Let $R$ be a principal ideal domain. Let $a\in R$ be a nonzero, non-unit element. Show that the following are equivalent.

(1) The ideal $(a)$ generated by $a$ is maximal.
(2) The ideal $(a)$ is prime.
(3) The element $a$ is irreducible.

Proof.

(1) $\implies$ (2)

Note that the ideal $(a)$ is maximal if and only if $R/(a)$ is a field. In particular $R/(a)$ is a domain and hence $(a)$ is a prime ideal.

(Note that this is true without assuming $R$ is a PID.)

(2) $\implies$ (3)

Now suppose that the ideal $(a)$ is prime.

Let $a=bc$ for some elements $b, c \in R$. Then the element $a=bc$ is in the prime ideal $(a)$, and thus we have either $b$ or $c$ is in $(a)$. Without loss of generality, we assume that $b\in (a)$.

Then we have $b=ad$ for some $d\in R$. It follows that we have
\[a=bc=adc\] and since $R$ is a domain, we have
\[1=dc\] and hence $c$ is a unit. Therefore the element $a$ is irreducible.

(3) $\implies$ (1)

Suppose that $a$ is an irreducible element.
Let $I$ be an ideal of $R$ such that
\[(a) \subset I \subset R.\]

Since $R$ is a PID, there exists $b\in R$ such that $I=(b)$.
Then since $(a)\subset (b)$, we have $a=bc$ for some $c\in R$.

The irreducibility of $a$ implies that either $b$ or $c$ is a unit.

If $b$ is a unit, then we have $I=R$. If $c$ is a unit, then we have $(a)=I$.
Therefore the ideal $(a)$ is maximal.

The post Three Equivalent Conditions for an Ideal is Prime in a PID first appeared on Problems in Mathematics.

Let $R$ be a finite commutative ring with identity $1$. Prove that every prime ideal of $R$ is a maximal ideal of $R$.

Proof.

We give two proofs. The first proof uses a result of a previous problem. The second proof is self-contained.

Proof 1.

Let $I$ be a prime ideal of the ring $R$. Then the quotient ring $R/I$ is an integral domain since $I$ is a prime ideal. Since $R$ is finite, $R/I$ is also finite.

By Problem Finite Integral Domain is a Field, any finite integral domain is a field. This yield that $R/I$ is a field, and hence $I$ is a maximal ideal.

Proof 2.

In this proof, we prove the problem from scratch.

Let $I$ be a prime ideal of the ring $R$. Then the quotient ring $R/I$ is an integral domain since $I$ is a prime ideal. Since $R$ is finite, $R/I$ is also finite.
We claim that $R/I$ is a field. For any nonzero element $a\in R/I$, define the map
\[f_a: R/I \to R/I\] by sending $x\in R/I$ to $ax \in R/I$.

We show that the map $f_a$ is injective.
If $ax=ay$ for $x, y \in R/I$, then we have $a(x-y)=0$, and we have $x-y=0$ as $R/I$ is an integral domain and $a\neq 0$. Thus $x=y$ and the map $f_a$ is injective.
Since $R/I$ is a finite set, the map $f_a$ is surjective as well. Hence there exists $b \in R/I$ such that $f_a(b)=1$, that is, $ab=1$. Thus $a$ is a unit in $R/I$.
Since $a$ is an arbitrary nonzero element of $R/I$, we conclude that $R/I$ is a field.

Since the quotient ring $R/I$ is a field, the ideal $I$ is maximal.

The post Every Prime Ideal of a Finite Commutative Ring is Maximal first appeared on Problems in Mathematics.

Let $R$ be a commutative ring with $1$.
Suppose that the localization $R_{\mathfrak{p}}$ is a Noetherian ring for every prime ideal $\mathfrak{p}$ of $R$.
Is it true that $A$ is also a Noetherian ring?

Proof.

The answer is no. We give a counterexample.
Let
\[R=\prod_{i=1}^{\infty}R_i,\] where $R_i=\Zmod{2}$.
As $R$ is not finitely generated, it is not Noetherian.

Note that every element $x\in R$ is idempotent, that is, we have $x^2=x$.

Let $\mathfrak{p}$ be a prime ideal in $R$.
Then $R/\mathfrak{p}$ is a domain and we have $x^2=x$ for any $x\in R/\mathfrak{p}$.
It follows that $x=0, 1$ and $R/\mathfrak{p} \cong \Zmod{2}$.
This also shows that every prime ideal in $R$ is maximal.

Now let us determine the localization $R_{\mathfrak{p}}$.

As the prime ideal $\mathfrak{p}$ does not contain any proper prime ideal (since every prime is maximal), the unique maximal ideal $\mathfrak{p}R_{\mathfrak{p}}$ of $R_{\mathfrak{p}}$ contains no proper prime ideals.

Recall that in general the intersection of all prime ideals is the ideal of all nilpotent elements.
Since $R_{\mathfrak{p}}$ does not have any nonzero nilpotent element, we see that $\mathfrak{p}R_{\mathfrak{p}}=0$.

(Remark: every Boolean ring has no nonzero nilpotent elements.)

It follows that $R_{\mathfrak{p}}$ is a filed, in particular an integral domain.

As before, since every element $x$ of $R_{\mathfrak{p}}$ satisfies $x^2=x$, we conclude that $x=0, 1$ and $R_{\mathfrak{p}} \cong \Zmod{2}$.

Since a field is Noetherian the localization $R_{\mathfrak{p}}$ is Noetherian for every prime ideal $\mathfrak{p}$ of $R$.

In summary, $R$ is not a Noetherian ring but the localization $R_{\mathfrak{p}}$ is Noetherian for every prime ideal $\mathfrak{p}$ of $R$.

The post If the Localization is Noetherian for All Prime Ideals, Is the Ring Noetherian? first appeared on Problems in Mathematics.

Let $R$ be a commutative ring with $1$.

Prove that if every proper ideal of $R$ is a prime ideal, then $R$ is a field.

Proof.

As the zero ideal $(0)$ of $R$ is a proper ideal, it is a prime ideal by assumption.
Hence $R=R/\{0\}$ is an integral domain.

Let $a$ be an arbitrary nonzero element in $R$.
We prove that $a$ is invertible.
Consider the ideal $(a^2)$ generated by the element $a^2$.

If $(a^2)=R$, then there exists $b\in R$ such that $1=a^2b$ as $1\in R=(a^2)$.
Hence we have $1=a(ab)$ and $a$ is invertible.

Next, if $(a^2)$ is a proper ideal, then $(a^2)$ is a prime ideal by assumption.
Since the product $a\cdot a=a^2$ is in the prime ideal $(a^2)$, it follows that $a\in (a^2)$.
Thus, there exists $b\in R$ such that $a=a^2b$.
Equivalently, we have $a(ab-1)=0$.

We have observed above that $R$ is an integral domain. As $a\neq 0$, we must have $ab-1=0$, and hence $ab=1$.
This implies that $a$ is invertible.

Therefore, every nonzero element of $R$ is invertible.
Hence $R$ is a field.

The post If Every Proper Ideal of a Commutative Ring is a Prime Ideal, then It is a Field. first appeared on Problems in Mathematics.

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 $ab=0$.

If $R$ contain no nonzero zero divisors, then $R$ is called an integral domain.

Solution.

The quotient ring $R/I$ of an integral domain is not necessarily an integral domain.

Consider, for example, the ring of integers $\Z$ and ideal $I=4Z$.
Note that $\Z$ is an integral domain.

We claim that the quotient ring $\Z/4\Z$ is not an integral domain.
In fact, the element $2+4\Z$ is a nonzero element in $\Z/4\Z$.

However, the product
\[(2+4\Z)(2+4\Z)=4+\Z=0+\Z\] is zero in $\Z/4\Z$.
This implies that $2+4\Z$ is a zero divisor, and thus $\Z/4\Z$ is not an integral domain.

Comment.

Note that in general, the quotient $R/I$ is an integral domain if and only if $I$ is a prime ideal of $R$.
In our above example, the ideal $I=4\Z$ is not a prime ideal of $\Z$.

The post Is the Quotient Ring of an Integral Domain still an Integral Domain? first appeared on Problems in Mathematics.

(a) Prove that every prime ideal of a Principal Ideal Domain (PID) is a maximal ideal.

(b) Prove that a quotient ring of a PID by a prime ideal is a PID.

Proof.

(a) Prove that every PID is a maximal ideal.

Let $R$ be a Principal Ideal Domain (PID) and let $P$ be a nonzero prime ideal of $R$.
Since $R$ is a PID, every ideal of $R$ is principal.

Hence there exists $p\in R$ such that $P=(p)$.
Because $P$ is a nonzero ideal, we see that $p\neq 0$.

Let $I=(a)$ be an ideal of $R$ such that $P \subset I\subset R$.
To show that $P$ is a maximal ideal, we must show that $I=P$ or $I=R$.

Since $p\in (p)\subset (a)$, we have $p=ra$ for some $r\in R$.
As $p=ra$ is in the prime ideal $(p)$, we have either $a\in (p)$ or $r\in (p)$.

If $a\in (p)$, then it follows that $(a)\subset (p)$, and hence $(a)=(p)$.
So, in this case, we have $I=P$.

If $r\in (p)$, then we have $r=sp$ for some $s\in R$.
It yields that
\begin{align*}
p=ra=spa \quad \Leftrightarrow \quad p(1-sa)=0.
\end{align*}

Since $R$ is an integral domain and $p\neq 0$, this gives $sa=1$.
It follows that $1\in (a)$ and thus $I=(a)=R$.

We have shown that if $P\subset I \subset R$ for some ideal $I$, then we have either $I=P$ or $I=R$.
Hence we conclude that $P$ is a maximal ideal of $R$.

(b) Prove that a quotient ring of a PID by a prime ideal is a PID.

Let $P$ be a prime ideal of a PID $R$.
It follows from part (a) that the ideal $P$ is maximal.
Thus the quotient $R/P$ is a field.

The only ideals of the field $R/P$ are the zero ideal $(0)$ and $R/P=(1)$ itself, which are principal.
Hence $R/P$ is a PID.

The post Every Prime Ideal in a PID is Maximal / A Quotient of a PID by a Prime Ideal is a PID first appeared on Problems in Mathematics.

Let $R$ be a commutative ring with identity $1\neq 0$. Suppose that for each element $a\in R$, there exists an integer $n > 1$ depending on $a$.

Then prove that every prime ideal is a maximal ideal.

Hint.

Let $R$ be a commutative ring with $1$ and $I$ be an ideal of $R$.

Recall the following facts:

• $I$ is a prime ideal if and only if $R/I$ is an integral domain.
• $I$ is a maximal ideal if and only if $R/I$ is a field.

Proof.

Let $I$ be a prime ideal of the ring $R$. To prove that $I$ is a maximal ideal, it suffices to show that the quotient $R/I$ is a field.

Let $\bar{a}=a+I$ be a nonzero element of $R/I$, where $a\in R$.
It follows from the assumption that there exists an integer $n > 1$ such that $a^n=a$.

Then we have
\[\bar{a}^n=a^n+I=a+I=\bar{a}.\] Thus we have
\[\bar{a}(\bar{a}^{n-1}-1)=0\] in $R/I$.

Note that $R/I$ is an integral domain since $I$ is a prime ideal.

Since $\bar{a}\neq 0$, the above equality yields that $\bar{a}^{n-1}-1=0$, and hence
\[\bar{a}\cdot \bar{a}^{n-2}=1.\] It follows that $\bar{a}$ has a multiplicative inverse $\bar{a}^{n-2}$.

This proves that each nonzero element of $R/I$ is invertible, hence $R/I$ is a field.
We conclude that $I$ is a field.

The post Every Prime Ideal is Maximal if $a^n=a$ for any Element $a$ in the Commutative Ring first appeared on Problems in Mathematics.

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:

1. An ideal $I$ of $R$ is prime if and only if $R/I$ is an integral domain.
2. An ideal $I$ of $R$ is maximal if and only if $R/I$ is a field.

Example 1: $\Z$ and $(0)$

The first example is the ring of integers $R=\Z$ and the zero ideal $I=(0)$.
Note that the quotient ring is $\Z/(0)\cong \Z$ and it is integral domain but not a field.
Thus the ideal $(0)$ is a prime ideal by Fact 1 but not a maximal ideal by Fact 2.

Remark

Note that $(0)$ is the only prime ideal of $\Z$ that is not a maximal ideal.
Nonzero ideals of $\Z$ are $(p)$ for some prime number $p$.

Example 2: $\Z[x]$ and $(x)$

The second example is the ring of polynomials $R=\Z[x]$ over $\Z$ and the principal ideal $I=(x)$ generated by $x\in \Z[x]$.
The quotient ring is $\Z[x]/(x)\cong \Z$, which is an integral domain but not a field.
Thus the ideal $(x)$ is prime but not maximal by Fact 1, 2.

Example 3: $\Q[x,y]$ and $(x)$

The third example is the ring of polynomials in two variables $R=\Q[x, y]$ over $\Q$ and the principal ideal $I=(x)$ generated by $x$.
The quotient ring $\Q[x,y]/(x)$ is isomorphic to $\Q[y]$.
(The proof of this isomorphism is given in the post Prove the Ring Isomorphism $R[x,y]/(x) \cong R[y]$.)

Note that $\Q[y]$ is an integral domain but it is not a field since, for instance, the element $y\in \Q[y]$ is not a unit.
Hence Fact 1, 2 implies that the ideal $(x)$ is prime but not maximal in the ring $\Q[x, y]$.

The post Examples of Prime Ideals in Commutative Rings that are Not Maximal Ideals first appeared on Problems in Mathematics.

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?

Solution.

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.

Related Question.

A direct proof that the ideal $P=(2, \sqrt{10})$ is prime in the ring $\Z[\sqrt{10}]$ is given in the post “A prime ideal in the ring $\Z[\sqrt{10}]$“.

The post Determine the Quotient Ring $\Z[\sqrt{10}]/(2, \sqrt{10})$ first appeared on Problems in Mathematics.

Let $f: R\to R’$ be a ring homomorphism. Let $P$ be a prime ideal of the ring $R’$.

Prove that the preimage $f^{-1}(P)$ is a prime ideal of $R$.

Proof.

The preimage of an ideal by a ring homomorphism is an ideal.
(See the post “The inverse image of an ideal by a ring homomorphism is an ideal” for a proof.)

Thus, $f^{-1}(P)$ is an ideal of $R$.

We prove that the ideal $f^{-1}(P)$ is prime.
Suppose that we have $ab\in f^{-1}(P)$ for $a, b\in R$. Then we have $f(ab) \in P$.
Since $f$ is a ring homomorphism, we obtain
\begin{align*}
f(a)f(b)=f(ab)\in P.
\end{align*}

Since $P$ is a prime ideal, it follows that either $f(a)\in P$ or $f(b)\in P$.
Hence we have either $a\in f^{-1}(P)$ or $b\in f^{-1}(P)$.
This proves that the ideal $f^{-1}(P)$ is prime.

The post The Preimage of Prime ideals are Prime Ideals first appeared on Problems in Mathematics.