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Let $\Bbb Z_3[i] = \{a + bi \mid a, b ∈ \Bbb Z_3, i^2 = −1\}$ and denote $·_3, +_3, −_3$ the multiplication, the addition and the subtraction $mod$ $3$. With these notations, the addition and the multiplication in $\Bbb Z_3[i]$ are defined as follows. For any $x = a + bi$ and $y = c + di$ with $a, b, c, d, ∈ \Bbb Z_3$ we have:

$x + y = (a +_3 c) + (b +_3 d)i$

$x · y = (a ·_3 c −_3 b ·_3 d) + (a ·_3 d +_3 b ·_3 c)i$

a). How many elements are in $\Bbb Z_3[i]$? Explain.

b). Find the units in $\Bbb Z_3[i]$ and compute their inverses

This is really confusing me but my attempt thus far: for $x = a+bi$, $a, b$ can assume values from $\Bbb Z_3: 0, 1, 2$. So there are $9$ possible values for $\forall x \in\Bbb Z_3[i] $. Under the operations, there are $81$ possible ways to perform $x+y$, $x-y$, and $xy$ each respectively. I'm sure there is some repetition among each of these sets which would reduce the number of elements in the set. Aside from listing the elements out, I don't really know where to begin.

Any help would be much appreciated.

Thomas Andrews
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TfwBear
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2 Answers2

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(a) is solved correctly by OP.

For (b), you first need to know what the multiplicative identity is, which is $1+0i$. Hence, you want to determine which of the nine choices for $a,b$ have a corresponding $c,d$ so that $x\cdot y=1+0i$. There are nine mini-problems.

Sample: $x=1+i$, i.e. $a=b=1$. Now we need $ac-bd=c-d=1$ and $ad+bc=c+d=0$. Adding, we get $2c=0$, so $c=0$ since we are in $\mathbb{Z}_3$. But now $c+d=0$ tells us $d=0$, and $c-d\neq 1$. Hence there is no choice of $y=c+di$ so that $x\cdot y=1+0i$.

Now, set $x=1+2i$ and do it all over again, then for the other seven elements.

vadim123
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  • Wouldn't there be any repeated elements in my method though? – TfwBear Nov 18 '15 at 03:16
  • Sorry, I do not understand the question in this comment. – vadim123 Nov 18 '15 at 03:17
  • I mean when I am listing the elements of the set. Would I not get the same value for some $x+y=xy$ or $xy= x-y$ for instance? – TfwBear Nov 18 '15 at 03:19
  • There certainly are $81$ pairs $(x,y)$ with $x,y \in \mathbb Z_3[i]$, on which you can perform addition, multiplication, and subtraction. But of course, by pigeon-hole, many of those pairs with their respective operations will give the same element of $\mathbb Z_3[i]$. – Dustan Levenstein Nov 18 '15 at 03:19
  • @DustanLevenstein so then would that count toward the cardinality? How would I eliminate any chance of repeating elements? Or is it ok for a set to have multiple elements of equal value? – TfwBear Nov 18 '15 at 03:21
  • If you're asking whether it's okay to have $x+_3 y = a \cdot_3 b$ for some values of $x, y, a, b$, then of course it's okay. Back in the more familiar territory of real numbers, consider $3+5 = 2 \cdot 4$. It's not a case of "repeating elements"; it's just the same element obtained from different arithmetic operations with different inputs. There's no problem here. – Dustan Levenstein Nov 18 '15 at 03:25
  • Ah ok. I'm new to abstract algebra. I keep thinking in terms of actual values rather than elements under operations. – TfwBear Nov 18 '15 at 03:29
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Once you get more experience with these matters, you’ll see that there are many ways of finding reciprocals of elements in finite fields.

In this case, since your field is quadratic over the field with three elements, which I’ll denote $\Bbb F_3$, you can do the same thing as you did in High School to find the reciprocal of, for instance $1+i$: $$ \frac1{1+i}=\frac1{1+i}\cdot\frac{1-i}{1-i}=\frac{1-i}{1-i^2}=\frac{1-i}2=-1+i\,, $$ since $2=-1$ in $\Bbb F_3$.

Another much better way, once you know the theorem that the set of nonzero elements in a finite field is a cyclic group, that is, all are powers of one well-chosen element, you can blunder about till you find one of these “primitive elements”, and then write down its powers in order. One choice here is $R=1-i$, and you get $R^2=i$, $R^3=1+i$, $R^4=-1$, and I’ll let you complete the list, going all the way to $R^8=1$. Then you see that since $1+i=R^3$, its reciprocal is $R^5$, which you have on your list.

Lubin
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