# Thread: Prime and irreducable U.F.D

1. ## Prime and irreducable U.F.D

This question is from Dummit and Foote Abstract Algebra pg 293 #5.

Let $\displaystyle R=\mathbb{Z}[\sqrt{-n}]$ where n is a square free integeter greater than 3.

(a) PRove that $\displaystyle 2, \sqrt{-n},1+\sqrt{-n}$ are irreducable in R.

(b) Prove that R is not a U.F.D Conclude that the quadratic inter ring is not a U.F.D for $\displaystyle n \equiv 2,3 \mod 4, n < -3$

Show that $\displaystyle \sqrt{-n} \mbox{ or } (1+\sqrt{-n})$ is not prime.

(c) Give an explicit ideal in R that is not principal. Hint: Use (b) and cosider a maximal ideal containing the non prime ideal $\displaystyle \sqrt{-n} \mbox{ or } 1+\sqrt{-n}$

For part (a) I have shown that none of the of them are units i.e they don't have a norm of 1 and they are non zero. so the only thing I have left to do is show if they do factor one of the factors is a unit

so for 2

suppose that there is a factorization then there exits $\displaystyle a,b \in \mathbb{Z}\sqrt{-n}$ such that

$\displaystyle a\cdot b =2$ Since the norm is a multiplicitive function I get

$\displaystyle N(a)\cdot N(b)=N(2) \iff N(a)\cdot N(b)=4$

Since a and b are not units and n(a) and N(b) are integers greater than 1

$\displaystyle N(a)=N(b)=2$

so let $\displaystyle a=\alpha +\beta \sqrt{-n}$

$\displaystyle N(a)=\alpha^2+\beta^2n=2$

but since n > 3 $\displaystyle \beta = 0$ and there does not exista an integer $\displaystyle \alpha$ such that $\displaystyle \alpha^2=2$

There for 2 is irreducable.

This is where I am stuck.

I don't see how to generalize this argument.

Thanks TES

2. Originally Posted by TheEmptySet
(a) PRove that $\displaystyle 2, \sqrt{-n},1+\sqrt{-n}$ are irreducable in R.
You proved that 2 is irreducible. You can show that the other two are irreducible by a similar argument. If $\displaystyle \sqrt{-n} = \alpha \beta$ for non-units, then by taking norms $\displaystyle n = N(\alpha)N(\beta)$. Notice that $\displaystyle 1 < N(\alpha) < n$. Let $\displaystyle \alpha = a + b\sqrt{-n}$ then $\displaystyle N(\alpha) = a^2 + nb^2$. For this to be less than $\displaystyle n$ it is necessary for $\displaystyle b=0$. Therefore, $\displaystyle \alpha$ must be a real number. By a similar argument $\displaystyle \beta$ must be a real number. However, it is impossible to have $\displaystyle \alpha \beta$ to be an imaginary number. Thus, it is not possible to write $\displaystyle \sqrt{-n} = \alpha \beta$ for non-units.

Say that $\displaystyle 1+\sqrt{-n} = \alpha \beta$ for non-units. By taking norms, $\displaystyle 1 + n = N(\alpha)N(\beta)$. Thus, we require that $\displaystyle 1 < N(\alpha) < n+1$. If $\displaystyle \alpha = a + b\sqrt{-n}$ then $\displaystyle 1< a^2 + nb^2 < n+1$. We cannot have $\displaystyle |b|\geq 2$ so $\displaystyle b^2=0,1$. Likewise for $\displaystyle \beta = c + d\sqrt{-n}$. If $\displaystyle b^2=1$ then $\displaystyle 1 < a^2 + n < 1 \implies 1 < a^2 < 1$ which is a contradiction. Therefore, $\displaystyle b=0$, and now repeat the same argument as above.

3. Originally Posted by ThePerfectHacker
You proved that 2 is irreducible. You can show that the other two are irreducible by a similar argument. If $\displaystyle \sqrt{-n} = \alpha \beta$ for non-units, then by taking norms $\displaystyle n = N(\alpha)N(\beta)$. Notice that $\displaystyle 1 < N(\alpha) < n$. Let $\displaystyle \alpha = a + b\sqrt{-n}$ then $\displaystyle N(\alpha) = a^2 + nb^2$. For this to be less than $\displaystyle n$ it is necessary for $\displaystyle b=0$. Therefore, $\displaystyle \alpha$ must be a real number. By a similar argument $\displaystyle \beta$ must be a real number. However, it is impossible to have $\displaystyle \alpha \beta$ to be an imaginary number. Thus, it is not possible to write $\displaystyle \sqrt{-n} = \alpha \beta$ for non-units.

Say that $\displaystyle 1+\sqrt{-n} = \alpha \beta$ for non-units. By taking norms, $\displaystyle 1 + n = N(\alpha)N(\beta)$. Thus, we require that $\displaystyle 1 < N(\alpha) < n+1$. If $\displaystyle \alpha = a + b\sqrt{-n}$ then $\displaystyle 1< a^2 + nb^2 < n+1$. We cannot have $\displaystyle |b|\geq 2$ so $\displaystyle b^2=0,1$. Likewise for $\displaystyle \beta = c + d\sqrt{-n}$. If $\displaystyle b^2=1$ then $\displaystyle 1 < a^2 + n < 1 \implies 1 < a^2 < 1$ which is a contradiction. Therefore, $\displaystyle b=0$, and now repeat the same argument as above.
Thanks

4. Originally Posted by TheEmptySet
(b) Prove that R is not a U.F.D Conclude that the quadratic inter ring is not a U.F.D for $\displaystyle n \equiv 2,3 \mod 4, n < -3$
Remember a result about irreducible elements in a UFD. If $\displaystyle \pi \in R$ is irreducible and $\displaystyle R$ is a UFD then $\displaystyle \pi$ is a prime element. Above we have shown that $\displaystyle 2$ is a prime element. Assume that $\displaystyle R$ is a UFD. Notice that $\displaystyle 2|(n+1)$ (in the case when $\displaystyle n$ is odd) and that $\displaystyle n+1 = (1 + \sqrt{-n})(1-\sqrt{-n})$. Therefore, $\displaystyle 2|(1+\sqrt{-n})$ or $\displaystyle 2|(1-\sqrt{-n})$ so $\displaystyle 2(a+b\sqrt{-n}) = 1 \pm \sqrt{-n}$ for some $\displaystyle a,b\in \mathbb{Z}$. This would imply $\displaystyle 2a = 1$ which is impossible.

In the even case then $\displaystyle 2|-n$ however $\displaystyle -n = \sqrt{-n}^2$ would would imply $\displaystyle 2|\sqrt{-n}$, which is again impossible.