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

(a) PRove that $2, \sqrt{-n},1+\sqrt{-n}$ 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 $n \equiv 2,3 \mod 4, n < -3$

Show that $\sqrt{-n} \mbox{ or } (1+\sqrt{-n})$ 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 $\sqrt{-n} \mbox{ or } 1+\sqrt{-n}$

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 $a,b \in \mathbb{Z}\sqrt{-n}$ such that

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

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

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

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

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

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

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

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

Say that $1+\sqrt{-n} = \alpha \beta$ for non-units. By taking norms, $1 + n = N(\alpha)N(\beta)$. Thus, we require that $1 < N(\alpha) < n+1$. If $\alpha = a + b\sqrt{-n}$ then $1< a^2 + nb^2 < n+1$. We cannot have $|b|\geq 2$ so $b^2=0,1$. Likewise for $\beta = c + d\sqrt{-n}$. If $b^2=1$ then $1 < a^2 + n < 1 \implies 1 < a^2 < 1$ which is a contradiction. Therefore, $b=0$, 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 $\sqrt{-n} = \alpha \beta$ for non-units, then by taking norms $n = N(\alpha)N(\beta)$. Notice that $1 < N(\alpha) < n$. Let $\alpha = a + b\sqrt{-n}$ then $N(\alpha) = a^2 + nb^2$. For this to be less than $n$ it is necessary for $b=0$. Therefore, $\alpha$ must be a real number. By a similar argument $\beta$ must be a real number. However, it is impossible to have $\alpha \beta$ to be an imaginary number. Thus, it is not possible to write $\sqrt{-n} = \alpha \beta$ for non-units.

Say that $1+\sqrt{-n} = \alpha \beta$ for non-units. By taking norms, $1 + n = N(\alpha)N(\beta)$. Thus, we require that $1 < N(\alpha) < n+1$. If $\alpha = a + b\sqrt{-n}$ then $1< a^2 + nb^2 < n+1$. We cannot have $|b|\geq 2$ so $b^2=0,1$. Likewise for $\beta = c + d\sqrt{-n}$. If $b^2=1$ then $1 < a^2 + n < 1 \implies 1 < a^2 < 1$ which is a contradiction. Therefore, $b=0$, 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 $n \equiv 2,3 \mod 4, n < -3$
Remember a result about irreducible elements in a UFD. If $\pi \in R$ is irreducible and $R$ is a UFD then $\pi$ is a prime element. Above we have shown that $2$ is a prime element. Assume that $R$ is a UFD. Notice that $2|(n+1)$ (in the case when $n$ is odd) and that $n+1 = (1 + \sqrt{-n})(1-\sqrt{-n})$. Therefore, $2|(1+\sqrt{-n})$ or $2|(1-\sqrt{-n})$ so $2(a+b\sqrt{-n}) = 1 \pm \sqrt{-n}$ for some $a,b\in \mathbb{Z}$. This would imply $2a = 1$ which is impossible.

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