Jtdylktuy

Consider the UFD $mathbb Z[sqrt{2}]$. Prime and irreducible elements there are the same. How do I show that $5$ is irreducible?

I tried to write $5=(a+bsqrt 2)(c+dsqrt 2)$ or $(2bd+ac-5)+(bc+ad)sqrt 2=0$. We have $bc+ad=0$. If $c=0$, then either $d=0$ (in which case $5=0$) or $a=0$ (in which case $2bd=5$). Either case gives a contradiction. Suppose $cne 0$. Then $b=-(da)/c$. So $-2d^2a/c+ac-5=0$. But I don't see how to proceed. Maybe it's the wrong path?

To answer this question it is best to define a norm on $mathbb{Z}[sqrt2]$ as this will give you information regarding the arithmetic structure in the given domain. You can define the following norm $nu: mathbb{Z}[sqrt2] rightarrow mathbb{Z}$ given by $nu(a+bsqrt2)=a^2 - 2b^2$. It is easy to check that this is a multiplicative norm on the given domain; you just need to check that it satisfies the following properties:

• $nu(alpha) = 0$ if and only if $alpha = 0$




• $nu(alpha beta) = nu(alpha) nu(beta)$ for all $alpha,beta in mathbb{Z}[sqrt2]$

Now to determine whether 5 is irreducible, first notice that the norm of 5 under $nu$ is 25 i.e. $nu(5)=25$. Then suppose that 5 factored into two elements $alpha, beta in mathbb{Z}[sqrt2]$ i.e. $5=alphabeta$. We must have that $nu(alphabeta)=nu(alpha)nu(beta)=25$.

For this to occur either $nu(alpha)=nu(beta)=pm5$ or $nu(alpha)=pm1$ (or alternatively $nu(beta)=pm1$). It is easy to see that there does not exist any element in $mathbb{Z}[sqrt2]$ that has norm $pm5$; that is we cannot have $nu(alpha)=nu(beta)=pm5$. But then the only other option is that either $nu(alpha)=pm1$ or $nu(beta)=pm1$. However, if either $nu(alpha)=pm1$ or $nu(beta)=pm1$, then 5 must be irreducible since $alpha$ or $beta$ is a unit.

Your equation
$$5=(a+bsqrt 2)(c+dsqrt 2) = (ac + 2bd) + (ad+bc)sqrt2$$
gives the system
$$begin{cases}
ac+2bd = 5\
ad+bc = 0
end{cases}$$
Multiplying the first equation by $d$, the second by $-c$ and adding them yields $$2bd^2 - bc^2 = 5d implies b(2d^2-c^2) = 5d$$
Hence $5 mid b$ or $5 mid (2d^2 - c^2)$. However, the second possibility is impossible because $x^2 equiv pm 1 pmod 5$.
Therefore $5 mid b$.

Similarly, multiplying the first equation by $c$, the second by $-2d$ and adding them gives
$$ac^2-2ad^2 = 5c implies a(c^2-2d^2) = 5c$$
As above we conclude $5 mid a$.

Therefore $exists hat{a}, hat{b} in mathbb{Z}$ such that $a = 5hat{a}$ and $b = 5hat{b}$.
We have

$$5 = (a+bsqrt 2)(c+dsqrt 2) = (5hat{a}+5hat{b}sqrt 2)(c+dsqrt 2) = 5(hat{a}+hat{b}sqrt 2)(c+dsqrt 2)$$

Dividing be $5$ gives

$$1 = (hat{a}+hat{b}sqrt 2)(c+dsqrt 2)$$

so $c + dsqrt{2}$ is invertible in $mathbb{Z}[sqrt{2}]$ with $(c + dsqrt{2})^{-1} = hat{a}+hat{b}sqrt 2$.