1. ## UFD questions

Do you know how to solve the below two questions? If you do , please teach me. Thank you very much.

Question 1
Prove that Z is a UFD.

Question 2
Suppose Q[sqrt(d)] is a UFD, and alpha is an integer in Q[sqrt(d)] so that alpha and alpha bar has no common factor, but N(alpha) is a perfect square in Z. Show that alpha is a perfect square in the quadratic integers in Q[sqrt(d)].

2. Originally Posted by beta12

Question 1
Prove that Z is a UFD.
That is the Fundamental theorem of Arithmetic.
It was first stated by Euclid, in,
The Elements Book IX Proposition 14

Assume we know that we can factor any number into primes we wish to show that such a representation is unique.

Let,
$\displaystyle p_1p_2\cdot...\cdot p_n=N=q_1q_2\cdot...\cdot q_m$
Be the primes (not necessarily distinct) in non-decreasing order.

We know the left hand side is divisible by $\displaystyle p_1$ thus the right hand. Since they are prime, $\displaystyle p_1 = q_i$. That means $\displaystyle p_1\geq q_1$. Similarly, the right hand side is divisble by $\displaystyle q_1$ thus, $\displaystyle q_1=p_k$ thus, $\displaystyle p_1\leq q_1$. Thus, $\displaystyle p_1=q_1$.
Dividing both sides by that we have,
$\displaystyle p_2\cdot...\cdot p_n=q_2\cdot...\cdot q_m$.
If, $\displaystyle n\not = m$ then using the above arugument we can keep cancelling the left most primes on both sides eventually reaching,
$\displaystyle 1=q_r\cdot...\cdot q_n$
or,
$\displaystyle p_s\cdot...\cdot p_m=1$
Which is an impossibility.
Thus,
$\displaystyle n=m$.
Thus,
$\displaystyle p_1\cdot ... \cdot p_n=q_1\cdot ... \cdot q_n$
Using the above argument we can then show,
$\displaystyle p_k=q_k$ for all $\displaystyle 1\leq k \leq n$.
Q.E.D.

I think this elegant proof using the basic property $\displaystyle a\leq b \mbox{ and }a\geq b\to a=b$. Was developed by Gauss.

3. Now there just happens to be a more abstract way to prove this. There is a theorem that every PID is a UFD.

The integral domain $\displaystyle \mathbb{Z}$ is a principal ideal domain. Because since the additive group is cyclic all subgroups are themselves cyclic. And thus must have the form $\displaystyle n\mathbb{Z}$. But this is a principal ideal for $\displaystyle \mathbb{Z}$. Therefore, the integers form a Unique factorization domain.

4. Hi Perfecthacker,

Thank you very much. I got question 1 now. Thanks again.