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Let $R = \{a+bi\mid a,b \in \mathbb Z, i^2=-1\}$, with addition and multiplication defined by $(a+bi)+(c+di)=(a+c)+(b+d)i$ and $(a+bi)(c+di)=(ac-bd)+(bc+ad)i$, respectively.

(a) Verify that $R$ is an integral domain.

(b) Determine all units in $R$.

Let $R$ be a commutative ring with unity. Then $R$ is an integral domain if $R$ has no proper divisors of zero.

From this definition, I know that I must verify that $R$ has no proper divisors of zero, a nonzero element whose product is the zero element of the ring. But I first had a concern that how can this be a ring when $R$ has elements of complex numbers while $a$ and $b$ are elements of integers. Could it not be closed, and therefore not a ring at all? I may be confusing things. Also, it was defined where $a$ and $b$ belong to, but where did $c$ and $d$ come from?

I need help solving part (a)..

For (b), the solution says $1,-1,i,-i$ but I don't know why.

Thanks.

Lynn
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VincentJ
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  • First you must verify that it is a ring, then (in either order) that it is commutative and has a unit. Then you verify it is an integral domain. To verify it is a ring, write down the list of things a ring must be and verify each of them. For instance, is this set closed under the defined addition? – Eric Towers Jul 09 '14 at 02:45
  • $a$ and $b$ are not elements of integers; rather $a$ and $b$ are integers. The notatin $a,b\in\mathbb Z$ means $a$ and $b$ are elements of $\mathbb Z$. The set $\mathbb Z$ is the set of all integers. So $a$ and $b$ are elements of the set of all integers. That means $a$ and $b$ are integers. It does not mean $a$ and $b$ are elements of integers. – Michael Hardy Jul 09 '14 at 02:51
  • An equivalent definition of an integral domain is a nonzero commutative ring in which the product of any two nonzero elements is nonzero. (If $ab=0$ with $a,b\neq 0$ then $a$ and $b$ are proper divisors of $0$.) – Lynn Jul 09 '14 at 02:55
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    The problem is ill-specified. Is $,R = \Bbb Z[i] := \Bbb Z[x]/(x^2+1),,$ or, instead, is $,R,$ meant to be considered as a subring of $,\Bbb C?,$ Given that the ring operations are explicitly defined, I would guess the former. In particular this must be true if your textbook has not already constructed $,\Bbb C.,$ or some other field (or domain) containing $,\Bbb Z[i].\ \ $ – Bill Dubuque Jul 09 '14 at 03:13

2 Answers2

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a) Clearly this is an integral domain, it's a subring of the field, $\mathbb{C}$ (checking this is direct, but should be done)

Edit: An alternative means if you're using

$$R =\mathbb{Z}[x]/(x^2+1)$$

In this case you can see that $R$ is a ring immediately as it is defined as a quotient ring. To see it is an integral domain I would recommend appealing to the fact that $x^2+1$ is irreducible, so by lifting $a+b\bar{x}$ and $c+d\bar{x}$ to $\mathbb{Z}[x]$ to any $f(x), g(x)$ then if $a+b\bar{x}\ne 0$ we have that $f(x)=(x^2+1)q_f(x)+r_f(x)$ and $g(x)=(x^2+1)q_g(x)+r_g(x)$ with a non-zero remainder by assumption. But then by irreducibility (no roots) we end up with a non-zero remainder for the product, hence a non-zero residue product, proving you have an integral domain.

b) It is possible to show that $R$ is actually a Euclidean domain with the norming function $N(a+bi)=a^2+b^2$. From there it is straightforward to show that $N$ is multiplicative since $N(a+bi)=|a+bi|^2$ and the usual absolute value and squaring functions are multiplicative. Then if your element, $u$, is a unit, it must have an inverse, $x$, so $N(xu)=N(x)N(u)=\pm 1$ since $\pm 1$ are the only invertible integers. So $a^2+b^2=\pm 1$ if you have a unit. From there it follows from the definition of the ring that either $a=\pm 1$ or $b=\pm 1$, hence the result.

Adam Hughes
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  • Comment for a), does it being a subring of a field have anything to do with R having no proper divisors of zero? I'm not known to this, so please elaborate – VincentJ Jul 09 '14 at 02:55
  • Yes, you know that no two non-zero field elements can multiply to $0$, so since all of your ring's non-zero elements are also non-zero elements in a field, this follows immediately. – Adam Hughes Jul 09 '14 at 02:56
  • The approach you suggest for (a) presumes that one has already constructed some field (or domain) containing $,R.,$ But, given the way $R$ is defined, with explicit definitions for addition and multiplication, I doubt that is the case here. – Bill Dubuque Jul 09 '14 at 03:04
  • I suppose it should be said that checking that $R\subseteq\mathbb{C}$ holds. I'll edit things. – Adam Hughes Jul 09 '14 at 03:05
  • @Adam The point is that the textbook may not have already constructed $,\Bbb C.,$ Further, if that way was intended then why would the problem define the ring operations? – Bill Dubuque Jul 09 '14 at 03:07
  • I'm making the reasonable assumption that someone in ring theory has seen $\mathbb{C}$ before. Meta-questions like "if that way was intended" are dangerous to ask IMO: I've seen students confused enough about doing things axiomatically that they don't realize things sit inside of familiar objects, because they are still confused on the notation, not because the exercise cannot be done that way correctly. – Adam Hughes Jul 09 '14 at 03:10
  • @BillDubuque Re: your edited comment: That's entirely possible, but in a vacuum I opt for the simplest answer and opt to do more complex ones if prompted by the op as a matter of personal preference with the philosophy that simplicity is usually best for comprehension when available. – Adam Hughes Jul 09 '14 at 03:14
  • @Adam I've elaborated in a comment to the question. Hopefully the OP will clarify. – Bill Dubuque Jul 09 '14 at 03:15
  • While we're waiting I opted to take the solution to both problems: put another option in the case he's using the quotient ring definition. – Adam Hughes Jul 09 '14 at 03:28
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The things you're trying to find are equations: you can solve the problem by simply solving the equations.

To be an integral domain, for example, means that the only solution to

$$ (a+bi)(c+di) = 0 +0i$$

is $a=b=c=d=0$. What do you get if you try to solve this?

Note in particular that if $u+vi = x+yi$, then $u=x$ and $v=y$.