Units of the Gaussian Integers, Z[i]

[Math Amateur]

In John Stillwell's book: Elements of Number Theory, Chapter 6 concerns the Gaussian integers, \(\displaystyle \mathbb{Z} = \{ a + bi \ | \ a, b \in \mathbb{Z} \}\).

Exercise 6.1.1 reads as follows:

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"Show that the units of \(\displaystyle \mathbb{Z} \) are \(\displaystyle \ \pm 1, \ \pm i \ \)."

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Now an element \(\displaystyle a\) of a ring or integral domain such as \(\displaystyle \mathbb{Z} \) is a unit if there exists an element \(\displaystyle b\) in \(\displaystyle \mathbb{Z} \) such that \(\displaystyle ab = ba = 1\).

So, then ... it is easy to demonstrate that \(\displaystyle \ \pm 1, \ \pm i \ \) are units of \(\displaystyle \mathbb{Z} \) ... ... BUT ... ... how do we rigorously demonstrate that they are the only units ... ... presumably we proceed as follows:\(\displaystyle (a_1 + b_1 i)\) is a unit of \(\displaystyle \mathbb{Z} \) if there exists an element \(\displaystyle (a_2 + b_2 i)\) such that:

\(\displaystyle (a_1 + b_1 i) (a_2 + b_2 i) = (a_2 + b_2 i) (a_1 + b_1 i) = 1 = 1 + 0 i \) ... ... in which case, of course, ... ...

... \(\displaystyle (a_2 + b_2 i)\) is also a unit ... ...So, I think, it follows that if we obtain all the solutions to the equation

\(\displaystyle (a_1 + b_1 i) (a_2 + b_2 i) = 1\)

we will have all the units ... and further will have demonstrated that they are the only units ... ...

Now ... ...

\(\displaystyle (a_1 + b_1 i) (a_2 + b_2 i) = 1 = 1 + 0 i \)

\(\displaystyle \Longrightarrow \ \ (a_1a_2 - b_1b_2) + (a_1b_2 + a_2b_1) i = 1 + 0 i
\)

\(\displaystyle \Longrightarrow \ \ a_1a_2 - b_1b_2 = 1 \ \) and \(\displaystyle \ a_1b_2 + a_2b_1 = 0\)

... ... ?BUT ... where to from here ...

Can someone please help with this exercise by showing how to complete my approach ... OR ... by critiquing my approach and showing a better approach ...

Peter

 

[Euge]

Take the sum of the squares of the last two equations to get

\(\displaystyle (a_1 a_2 - b_1 b_2)^2 + (a_1 b_2 + a_2 b_1)^2 = 1.\)

Expanding the left hand side yields the expression

\(\displaystyle a_1^2 a_2^2 + b_1^2 b_2^2 + a_1^2 b_2^2 + a_2^2 b_1^2,\)

which factors as

\(\displaystyle (a_1^2 + b_1^2)(a_2^2 + b_2^2).\)

Thus $(a_1^2 + b_1^2)(a_2^2 + b_2^2) = 1$. Since $a_1^2 + b_1^2$ and $a_2^2 + b_2^2$ are non-negative integers, the latter equation implies that $a_1^2 + b_1^2 = 1 = a_2^2 + b_2^2$. In particular, none of $a_1, a_2, b_1$ or $b_2$ can be greater than $1$. Hence, the only solutions for the pairs $(a_1, b_1)$ and $(a_2, b_2)$ are $(\pm 1, 0)$ and $(0, \pm 1)$. What does this mean for $\Bbb Z$?

 

[Math Amateur]

Take the sum of the squares of the last two equations to get

\(\displaystyle (a_1 a_2 - b_1 b_2)^2 + (a_1 b_2 + a_2 b_1)^2 = 1.\)

Expanding the left hand side yields the expression

\(\displaystyle a_1^2 a_2^2 + b_1^2 b_2^2 + a_1^2 b_2^2 + a_2^2 b_1^2,\)

which factors as

\(\displaystyle (a_1^2 + b_1^2)(a_2^2 + b_2^2).\)

Thus $(a_1^2 + b_1^2)(a_2^2 + b_2^2) = 1$. Since $a_1^2 + b_1^2$ and $a_2^2 + b_2^2$ are non-negative integers, the latter equation implies that $a_1^2 + b_1^2 = 1 = a_2^2 + b_2^2$. In particular, none of $a_1, a_2, b_1$ or $b_2$ can be greater than $1$. Hence, the only solutions for the pairs $(a_1, b_1)$ and $(a_2, b_2)$ are $(\pm 1, 0)$ and $(0, \pm 1)$. What does this mean for $\Bbb Z$?

Thanks for the help Euge ...

You ask:

"Hence, the only solutions for the pairs $(a_1, b_1)$ and $(a_2, b_2)$ are $(\pm 1, 0)$ and $(0, \pm 1)$. What does this mean for $\Bbb Z$?"

Since \(\displaystyle (a_1, b_1) = (a_1 + b_1 i)\) ... ...

... the solution \(\displaystyle (a_1, b_1) = (\pm 1, 0)\) ..

... means that elements \(\displaystyle \pm 1 = \pm 1 + 0\) are units of $\Bbb Z$ ...while, in the other case ...

Since \(\displaystyle (a_2, b_2) = (a_2 + b_2 i)\) ... ...

... the solution \(\displaystyle (a_2, b_2) = (0, \pm 1)\) ..

... means that elements \(\displaystyle \pm i = 0 + \pm i \) are units of $\Bbb Z$ ...

Is that the interpretation you were looking for?
Thanks once again for your help!

Peter

 

[Euge]

Thanks for the help Euge ...

You ask:

"Hence, the only solutions for the pairs $(a_1, b_1)$ and $(a_2, b_2)$ are $(\pm 1, 0)$ and $(0, \pm 1)$. What does this mean for $\Bbb Z$?"

Since \(\displaystyle (a_1, b_1) = (a_1 + b_1 i)\) ... ...

... the solution \(\displaystyle (a_1, b_1) = (\pm 1, 0)\) ..

... means that elements \(\displaystyle \pm 1 = \pm 1 + 0\) are units of $\Bbb Z$ ...while, in the other case ...

Since \(\displaystyle (a_2, b_2) = (a_2 + b_2 i)\) ... ...

... the solution \(\displaystyle (a_2, b_2) = (0, \pm 1)\) ..

... means that elements \(\displaystyle \pm i = 0 + \pm i \) are units of $\Bbb Z$ ...

Is that the interpretation you were looking for?
Thanks once again for your help!

Peter

Well, not exactly. The pair $(a_1, b_1)$ is not equal to $a_1 + ib_1$, but corresponds to $a_1 + ib_1$. Likewise $(a_2,b_2)$ is not equal to $a_2 + ib_2$, but corresponds to $a_2 + ib_2$. So $(\pm 1, 0)$ corresponds to $\pm 1$ and $(0, \pm 1)$ corresponds to $\pm i$. The possible units in $\Bbb Z$ are therefore $\pm 1$ and $\pm i$. You've shown that $\pm 1$ and $\pm i$ are in fact units in $\Bbb Z$. Hence, these are the only units in $\Bbb Z$.

 

[Math Amateur]

Well, not exactly. The pair $(a_1, b_1)$ is not equal to $a_1 + ib_1$, but corresponds to $a_1 + ib_1$. Likewise $(a_2,b_2)$ is not equal to $a_2 + ib_2$, but corresponds to $a_2 + ib_2$. So $(\pm 1, 0)$ corresponds to $\pm 1$ and $(0, \pm 1)$ corresponds to $\pm i$. The possible units in $\Bbb Z$ are therefore $\pm 1$ and $\pm i$. You've shown that $\pm 1$ and $\pm i$ are in fact units in $\Bbb Z$. Hence, these are the only units in $\Bbb Z$.

Thanks for the help Euge ...

Late here in Tasmania now ... will work through your post in the morning ...

Peter

 

[Math Amateur]

Thanks for the help Euge ...

Late here in Tasmania now ... will work through your post in the morning ...

Peter

OK I think I understand ... we are dealing with an isomorphism, not an equality ...

Thank you for that, Euge ... these points are VERY important to me as I wish to gain an understanding of the mathematics involved ... not just get a 'solution' to an exercise ...

Thanks again,

Peter