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 $R$ and $R’$ be commutative rings and let $f:R\to R’$ be a ring homomorphism.
Let $I$ and $I’$ be ideals of $R$ and $R’$, respectively.

(a) Prove that $f(\sqrt{I}\,) \subset \sqrt{f(I)}$.

(b) Prove that $\sqrt{f^{-1}(I’)}=f^{-1}(\sqrt{I’})$

(c) Suppose that $f$ is surjective and $\ker(f)\subset I$. Then prove that $f(\sqrt{I}\,) =\sqrt{f(I)}$

Proof.

(a) Prove that $f(\sqrt{I}\,) \subset \sqrt{f(I)}$.

Let $x\in f(\sqrt{I}\,)$ be an arbitrary element. Then there is $a\in \sqrt{I}$ such that $f(a)=x$. As $a\in \sqrt{I}$, there exists a positive integer $n$ such that $a^n\in I$.

It follows that we have
\begin{align*}
x^n=f(a)^n=f(a^n)\in f(I).
\end{align*}

This implies that $x\in \sqrt{f(I)}$.
Hence we have $f(\sqrt{I}\,) \subset \sqrt{f(I)}$.

(b) Prove that $\sqrt{f^{-1}(I’)}=f^{-1}(\sqrt{I’})$

$(\subset)$ Let $x\in \sqrt{f^{-1}(I’)}$. Then there is a positive integer $n$ such that $x^n\in f^{-1}(I’)$ and thus $f(x^n)\in I’$.
As $f$ is a ring homomorphism, it follows that $f(x)^n=f(x^n)\in I’$.

Hence $f(x)\in \sqrt{I’}$, and then $x\in f^{-1}(\sqrt{I’})$.
This proves that $\sqrt{f^{-1}(I’)} \subset f^{-1}(\sqrt{I’})$.

$(\supset)$ Let $x\in f^{-1}(\sqrt{I’})$. Then $f(x)\in \sqrt{I’}$. It follows that there exists a positive integer $n$ such that $f(x^n)=f(x)^n\in I’$.
Hence $x^n\in f^{-1}(I’)$, and we deduce that $x\in \sqrt{f^{-1}(I’)}$.

This proves that $f^{-1}(\sqrt{I’}) \subset \sqrt{f^{-1}(I’)}$.
Combining this with the previous inclusion yields that $\sqrt{f^{-1}(I’)}=f^{-1}(\sqrt{I’})$.

(c) Suppose that $f$ is surjective and $\ker(f)\subset I$. Then prove that $f(\sqrt{I}\,) =\sqrt{f(I)}$

We now suppose that $f$ is surjective and $\ker(f)\subset I$. We proved $f(\sqrt{I}\,) \subset \sqrt{f(I)}$ in part (a). To show the reverse inclusion, let $x\in \sqrt{f(I)}\subset R’$.
Then there is a positive integer $n$ such that $x^n\in f(I)$.
So there exists $a\in I$ such that $f(a)=x^n$.

Since $f:R\to R’$ is surjective, there exists $y\in R$ such that $f(y)=x$.
Then we have
\begin{align*}
f(a)=x^n=f(y)^n=f(y^n),
\end{align*}
and hence $f(a-y^n)=0$.
Thus $a-y^n\in \ker(f) \subset I$ by assumption.
As $a\in I$, it follows that $y^n\in I$ as well.

We deduce that $y\in \sqrt{I}$ and
\[x=f(y)\in f(\sqrt{I}),\] which completes the proof that $\sqrt{f(I)} \subset f(\sqrt{I})$.

Putting together this inclusion and the inclusion in (a) yields the required equality $f(\sqrt{I}\,) =\sqrt{f(I)}$.

The post Ring Homomorphisms and Radical Ideals first appeared on Problems in Mathematics.

Let $I=(x, 2)$ and $J=(x, 3)$ be ideal in the ring $\Z[x]$.

(a) Prove that $IJ=(x, 6)$.

(b) Prove that the element $x\in IJ$ cannot be written as $x=f(x)g(x)$, where $f(x)\in I$ and $g(x)\in J$.

Hint.

If $I=(a_1,\dots, a_m)$ and $J=(b_1, \dots, b_n)$ are ideals in a commutative ring, then we have
\[IJ=(a_ib_j),\] where $1\leq i \leq m$ and $1\leq j \leq n$.

Proof.

(a) Prove that $IJ=(x, 6)$.

Note that the product ideal $IJ$ is generated by the products of generators of $I$ and $J$, that is, $x^2, 2x, 3x, 6$. That is, $IJ=(x^2, 2x, 3x, 6)$.

It follows that $IJ$ contains $x=3x-2x$ as well. As the first three generators can be generated by $x$, we deduce that $IJ=(x, 6)$.

(b) Prove that the element $x\in IJ$ cannot be written as $x=f(x)g(x)$, where $f(x)\in I$ and $g(x)\in J$.

Assume that $x=f(x)g(x)$ for some $f(x)\in I$ and $g(x) \in J$.
As $\Z[x]$ is a UFD, we have either
\[f(x)=\pm x, g(x)=\pm 1, \text{ or } f(x)=\pm 1, g(x)=\pm x.\]

In the former case, we have $1\in J$ and hence $J=\Z[x]$, which is a contradiction.
Similarly, in the latter case, we have $1\in I$ and hence $I=\Z[x]$, which is a contradiction.
Thus, in either case, we reached a contradiction.

Hence, $x$ cannot be written as the product of elements in $I$ and $J$.

Comment.

Let $I$ and $J$ be an ideal of a commutative ring $R$.
Then the product of ideals $I$ and $J$ is defined to be
\[IJ:=\{\sum_{i=1}^k a_i b_i \mid a_i\in I, b_i\in J, k\in \N\}.\]

The above problems shows that in general, there are elements in the product $IJ$ that cannot be expressed simply as $ab$ for $a\in I$ and $b\in J$.

The post Example of an Element in the Product of Ideals that Cannot be Written as the Product of Two Elements 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.

Let $R$ and $S$ be rings with $1\neq 0$.

Prove that every ideal of the direct product $R\times S$ is of the form $I\times J$, where $I$ is an ideal of $R$, and $J$ is an ideal of $S$.

Proof.

Let $K$ be an ideal of the direct product $R\times S$.
Define
\[I=\{a\in R \mid (a,b)\in K \text{ for some } b\in S\}\] and
\[J=\{b\in S \mid (a, b)\in K \text{ for some } a\in R\}.\]

We claim that $I$ and $J$ are ideals of $R$ and $S$, respectively.

Let $a, a’\in I$. Then there exist $b, b’\in S$ such that $(a, b), (a’, b’)\in K$.
Since $K$ is an ideal we have
\[(a,b)+(a’,b’)=(a+a’, b+b)\in k.\]

It follows that $a+a’\in I$.
Also, for any $r\in R$ we have
\[(r,0)(a,b)=(ra,0)\in K\] because $K$ is an ideal.

Thus, $ra\in I$, and hence $I$ is an ideal of $R$.
Similarly, $J$ is an ideal of $S$.

Next, we prove that $K=I \times J$.
Let $(a,b)\in K$. Then by definitions of $I$ and $J$ we have $a\in I$ and $b\in J$.
Thus $(a,b)\in I\times J$. So we have $K\subset I\times J$.

On the other hand, consider $(a,b)\in I \times J$.
Since $a\in I$, there exists $b’\in S$ such that $(a, b’)\in K$.
Also since $b\in J$, there exists $a’\in R$ such that $(a’, b)\in K$.

As $K$ is an ideal of $R\times S$, we have
\[(1,0)(a,b’)=(a,0)\in K \text{ and } (0, 1)(a’,b)=(0, b)\in K.\] It yields that
\[(a,b)=(a,0)+(0,b)\in K.\] Hence $I\times J \subset K$.

Putting these inclusions together gives $k=I\times J$ as required.

Remark.

The ideals $I$ and $J$ defined in the proof can be alternatively defined as follows.
Consider the natural projections
\[\pi_1: R\times S \to R \text{ and } \pi_2:R\times S \to S.\] Define
\[I=\pi_1(K) \text{ and } J=\pi_2(K).\]

Since the natural projections are surjective ring homomorphisms, the images $I$ and $J$ are ideals in $R$ and $S$, respectively.
(see the post The Image of an Ideal Under a Surjective Ring Homomorphism is an Ideal.)

The post Every Ideal of the Direct Product of Rings is the Direct Product of Ideals first appeared on Problems in Mathematics.

Let $R$ and $S$ be rings. Suppose that $f: R \to S$ is a surjective ring homomorphism.

Prove that every image of an ideal of $R$ under $f$ is an ideal of $S$.
Namely, prove that if $I$ is an ideal of $R$, then $J=f(I)$ is an ideal of $S$.

Proof.

As in the statement of the problem, let $I$ be an ideal of $R$.
Our goal is to show that the image $J=f(I)$ is an ideal of $S$.

For any $a,b\in J$ and $s\in S$, we need to show that

Since $a, b\in J=f(I)$, there exists $a’, b’\in I$ such that
\[f(a’)=a \text{ and } f(b’)=b.\]

Then we have
\begin{align*}
a+b=f(a’)+f(b’)=f(a’+b’)
\end{align*}
since $f$ is a homomorphism.

Since $I$ is an ideal, the sum $a’+b’$ is in $I$.
This yields that $a+b\in f(I)=J$, which proves (1).

Since $f:R\to S$ is surjective, there exists $r\in R$ such that $f(r)=s$.
It follows that
\[sa=f(r)f(a’)=f(ra’)\] since $f$ is a homomorphism.

Since $I$ is an ideal of $R$, the product $ra’$ is in $I$.
Hence $sa\in f(I)=J$, and (2) is proved.

Therefore the image $J=f(I)$ is an ideal of $S$.

The post The Image of an Ideal Under a Surjective Ring Homomorphism is an Ideal 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.

A ring is called local if it has a unique maximal ideal.

(a) Prove that a ring $R$ with $1$ is local if and only if the set of non-unit elements of $R$ is an ideal of $R$.

(b) Let $R$ be a ring with $1$ and suppose that $M$ is a maximal ideal of $R$.
Prove that if every element of $1+M$ is a unit, then $R$ is a local ring.

Proof of (a).

$(\implies)$: If $R$ is a local ring then the set of non-units is an ideal

Suppose that $R$ is a local ring and let $M$ be the unique maximal ideal of $R$.

We denote by $I$ the set of non-unit elements of $R$.

Let $a, b\in I$.
Since $a, b$ are non-unit elements, the ideals $(a)$ and $(b)$ generated by $a$ and $b$, respectively, are proper ideals of $R$.
Since $M$ is the only maximal ideal of $R$, it follows that
\[(a) \subset M \text{ and } (b) \subset M.\]

It yields that $a-b\in M$ since $a, b\in M$ and $M$ is an ideal.
As $M$ is a proper ideal, $a-b$ is a non-unit, hence $a-b\in I$.

Also for any $r\in R$, we have $ra\in M$ since $a\in M$ and $M$ is an ideal of $R$.
It follows that $ra$ is a non-unit because $M$ is a proper ideal.
Hence $ra\in I$.

Therefore the set $I$ is an ideal of $R$.

$(\impliedby)$: If the set of non-units is an ideal, then $R$ is a local ring

Suppose that the set $I$ of non-units elements in $R$ is an ideal of $R$.
Since $1\in R$ is a unit, $I$ is a proper ideal.

Let $M$ be an arbitrary maximal ideal of $R$.
Note that every element of $M$ is a non-unit element of $R$ since $M$ is a proper ideal.
Thus we have $M\subset I$.
Since $M$ is a maximal ideal, it yields that $M=I$.

Therefore $I$ is the unique maximal ideal of $R$, and hence $R$ is a local ring.

Proof of (b).

We prove that the maximal ideal $M$ is the set of non-units elements in $R$.
Then the result follows from part (a).

Take any $a\in R\setminus M$.
Then the ideal $(a)+M$ generated by $a$ and $M$ is strictly larger than $M$.
Hence
\[(a)+M=R\] by the maximality of $M$.

Then there exists $r\in R$ and $m\in M$ such that
\[ra+m=1.\] Since $ra=1-m\in 1+M$, it follows from the assumption that $ra$ is a unit.
It yield that $a$ is a unit.

Since $M$ contains no unit elements, we see that $M$ consists of non-unit elements of $R$.
Thus, by part (a) we conclude that $R$ is a local ring.

The post A ring is Local if and only if the set of Non-Units is an Ideal first appeared on Problems in Mathematics.