# Thread: can the following elements be decomposed into primary factors?

1. ## can the following elements be decomposed into primary factors?

In the ring $\displaystyle \mathbb{Z}[\sqrt{2}]$, determine can the following elements be decomposed into primary factors and is the decomposition unique:
$\displaystyle 5, 2+\sqrt{2}, 1+\sqrt{2}$.

I can't find anything in my notes on this, could someone please help me out, or maybe direct me to some online notes where it is explained?

Thank you!

2. Originally Posted by marianne
In the ring $\displaystyle \mathbb{Z}[\sqrt{2}]$, determine can the following elements be decomposed into primary factors and is the decomposition unique:
$\displaystyle 5, 2+\sqrt{2}, 1+\sqrt{2}$.

I can't find anything in my notes on this, could someone please help me out, or maybe direct me to some online notes where it is explained?

Thank you!
$\displaystyle 1+\sqrt{2}$ is a unit and so it has neither prime nor primary factors. $\displaystyle 2+\sqrt{2}$ is associate to $\displaystyle \sqrt{2}$ because $\displaystyle 2+\sqrt{2}=\sqrt{2}(1+\sqrt{2})$ and $\displaystyle \sqrt{2}$ can be easily seen to be prime.

finally $\displaystyle 5$ is also a prime: suppose $\displaystyle 5 \mid (a+b\sqrt{2})(c+d\sqrt{2})$ in $\displaystyle \mathbb{Z}[\sqrt{2}].$ then: $\displaystyle 5 \mid ac+2bd$ and $\displaystyle 5 \mid ad + bc$ in $\displaystyle \mathbb{Z}.$ thus: $\displaystyle 5 \mid b(ac+2bd)-a(ad+bc)=d(2b^2-a^2).$ now if

$\displaystyle 5 \mid d,$ then $\displaystyle 5 \mid ac$ and $\displaystyle 5 \mid bc$ and so either $\displaystyle 5 \mid c,$ which gives us $\displaystyle 5 \mid c + d \sqrt{2},$ or $\displaystyle 5 \mid a$ and $\displaystyle 5 \mid b,$ which gives us $\displaystyle 5 \mid a+b \sqrt{2}.$ finally if $\displaystyle 5 \mid 2b^2 - a^2,$ then $\displaystyle 5 \mid a$ & $\displaystyle 5 \mid b.$ why?

3. Originally Posted by marianne
In the ring $\displaystyle \mathbb{Z}[\sqrt{2}]$, determine can the following elements be decomposed into primary factors and is the decomposition unique:
$\displaystyle 5, 2+\sqrt{2}, 1+\sqrt{2}$.

I can't find anything in my notes on this, could someone please help me out, or maybe direct me to some online notes where it is explained?

Thank you!
Define $\displaystyle N(a+b\sqrt{2}) = |a^2 - 2b^2|$. Notice that $\displaystyle N(\alpha) = 1$ if and only if $\displaystyle \alpha$ is a unit in this ring.
Also notice that $\displaystyle N(\alpha \beta) = N(\alpha) N(\beta)$.

If $\displaystyle N(\alpha) = p$, a prime, then $\displaystyle \alpha$ must be irreducible in this ring. Because if it was then we can write $\displaystyle \alpha = \beta,\gamma$ for non-units and so by taking norms, $\displaystyle p = N(\beta)N(\gamma)$ but the $\displaystyle p$ factors non-trivially since $\displaystyle N(\beta),N(\gamma)>1$ for they are non-units. This is a contradiction.

Since $\displaystyle N(2+\sqrt{2}) = |4 - 1| = 3$ it means it is irreducible.
Since $\displaystyle N(1+\sqrt{2}) = |1 - 2| = 1$ it means it is a unit.

I see that NonCommAlg already replied with the case involving $\displaystyle 5$.

4. A huge thank you to both of you!

5. I'm having similar problems and I don't understand this solution.

finally $\displaystyle 5$ is also a prime: suppose $\displaystyle 5 \mid (a+b\sqrt{2})(c+d\sqrt{2})$ in $\displaystyle \mathbb{Z}[\sqrt{2}].$
We're trying to prove that 5 does not divide $\displaystyle a+b\sqrt{2}$ or $\displaystyle c+d\sqrt{2}$, right?

now if
$\displaystyle 5 \mid d,$ then $\displaystyle 5 \mid ac$ and $\displaystyle 5 \mid bc$
I don't understand this, and can't go on.
We're trying to prove that 5 does not divide $\displaystyle a+b\sqrt{2}$ or $\displaystyle c+d\sqrt{2}$, right?
no! we need to show that if 5 divides $\displaystyle (a+b\sqrt{2})(c+d\sqrt{2}),$ then 5 divides either $\displaystyle a+b\sqrt{2}$ or $\displaystyle c+d\sqrt{2},$ which will prove that 5 is prime in $\displaystyle \mathbb{Z}[\sqrt{2}].$
because we also have $\displaystyle 5 \mid ac+2bd$ and $\displaystyle 5 \mid ad + bc.$ (see the second line of my previous post!) i hope it's more clear now.