# Showing that a ring is an integral domain

1. Apr 12, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
Show that $\mathbb{Z} [\sqrt{d}] = \{a+b \sqrt{d} \ | \ a,b,d \in \mathbb{Z} \}$ is an integral domain

2. Relevant equations

3. The attempt at a solution
Do I have to go through all of the axioms to do this? For example, do I have to show that it is an abelian group under addition, that multiplication is associative, and that the distributive property holds, on top of showing that multiplication is commutative and that there is a multiplicative identity and that there are no zero divisors?
This would seem like a lot of unnecessary, but easy, work, so I was wondering if there is a faster way.

2. Apr 13, 2017

### Staff: Mentor

Well, most of these attributes are quite obvious by the arithmetic rules in this ring. The essential part here is to see, that you don't get zero divisors from such extensions. If you want to go the long way, then you have to do it, but only for terms with $\sqrt{d}$ in it, the rest is inherited from $\mathbb{Z}$. Since $d \in \mathbb{Z}$, the shortest answer is probably $\mathbb{Z}[\sqrt{d}] \subseteq \mathbb{R}$. But it makes sense to convince yourself that $\mathbb{Z}[\sqrt{d}]$ hasn't zero divisors, which is short enough to do and shows which properties of $\mathbb{Z}$ are actually used, in case a ring isn't the integers.

3. Apr 13, 2017

### Mr Davis 97

If I am trying to show that $\mathbb{Z}[\sqrt{d}]$ has no divisors, do I proceed by a contradiction argument, such as, assume that $(a+b \sqrt{d})(c + e \sqrt{d}) = 0$, and show that $a=b=c=e=0$?

4. Apr 13, 2017

### Staff: Mentor

Yes. And if you do it step by step, you see the properties of the integers which are needed, e.g. it isn't true for $\mathbb{Z}_6[\sqrt{d}]$.

5. Apr 13, 2017

### Mr Davis 97

Well if I expand it out and compare the two sides I get the two equations ac + bed = 0 and ae + bc = 0, but I don't see how this shows that a = b = c = e = 0

6. Apr 13, 2017

### Dick

This integral domain is actually a field. You should be able to show every non-zero element has a multiplicative inverse. That would show that it's an integral domain.

7. Apr 13, 2017

### Mr Davis 97

I don't see how every element has an inverse. For example, what is the inverse of $2+\sqrt{2}$? It can't be $1-\frac{1}{2} \sqrt{2}$, because $1/2$ is not an integer.

8. Apr 13, 2017

### Dick

Oh right, sorry. I forgot you were working over the integers, not the rationals. Nevertheless, if there are no zero divisors over the rationals, there won't be over the integers. Correct?

9. Apr 13, 2017

### Mr Davis 97

So how can I show that there are no zero divisors?

10. Apr 13, 2017

### Staff: Mentor

You have $ac + bed = 0$ and $ae + bc = 0$. What does this mean for $ace$?

11. Apr 13, 2017

### Dick

Pick an arbitrary nonzero element of the ring and write down its inverse. Say why it's always well-defined.

Last edited: Apr 13, 2017