Finding the units of the ring Z[sqrt(-3)]

[Andrusko]

Homework Statement

In the ring $$Z[\sqrt{-3}]$$ find all units and prove that 2 is irreducible but 7 is not.

Homework Equations

The Attempt at a Solution

Well a unit is a non-zero element of the ring that when multiplied by some other non-zero element of the ring gives the unity of the ring.

ie; $$ab = 1\;a,b \in Z[\sqrt{-3}]$$

in otherwords b is the multiplicative inverse of a.

a is of the form $$a = x + y\sqrt{3} i$$ which just has the inverse of $$\frac{x - y\sqrt{3}i}{x^{2} + 3y^{2}}$$ (ie; b has this form)

So everything with the form of b with integer coefficients is a unit and 0 is not a unit.

I really don't understand what this question is asking, because I know what I've written down is completely stupid.

How would one go about finding the units of this ring?

 

[HallsofIvy]

Homework Statement

In the ring $$Z[\sqrt{-3}]$$ find all units and prove that 2 is irreducible but 7 is not.

Homework Equations

The Attempt at a Solution

Well a unit is a non-zero element of the ring that when multiplied by some other non-zero element of the ring gives the unity of the ring.

ie; $$ab = 1\;a,b \in Z[\sqrt{-3}]$$

in otherwords b is the multiplicative inverse of a.

a is of the form $$a = x + y\sqrt{3} i$$ which just has the inverse of $$\frac{x - y\sqrt{3}i}{x^{2} + 3y^{2}}$$ (ie; b has this form)

So everything with the form of b with integer coefficients is a unit and 0 is not a unit.

Which means that $x^2+ 3y^2$ is a factor of both x and y. For what x and y is that true?

Another way to do this is to look at $(x+ y\sqrt{3})(a+ b\sqrt{3})$$= (ax+ 3by)+ (ay+bx)\sqrt{3}= 1$. What must x, y, a, and b be?

I really don't understand what this question is asking, because I know what I've written down is completely stupid.

How would one go about finding the units of this ring?