MHB Is $2 + 8\sqrt{-5}$ a Unit or Irreducible in $\mathbb{Z} + \mathbb{Z}\sqrt{-5}$?

abs1
Messages
4
Reaction score
0
prove that $2+8{\sqrt{-5}}$ is unit and irreducible or not in $\mathbb Z+\mathbb Z{\sqrt{-5}}$.
 
Last edited by a moderator:
Physics news on Phys.org
abs said:
prove that $2+8{\sqrt{-5}}$ is unit and irreducible or not in $\mathbb Z+\mathbb Z{\sqrt{-5}}$.

Hint: we can write $2+8{\sqrt{-5}}=2(1+4\sqrt{-5})$.
 
Klaas van Aarsen said:
Hint: we can write $2+8{\sqrt{-5}}=2(1+4\sqrt{-5})$.

please explain in detail if possible
 
abs said:
please explain in detail if possible

What is the definition of a unit?
 
an element alpha belong to k ia called a unit if alpha divisible by 1.
dear it is my question if u not solved it then no problem its ok .if u solved it then give me complete explanation thank u so much
irreducible element:a non zero non unit element alpha belong to k is said to be irreducible if aplha=ab.
either a is unit or b is unit.
i give u both def. of unit and irreducible thank u so much
 
Last edited by a moderator:
abs said:
an element alpha belong to k ia called a unit if alpha divisible by 1.

Not quite.
From wiki:

a unit in a ring with identity $R$ is any element $u$ that has an inverse element in the multiplicative monoid of $R$, i.e. an element $v$ such that
$$uv = vu = 1_R,$$
where $1_R$ is the multiplicative identity​

abs said:
dear it is my question if u not solved it then no problem its ok .if u solved it then give me complete explanation thank u so much

Sorry, we are a math help site.
We do not usually give complete solutions.
Instead we give hints or similar to help people to learn math.

abs said:
irreducible element:a non zero non unit element alpha belong to k is said to be irreducible if aplha=ab.
either a is unit or b is unit.

If you're up to it...

The hint I gave showed that we can split the expression in two factors that we might call $a$ and $b$.
Let's start with $2$.
Is it a unit? That is, does it have a multiplicative inverse in the given ring?
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 1 ·
Replies
1
Views
750
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
48
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
21
Views
1K
Replies
15
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K