Question #147956
Show that R = {a + b√-3 a, b belongs to Z} i s not a
unique factorization domain by expressing
4 as a product of two irreducible elements in
R in two different ways.
1
Expert's answer
2020-12-02T01:43:33-0500

We can clearly see that

4=22=(1+13)(113)4 = 2\cdot 2 = (1+1\sqrt{-3})\cdot (1-1\sqrt{-3})

First let's prove that they are irreducible.

We can view elements of RR as complex numbers (obviously) and thus we know that if 2=(a+b3)(c+d3)2=(a+b\sqrt{-3})(c+d\sqrt{-3}) we have (a2+3b2)(c2+3d2)=4(a^2+3b^2)(c^2+3d^2)=4 and as the last expression is an equation in Z\mathbb{Z} we can easily conclude that either a2+3b2=1,a^2+3b^2=1, or c2+3d2=1c^2+3d^2=1 and thus we have 2=(±2)(±1)2= (\pm2) \cdot (\pm1) and 22 is irreducible.

Now let's prove that 1±31\pm\sqrt{-3} are irreducible. By the same argument of absolute values in C\mathbb{C} we find that 1+3=(±1)(±1(1+3))1+\sqrt{-3} = (\pm1)\cdot (\pm1 \cdot (1+\sqrt{-3})) and same for 131-\sqrt{-3} . Therefore they are also irreducible.

In addition, while proving this we have studied all unit elements : ±1\pm1 and so 1+31+\sqrt{-3} and 22 are not associated. Thus we conclude that RR is not a unique factorization domain.



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS