Prove that if $R$ is a commutative ring and $\frakN(R)$ is its nilradical, then the zero is the only nilpotent element of $R/\frakN(R)$. That is, show that $\frakN(R/\frakN(R))=0$.

Proof.

Let $r\in R$ and if $x:=r+\frakN(R) \in R/\frakN(R)$ is a nilpotent element of $R/\frakN(R)$, then there exists an integer $n$ such that
\[x^n=(r+\frakN(R))^n=r^n+\frakN(R)=\frakN(R).\] Thus we have
\[r^n\in \frakN(R).\]

This means that $r^n$ is an nilpotent element of $R$, and hence there exists an integer $m$ such that
\[r^{nm}=(r^n)^m=0.\]

Therefore $r$ is an nilpotent element of $R$, that is $r\in \frakN(R)$ and we have $x=\frakN(R)$, which is the zero element in $R/\frakN(R)$.

The post The Zero is the only Nilpotent Element of the Quotient Ring by its Nilradical 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.

Is it true that a set of nilpotent elements in a ring $R$ is an ideal of $R$?

If so, prove it. Otherwise give a counterexample.

Proof.

We give a counterexample.
Let $R$ be the noncommutative ring of $2\times 2$ matrices with real coefficients.
Consider the following matrices $A, B$ in $R$.
\[A=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix} \text{ and } B=\begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix}.\] Direct computation shows that $A^2$ and $B^2$ are the zero matrix, hence $A, B$ are nilpotent elements.

However, the sum $A+B=\begin{bmatrix}
0 & 1\\
1& 0
\end{bmatrix}$ is not nilpotent as we have
\begin{align*}
\begin{bmatrix}
0 & 1\\
1& 0
\end{bmatrix}^n
=\begin{cases} \begin{bmatrix}
0 & 1\\
1& 0
\end{bmatrix} & \text{if $n$ is odd}\\[10pt] \begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix} & \text{ if $n$ is even}.
\end{cases}
\end{align*}
Hence the set of nilpotent elements in $R$ is not an ideal as it is not even an additive abelian group.

Comment.

If a ring $R$ is commutative, then it is true that the set of nilpotent elements form an ideal, which is called the nilradical of $R$.

The post Is the Set of Nilpotent Element an Ideal? first appeared on Problems in Mathematics.

Let $R$ be a commutative ring with $1$ such that every element $x$ in $R$ is idempotent, that is, $x^2=x$. (Such a ring is called a Boolean ring.)

(a) Prove that $x^n=x$ for any positive integer $n$.

(b) Prove that $R$ does not have a nonzero nilpotent element.

Proof.

(a) Prove that $x^n=x$ for any positive integer $n$.

By assumption, $x^n=x$ is true for $n=1, 2$.

Suppose that $x^k=x$ for some $k \geq 2$ (induction hypothesis).
Then we have
\begin{align*}
x^{k+1}&=xx^k\\
&=xx &&\text{by induction hypothesis}\\
&=x^2=x &&\text{by assumption.}
\end{align*}

Thus, we conclude that $x^n=x$ for any positive integer $n$ by induction.

(b) Prove that $R$ does not have a nonzero nilpotent element.

Let $x$ be a nilpotent element in $R$. That is, there is a positive integer $n$ such that $x^n=0$.

It follows from part (a) that $x=x^n=0$.
Thus every nilpotent element in $R$ is $0$.

The post Boolean Rings Do Not Have Nonzero Nilpotent Elements first appeared on Problems in Mathematics.

Let $R$ be a ring and assume that whenever $ab=ca$ for some elements $a, b, c\in R$, we have $b=c$.

Then prove that $R$ is a commutative ring.

Proof.

Let $x, y$ be arbitrary elements in $R$. We want to show that $xy=yx$.
Consider the identity
\[y(xy)=(yx)y.\] This can be written as $ab=ca$ if we put $a=y, b=xy, c=yx$.

It follows from the assumption that we have $b=c$.
Equivalently, we have $xy=yx$.

As this is true for any $x, y\in R$, we conclude that $R$ is a commutative ring.

The post A Ring is Commutative if Whenever $ab=ca$, then $b=c$ 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 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.