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.

Suppose that $p$ is a prime number greater than $3$.
Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$.

(a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$.

(b) Determine the index $[G : S]$.

(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$.

Proof.

(a) Prove that $S=\{x^2\mid x\in G\}$ is a subgroup of $G$.

Consider the map $\phi:G \to G$ defined by $\phi(x)=x^2$ for $x\in G$.
Then $\phi$ is a group homomorphism. In fact, for any $x, y \in G$, we have
\begin{align*}
\phi(xy)=(xy)^2=x^2y^2=\phi(x)\phi(y)
\end{align*}
as $G$ is an abelian group.

By definition of $\phi$, the image is $\im(\phi)=S$.
Since the image of a group homomorphism is a group, we conclude that $S$ is a subgroup of $G$.

(b) Determine the index $[G : S]$.

By the first isomorphism theorem, we have
\[G/\ker(\phi)\cong S.\]

If $x\in \ker(\phi)$, then $x^2=1$.
It follows that $(x-1)(x+1)=0$ in $\Zmod{p}$.
Since $\Zmod{p}$ is an integral domain, it follows that $x=\pm 1$ and $\ker(\phi)=\{\pm 1\}$.

Thus, $|S|=|G/\ker(\phi)|=(p-1)/2$ and hence the index is
\[[G:S]=|G|/|S|=2.\]

(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$.

Since $-1\notin S$ and $[G:S]=2$, we have the decomposition
\[G=S\sqcup (-S).\] Suppose that an element $a$ in $G$ is not in $S$.

Then, we have $a\in -S$.
Thus, there exists $b\in S$ such that $a=-b$.
It follows that $-a=b\in S$. Therefore, we have either $a\in S$ or $-a\in S$.

The post The Set of Square Elements in the Multiplicative Group $(\Zmod{p})^*$ 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) Let $F$ be a field. Show that $F$ does not have a nonzero zero divisor.

(b) Let $R$ and $S$ be nonzero rings with identities.
Prove that the direct product $R\times S$ cannot be a field.

Proof.

(a) Show that $F$ does not have a nonzero zero divisor.

Seeking a contradiction, suppose that $x$ is a nonzero zero divisor of the field $F$. This means that there exists a nonzero element $y\in F$ such that
\[yx=0.\] Since $y$ is a nonzero element in $F$, we have the inverse $y^{-1}$ in $F$.

Hence we have
\begin{align*}
0=y^{-1}\cdot 0=y^{-1}(yx)=(y^{-1}y)x=x.
\end{align*}
This is a contradiction because $x$ is a nonzero element.

We conclude that the field $F$ does not have a nonzero zero divisor.

(Remark that it follows that a field is an integral domain.)

(b) Prove that the direct product $R\times S$ cannot be a field.

Since $R$ and $S$ have identities, the direct product $R\times S$ contains nonzero elements $(1,0)$ and $(0,1)$.

The product of these elements is
\[(1,0)\cdot (0,1)=(1\cdot 0, \, 0\cdot 1)=(0,0).\] Similarly we also have
\[(0,1)\cdot (1,0)=(0,0).\]

It follows that $(1,0)$ is a nonzero zero divisor of $R\times S$. By part (a), a field does not have a nonzero zero divisor.
Hence $R\times S$ is never a field.

The post No Nonzero Zero Divisor in a Field / Direct Product of Rings is Not a Field 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.

Let $R$ be a ring with $1$. Suppose that $R$ is an integral domain and an Artinian ring.
Prove that $R$ is a field.

Definition (Artinian ring).

A ring $R$ is called Artinian if it satisfies the defending chain condition on ideals.
That is, whenever we have ideals $I_n$ of $R$ satisfying
\[I_1\supset I_2 \supset \cdots \supset I_n \supset \cdots,\] there is an integer $N$ such that
\[I_N=I_{N+1}=I_{N+2}=\cdots.\]

Proof.

Let $x\in R$ be a nonzero element. To prove $R$ is a field, we show that the inverse of $x$ exists in $R$.
Consider the ideal $(x)=xR$ generated by the element $x$. Then we have a descending chain of ideals of $R$:
\[(x) \supset (x^2) \supset \cdots \supset (x^i) \supset (x^{i+1})\supset \cdots.\]

In fact, if $r\in (x^{i+1})$, then we write it as $r=x^{i+1}s$ for some $s\in R$.
Then we have
\[r=x^i\cdot xs\in (x^i)\] since $(x^i)$ is an ideal and $xs\in R$.
Hence $(x^{i+1})\subset (x^i)$ for any positive integer $i$.

Since $R$ is an Artinian ring by assumption, the descending chain of ideals terminates.
That is, there is an integer $N$ such that we have
\[(x^N)=(x^{N+1})=\cdots.\]

It follows from the equality $(x^N)=(x^{N+1})$ that there is $y\in R$ such that
\[x^N=x^{N+1}y.\] It yields that
\[x^N(1-xy)=0.\]

Since $R$ is an integral domain, we have either $x^N=0$ or $1-xy=0$.
Since $x$ is a nonzero element and $R$ is an integral domain, we know that $x^N\neq 0$.

Thus, we must have $1-xy=0$, or equivalently $xy=1$.
This means that $y$ is the inverse of $x$, and hence $R$ is a field.

The post Every Integral Domain Artinian Ring is a Field first appeared on Problems in Mathematics.

(a) Let $R$ be an integral domain and let $M$ be a finitely generated torsion $R$-module.
Prove that the module $M$ has a nonzero annihilator.
In other words, show that there is a nonzero element $r\in R$ such that $rm=0$ for all $m\in M$.
Here $r$ does not depend on $m$.

(b) Find an example of an integral domain $R$ and a torsion $R$-module $M$ whose annihilator is the zero ideal.

Proof.

(a) Prove that the module $M$ has a nonzero annihilator.

Since $M$ is a finitely generated $R$-module, there is a finite set
\[A:=\{a_1, a_2, \dots, a_n\} \subset M\] such that $M=RA$

As $M$ is a torsion $R$-module, for each $a_i\in A\subset M$ there is a nonzero element $r_i\in R$ such that
\[r_ia_i=0.\] Let us put $r\in R$ to be the product of these $r_i$:
\[r:=r_1 r_2 \cdots r_n.\] Note that $r$ is a nonzero element of $R$ since each $r_i$ is nonzero and $R$ is an integral domain.

We claim that the element $r$ annihilates the module $M$.
Let $m$ be an arbitrary element in $M$. Since $M$ is generated by the set $A$, we can write
\[m=s_1a_1+s_2a_2+\cdots +s_n a_n\] for some elements $s_1, s_2, \dots, s_n\in R$.

Note that since $R$ is an integral domain, it is commutative by definition.
Hence we can change the order of the product in $r$ freely. Thus for each $i$ we can write
\[r=p_ir_i,\] where $p_i$ is the product of all $r_j$ except $r_i$.

Then it follows that we have
\begin{align*}
ra_i&=p_ir_ia_i=p_i0=0 \tag{*}
\end{align*}
for each $i$.
Using this, we obtain
\begin{align*}
rm&=r(s_1a_1+s_2a_2+\cdots +s_n a_n)\\
&=rs_1a_1+rs_2a_2+\cdots +rs_n a_n\\
&=s_1ra_1+s_2ra_2+\cdots +s_n ra_n && \text{as $R$ is commutative}\\
&=s_10+s_20+\cdots +s_n 0 && \text{by (*)}\\
&=0.
\end{align*}
Therefore, for any element $m\in M$ we have proved that $rm=0$.
Thus the nonzero element $r$ annihilates the module $M$.

(b) Find an example of an integral domain $R$ and a torsion $R$-module $M$ whose annihilator is the zero ideal.

Let $R=\Z$ be the ring of integers. Then $R=\Z$ is an integral domain.
Consider the $\Z$-module
\[M=\oplus_{i=1}^{\infty}\Zmod{2^i}.\]

Then each element $a\in M$ can be written as
\[a=(a_1+\Zmod{2}, a_2+\Zmod{2^2}, \dots, a_k+\Zmod{2^k}, 0, 0, \dots)\] for some $a_1, a_2, \dots, a_k\in \Z$.
(Here $k$ depends on $a$.)

It follows that we have
\[2^ka=0,\] and thus $M$ is a torsion $\Z$-module.

We now prove that any annihilator of $M$ must be the zero element of $R=\Z$.
Let $r\in \Z$ be an annihilator of $M$.
Choose an integer $k$ so that $r < 2^k$. Consider the element \[a=(0, 0, \dots, 1+\Zmod{2^k}, 0, 0, \dots)\] in $M$. The only nonzero entry of $a$ is at the $k$-th place. Since $r$ is an annihilator, we have \begin{align*} 0=ra=(0, 0, \dots, r+\Zmod{2^k}, 0, 0, \dots) \end{align*} and this implies that $r=0$ because $r < 2^k$. We conclude that the annihilator is the zero ideal.

The post Finitely Generated Torsion Module Over an Integral Domain Has a Nonzero Annihilator first appeared on Problems in Mathematics.

Let $R$ be a ring with $1$. An element of the $R$-module $M$ is called a torsion element if $rm=0$ for some nonzero element $r\in R$.
The set of torsion elements is denoted
\[\Tor(M)=\{m \in M \mid rm=0 \text{ for some nonzero} r\in R\}.\]

(a) Prove that if $R$ is an integral domain, then $\Tor(M)$ is a submodule of $M$.
(Remark: an integral domain is a commutative ring by definition.) In this case the submodule $\Tor(M)$ is called torsion submodule of $M$.

(b) Find an example of a ring $R$ and an $R$-module $M$ such that $\Tor(M)$ is not a submodule.

(c) If $R$ has nonzero zero divisors, then show that every nonzero $R$-module has nonzero torsion element.

 

>

Proof.

(a) Prove that if $R$ is an integral domain, then $\Tor(M)$ is a submodule of $M$.

To prove $\Tor(M)$ is a submodule of $M$, we check the following submodule criteria:

1. $\Tor(M)$ is not empty.
2. For any $m, n\in \Tor(M)$ and $t\in R$, we have $m+tn\in M$.

It is clear that the zero element $0$ in $M$ is in $\Tor(M)$, hence condition 1 is met.

To prove condition 2, let $m, n \in \Tor(M)$ and $t\in R$.
Since $m, n$ are torsion elements, there exist nonzero elements $r, s\in R$ such that $rm=0, sn=0$.

Since $R$ is an integral domain, the product $rs$ of nonzero elements is nonzero.
We have
\begin{align*}
rs(m+tn)&=rsm+rstn\\
&=s(rm)+rt(sn) &&\text{($R$ is commutative)}\\
&=s0+rt0=0.
\end{align*}
This yields that $m+tn$ is a torsion elements, hence $m+tn\in \Tor(M)$.
This condition 2 is met as well, and we conclude that $\Tor(M)$ is a submodule of $M$.

(b) Find an example of a ring $R$ and an $R$-module $M$ such that $\Tor(M)$ is not a submodule.

Let us consider $R=\Zmod{6}$ and let $M$ be the $R$-module $R$.
We just simply write $n$ for the element $n+6\Z$ in $R=\Zmod{6}$.
Then we have
\[3\cdot 2=0 \text{ and } 2\cdot 3=0.\] This implies that $2$ and $3$ are torsion elements of the module $M$.

However, the sum $5=2+3$ is not a torsion element in $M$ since if $r\cdot 5=0$ in $\Zmod{6}$, then $r=0$.
Thus, $\Tor(M)$ is not closed under addition. Hence it is not a submodule of $M$.

(c) If $R$ has nonzero zero divisors, then show that every nonzero $R$-module has nonzero torsion element.

Let $r$ be nonzero zero divisors of $R$. That is, there exists a nonzero element $s\in R$ such that $rs=0$. Let $M$ be a nonzero $R$-module and let $m$ be a nonzero element in $M$.
Put $n=sm$.

If $n=0$, then this implies $m$ is a nonzero torsion element of $M$, and we are done.
If $n\neq 0$, then we have
\begin{align*}
rn=r(sm)=(rs)m=0m=0.
\end{align*}
This yields that $n$ is a nonzero torsion element of $M$.

Hence, in either case, we obtain a nonzero torsion element of $M$. This completes the proof.

The post Torsion Submodule, Integral Domain, and Zero Divisors first appeared on Problems in Mathematics.

Let $R$ be a commutative ring with unity.
Then show that every maximal ideal of $R$ is a prime ideal.

 

We give two proofs.

Proof 1.

The first proof uses the following facts.

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

Let $M$ be a maximal ideal of $R$. Then by Fact 2, $R/M$ is a field.
Since a field is an integral domain, $R/M$ is an integral domain. Thus by Fact 1, $M$ is a prime ideal.

Proof 2.

In this proof, we solve the problem without using Fact 1, 2.
Let $M$ be a maximal ideal of $R$.

Seeking a contradiction, let us assume that $M$ is not a prime ideal.

Then there exists $a, b\in R$ such that the product $ab$ is in $M$ but $a \not \in M$ and $b \not \in M$.

Consider the ideal $(a)+M$ generated by $a$ and $M$.
Since the ideal $(a)+M$ is strictly larger than $M$, we must have $R=(a)+M$ by the maximality of $M$.

Since $1\in R=(a)+M$, we have
\[1=ra+m,\] where $r\in R$ and $m\in M$.

Similarly, we have $R=(b)+M$ and
\[1=sb+n,\] where $s\in R$ and $n\in M$.

From these equalities, we obtain
\begin{align*}
1&=1\cdot 1\\
&=(ra+m)(sb+n)\\
&=rsab+ran+msb+mn.
\end{align*}

Since $ab, m, n$ are elements in the ideal $M$, the last expression is in $M$.

This yields that $1\in M$, and hence $M=R$. Since a maximal ideal is a proper ideal by definition, this is a contradiction.
Thus, $R$ must be a prime ideal.

Prime ideals are Not Necessarily Maximal

We just have shown that every maximal ideal is a prime ideal.

The converse, however, is not true.

That is, some prime ideals are not maximal ideals.

See the post ↴
Examples of Prime Ideals in Commutative Rings that are Not Maximal Ideals
for examples of rings and prime ideals that are not maximal ideals.

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