Let
\[P=(2, \sqrt{10})=\{a+b\sqrt{10} \mid a, b \in \Z, 2|a\}\]
be an ideal of the ring
\[\Z[\sqrt{10}]=\{a+b\sqrt{10} \mid a, b \in \Z\}.\]
Then determine the quotient ring $\Z[\sqrt{10}]/P$.
Is $P$ a prime ideal? Is $P$ a maximal ideal?

We prove that the ring $\Z[\sqrt{10}]/P$ is isomorphic to the ring $\Zmod{2}$.

We define the map $\Psi:\Z[\sqrt{10}] \to \Zmod{2}$ by sending $a+b\sqrt{10}$ to $\bar{a}=a \pmod 2 \in \Zmod{2}$.
The map $\Psi$ is a ring homomorphism. To see this,
let $a+b\sqrt{10}, c+d\sqrt{10} \in \Z[\sqrt{10}]$ .

We have
\begin{align*}
\Psi\left( (a+b\sqrt{10})(c+d\sqrt{10}) \right) &=\Psi\left(ac+10bd+(ad+bc)\sqrt{10}\right)\\
&=ac+10bd \pmod{2}=ac \pmod{2}\\
&=\Psi(a+b\sqrt{10}) \Psi(c+d\sqrt{10}).
\end{align*}

We also have
\begin{align*}
\Psi\left( (a+b\sqrt{10})+(c+d\sqrt{10}) \right) &=\Psi\left( a+c+(b+d)\sqrt{10}) \right) \\
&=a+c \pmod{2}\\
&=\Psi(a+b\sqrt{10})+\Psi(c+d\sqrt{10}).
\end{align*}

Therefore the map $\Psi$ is a ring homomorphism.

Since $\Psi(0)=\bar{0}$ and $\Psi(1)=\bar{1}$, the map $\Psi:\Z[\sqrt{10}] \to \Zmod{2}$ is surjective.

We have $\Psi(a+b\sqrt{10})=\bar{0}$ if and only if $a$ is even.
Thus, the kernel of the homomorphism $\Psi$ is
\[\ker(\Psi)=\{a+b\sqrt{10} \mid a, b \in \Z, 2|a\}=P.\]

In summary the map $\Psi:\Z[\sqrt{10}] \to \Zmod{2}$ is a surjective ring homomorphism with the kernel $P$. Hence by the first isomorphism theorem, we have
\[\Z[\sqrt{10}] /P \cong \Zmod{2}\]
as we claimed.

Since $\Zmod{2}$ is a field, the ideal $P$ is a maximal ideal, and in particular $P$ is a prime ideal.

A Maximal Ideal in the Ring of Continuous Functions and a Quotient Ring
Let $R$ be the ring of all continuous functions on the interval $[0, 2]$.
Let $I$ be the subset of $R$ defined by
\[I:=\{ f(x) \in R \mid f(1)=0\}.\]
Then prove that $I$ is an ideal of the ring $R$.
Moreover, show that $I$ is maximal and determine […]

Prove the Ring Isomorphism $R[x,y]/(x) \cong R[y]$
Let $R$ be a commutative ring. Consider the polynomial ring $R[x,y]$ in two variables $x, y$.
Let $(x)$ be the principal ideal of $R[x,y]$ generated by $x$.
Prove that $R[x, y]/(x)$ is isomorphic to $R[y]$ as a ring.
Proof.
Define the map $\psi: R[x,y] \to […]

Examples of Prime Ideals in Commutative Rings that are Not Maximal Ideals
Give an example of a commutative ring $R$ and a prime ideal $I$ of $R$ that is not a maximal ideal of $R$.
Solution.
We give several examples. The key facts are:
An ideal $I$ of $R$ is prime if and only if $R/I$ is an integral domain.
An ideal $I$ of […]

The Quotient Ring by an Ideal of a Ring of Some Matrices is Isomorphic to $\Q$.
Let
\[R=\left\{\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix} \quad \middle | \quad a, b\in \Q \,\right\}.\]
Then the usual matrix addition and multiplication make $R$ an ring.
Let
\[J=\left\{\, \begin{bmatrix}
0 & b\\
0& 0
\end{bmatrix} […]

$(x^3-y^2)$ is a Prime Ideal in the Ring $R[x, y]$, $R$ is an Integral Domain.
Let $R$ be an integral domain. Then prove that the ideal $(x^3-y^2)$ is a prime ideal in the ring $R[x, y]$.
Proof.
Consider the ring $R[t]$, where $t$ is a variable. Since $R$ is an integral domain, so is $R[t]$.
Define the function $\Psi:R[x,y] \to R[t]$ sending […]

Three Equivalent Conditions for a Ring to be a Field
Let $R$ be a ring with $1$. Prove that the following three statements are equivalent.
The ring $R$ is a field.
The only ideals of $R$ are $(0)$ and $R$.
Let $S$ be any ring with $1$. Then any ring homomorphism $f:R \to S$ is injective.
Proof. […]

Is the Quotient Ring of an Integral Domain still an Integral Domain?
Let $R$ be an integral domain and let $I$ be an ideal of $R$.
Is the quotient ring $R/I$ an integral domain?
Definition (Integral Domain).
Let $R$ be a commutative ring.
An element $a$ in $R$ is called a zero divisor if there exists $b\neq 0$ in $R$ such that […]

## 1 Response

[…] For a proof, see the post “Determine the Quotient Ring $Z[sqrt{10}]/(2, sqrt{10})$“. […]