset of $\mathfrak{p}$ contains an integer.
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    $\begingroup$ am I correct in assuming that a generalization is $a+\sqrt{5}$ is prime provided $a^2-5$ is. $\endgroup$ – Rene Schipperus Dec 11 '16 at 15:08
  • $\begingroup$ It most certainly seems that way, doesn't it, @ReneSchipperus. I didn't think that far. Just about this specific case :-) $\endgroup$ – Jyrki Lahtonen Dec 11 '16 at 15:11
  • $\begingroup$ I can prove the isomorphism. But it's just because you told me it is true. I don't know why your hints work $\endgroup$ – chí trung châu Dec 11 '16 at 15:18
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    $\begingroup$ Those were something I cooked up for the occasion. But they come to mind quite naturally I think. Anyway, when needing to prove something is a prime ideal, one of the natural things to try is to check whether the quotient ring is an integral domain. Admittedly that only became natural to me after studying a bit more algebra. It is great that you want to learn more, but I really don't know if I had been able to come up with this argument during my first course in algebraic number theory. $\endgroup$ – Jyrki Lahtonen Dec 11 '16 at 15:25
  • $\begingroup$ @chítrungchâu I explain another way to view this in my answer, using a generalization of the division algorithm. $\endgroup$ – Bill Dubuque Dec 11 '16 at 17:23
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$a+b\sqrt 5$ is a multiple of $4+\sqrt5$ if and only if $4a-5b$ and $-a+4b$ are multiples of $11$. It's easy to prove that $11\mid4a-5b\iff11\mid-a+4b$

Now, suppose that $(a+b\sqrt 5)(c+d\sqrt 5)=(ac+5bd)+(ad+bc)\sqrt 5$ is a multiple of $4+\sqrt 5$. Then $$11\mid -ac-5bd+4ad+4bc$$ or $$11\mid a(4d-c)+b(4c-5d)$$ But $$a(4d-c)+b(4c-5d)\equiv a(4d-c)-4b(-12c+15d)\equiv (a-4b)(4d-c)\pmod{11}$$ This implies that $(a+b\sqrt5)(c+d\sqrt 5)$ is a multiple of $4+\sqrt 5$ only if one of the factors is. That is, $4+\sqrt 5$ is prime.

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  • $\begingroup$ How do you get the first sentence, it is Ok to verify but in general, Is it true that, $a+b\sqrt p$ is a multiple of $x+y\sqrt p$ iff $ax-pby$ and $-ya+xb$ is a multiple of $x^2-py^2$ ? $\endgroup$ – user257 Dec 11 '16 at 16:39
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    $\begingroup$ @user314 I explain this viewpoint further in my answer. $\endgroup$ – Bill Dubuque Dec 11 '16 at 17:21
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    $\begingroup$ $(a+b\sqrt 5)/(4+\sqrt 5)=(4a-5b)/11+\sqrt5(-a+4b)/11$ $\endgroup$ – ajotatxe Dec 11 '16 at 17:21
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Here is another proof. Althought @Jyrki Lahtonen 's in quicker, this uses more of the theory, and it generalizes.

The ring of integers $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$ is an integral domain and $4+\sqrt{5}$ is prime there since $N(4+\sqrt{5})=11$ is prime.

This means that if $a,b\in \mathbb{Z}[\sqrt{5}]$ then $(4+\sqrt{5})|a$ or $(4+\sqrt{5})|b$ in $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$.

Take the former, that means we have a relation of the form

$$a=(4+\sqrt{5})\frac{x+y\sqrt{5}}{2}$$

this implies that

$$4x+5y$$ and $$x+4y$$ are even and since the determinant of this matrix is $11$, we have that $x$ and $y$ are even.

Thus we have a relation

$$a=(4+\sqrt{5})(x_1+y_1\sqrt{5})$$ and so $(4+\sqrt{5})| a$ in $\mathbb{Z}[\sqrt{5}]$.

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Below I explain how to prove that $\,w = 4+\sqrt5\,$ is prime using a direct proof analogous to the classical proof for integers. This will also reveal the relationship between some of the other answers. First let's recall the classical Euclidean proof that atoms (irreducibles) $p$ are prime in $\,\Bbb Z$

$$ p\nmid a,\ p\mid ab\,\Rightarrow\,p\mid ab,pb\,\Rightarrow\,p\mid (ab,pb)=\color{#0a0}{(a,p)}b = b$$

since, applying Bezout's gcd identity we deduce that $\,p\nmid a\,\Rightarrow\, \color{#0a0}{(a,p) = 1}\, $ since $\,p\,$ is an atom.

By a short simple Lemma, $\,ww'=11\,\Rightarrow (w) = (w,11) = [w, 11] := w\Bbb Z + 11\Bbb Z,\,$ i.e. the ideal generators $\,w,11\,$ are actually module generators. Now we can use division with remainder to reduce any $\, \alpha = c+d\sqrt 5\,$ to a unique integer remainder $\ \alpha \bmod w \equiv n\pmod{11} $ as follows

${\rm mod}\ (w) = [4\!+\!{\sqrt 5},\, 11]\!:\,\ \color{#c00}{\sqrt5\equiv -4}\ $ $\Rightarrow$ $\ c\!+\!d\color{#c00}{\sqrt 5}\,\equiv\, (c\!\color{#c00}{-\!4}d)\bmod 11,\ $ by $\ 11\equiv 0$

For uniqueness: $\,n'\equiv \alpha\bmod w\equiv n\,$ $\Rightarrow$ $\,k = n'\!-\!n\in (w)\Rightarrow n'\equiv n\pmod{\!11}\, $ by

$$\begin{align} {\rm integer}\ k \in (w) = w\Bbb Z + 11\Bbb Z &\iff\! k = w\,j + 11\,i\ \ \ {\rm for\ some}\ \ i,j\in\Bbb Z\\ &\iff\! k = 11\,i\ \ \ {\rm for\ some}\,\ i\in\Bbb Z,\, \ {\rm by}\ \ w\not\in\Bbb Q\,\Rightarrow j=0\end{align}$$

Thus if $\, w\nmid \alpha \,$ then $\, 11\nmid n:= \alpha \bmod w,\,$ so $\,(11,n)= \color{#0a0}{1\in (\alpha,w)}$

Hence $\,w\nmid \alpha,\ w\mid \alpha\beta\,\Rightarrow\,w\mid \alpha\beta,w\beta\,\Rightarrow\, w\mid (\alpha\beta,w\beta) = \color{#0a0}{(\alpha,w)}\beta = \beta,\,$ as in the classical proof.

Remark $\ $ This explains the calculations in ajotatxe's answer, i.e. we are essentially doing remainder calculations modulo a triangular basis (something that will be clarified when one studies Hermite / Smith normal forms). It also gives another view on the hint in Jyrki's answer.

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    $\begingroup$ See also this answer where I present Euclid's lemma in Bezout, gcd, and ideal form. $\endgroup$ – Bill Dubuque Dec 11 '16 at 17:29

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