ring of Gaussian integers – Problems in Mathematics https://yutsumura.com Wed, 09 Aug 2017 15:30:39 +0000 en-US hourly 1 https://wordpress.org/?v=5.3.4 https://i2.wp.com/yutsumura.com/wp-content/uploads/2016/12/cropped-question-logo.jpg?fit=32%2C32&ssl=1 ring of Gaussian integers – Problems in Mathematics https://yutsumura.com 32 32 114989322 The Quotient Ring $\Z[i]/I$ is Finite for a Nonzero Ideal of the Ring of Gaussian Integers https://yutsumura.com/the-quotient-ring-zii-is-finite-for-a-nonzero-ideal-of-the-ring-of-gaussian-integers/ https://yutsumura.com/the-quotient-ring-zii-is-finite-for-a-nonzero-ideal-of-the-ring-of-gaussian-integers/#respond Tue, 08 Aug 2017 15:31:05 +0000 https://yutsumura.com/?p=4386 Let $I$ be a nonzero ideal of the ring of Gaussian integers $\Z[i]$. Prove that the quotient ring $\Z[i]/I$ is finite. Proof. Recall that the ring of Gaussian integers is a Euclidean Domain with...

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]]> Problem 534

Let $I$ be a nonzero ideal of the ring of Gaussian integers $\Z[i]$.

Prove that the quotient ring $\Z[i]/I$ is finite.

Proof.

Recall that the ring of Gaussian integers is a Euclidean Domain with respect to the norm
\[N(a+bi)=a^2+b^2\] for $a+bi\in \Z[i]$.
In particular, $\Z[i]$ is a Principal Ideal Domain (PID).

Since $I$ is a nonzero ideal of the PID $\Z[i]$, there exists a nonzero element $\alpha\in \Z[i]$ such that $I=(\alpha)$.
Let $a+bi+I$ be an arbitrary element in the quotient $\Z[i]/I$.
The Division Algorithm yields that
\[a+bi=q\alpha+r,\] for some $q, r\in \Z[i]$ and $N(r) < N(\alpha)$.

Since $a+bi-r=q\alpha \in I$, we have
\[a+bi+I=r+I.\] It follows that every element of $\Z[i]/I$ is represented by an element $r$ whose norm is less than $N(\alpha)$.

There are only finitely many elements in $\Z[i]$ whose norm is less than $N(\alpha)$.

(There are only finitely many integers $a, b$ satisfying $a^2+b^2 < N(\alpha)$.)

 

Hence the quotient ring $\Z[i]/I$ is finite.

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]]> https://yutsumura.com/the-quotient-ring-zii-is-finite-for-a-nonzero-ideal-of-the-ring-of-gaussian-integers/feed/ 0 4386 Ring of Gaussian Integers and Determine its Unit Elements https://yutsumura.com/ring-of-gaussian-integers-and-determine-its-unit-elements/ https://yutsumura.com/ring-of-gaussian-integers-and-determine-its-unit-elements/#respond Sat, 19 Nov 2016 16:23:12 +0000 https://yutsumura.com/?p=1421 Denote by $i$ the square root of $-1$. Let \[R=\Z[i]=\{a+ib \mid a, b \in \Z \}\] be the ring of Gaussian integers. We define the norm $N:\Z[i] \to \Z$ by sending $\alpha=a+ib$ to \[N(\alpha)=\alpha...

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]]> Problem 188

Denote by $i$ the square root of $-1$.
Let
\[R=\Z[i]=\{a+ib \mid a, b \in \Z \}\] be the ring of Gaussian integers.
We define the norm $N:\Z[i] \to \Z$ by sending $\alpha=a+ib$ to
\[N(\alpha)=\alpha \bar{\alpha}=a^2+b^2.\]

Here $\bar{\alpha}$ is the complex conjugate of $\alpha$.
Then show that an element $\alpha \in R$ is a unit if and only if the norm $N(\alpha)=\pm 1$.
Also, determine all the units of the ring $R=\Z[i]$ of Gaussian integers.

 

Proof.

Suppose that an element $\alpha$ is a unit of $R$.
Then there exists $\beta \in R$ such that $\alpha \beta=1$.

Then the norm of $\alpha \beta$ is
\begin{align*}
N(\alpha \beta)&=(\alpha \beta)(\overline{\alpha \beta})\\
&=\alpha \bar{\alpha} \beta \bar{\beta}\\
&=N(\alpha)N(\beta).
\end{align*}

Since the norm $N(1)=1$, we obtain
\[1=N(\alpha)N(\beta)\] in the ring $\Z$. Since $N(\alpha)$ and $N(\beta)$ are both integers, it follows that we have
\[N(\alpha)=\pm 1 \text{ and } N(\beta)=\pm 1.\]

On the other hand, suppose that $N(\alpha)=\pm 1$ for an element $\alpha\in R$.
Then let $\beta:=N(\alpha)^{-1}\bar{\alpha}$.

Since $N(\alpha)^{-1}=\pm 1$, the element $\beta \in R$. We have
\begin{align*}
\beta \alpha&=N(\alpha)^{-1}\bar{\alpha}\cdot \alpha\\
&=N(\alpha)^{-1}N(\alpha)=1.
\end{align*}
Thus the element $\alpha$ is a unit in $R$.

Using this result, let us determine all units of the ring $R$ of Gaussian integers.
An element $\alpha=a+ib \in R$ is a unit if and only if
\[N(\alpha)=a^2+b^2=1,\] where $a, b \in \Z$. Thus only solutions are
\[(a,b)=(\pm 1, 0), (0, \pm 1).\] Therefore the units of $R=\Z[i]$ are
\[\pm 1, \pm i.\]

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