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 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 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 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 $R$ be a ring with $1$.
Suppose that $a, b$ are elements in $R$ such that
\[ab=1 \text{ and } ba\neq 1.\]

(a) Prove that $1-ba$ is idempotent.

(b) Prove that $b^n(1-ba)$ is nilpotent for each positive integer $n$.

(c) Prove that the ring $R$ has infinitely many nilpotent elements.

Proof.

(a) Prove that $1-ba$ is idempotent.

We compute
\begin{align*}
(1-ba)^2&=(1-ba)(1-ba)=1-ba-ba+b\underbrace{ab}_{=1}a\\
&=1-ba-ba+ba=1-ba.
\end{align*}
Thus, we have $(1-ba)^2=1-ba$, and hence $1-ba$ is idempotent.

(b) Prove that $b^n(1-ba)$ is nilpotent for each positive integer $n$.

As a lemma, we show that $(1-ba)b=0$.
To see this, we calculate
\begin{align*}
(1-ba)b=b-b\underbrace{ab}_{=1}=b-b=0.
\end{align*}

Now we compute
\begin{align*}
b^n(1-ba)\cdot b^n(1-ba)&=b^n\underbrace{(1-ba)b}_{=0 \text{ by lemma}}b^{n-1}(1-ba)=0.
\end{align*}
This proves that $b^n(1-ba)$ is nilpotent.

(c) Prove that the ring $R$ has infinitely many nilpotent elements.

In part (a), we showed that the element $b^n(1-ba)$ is a nilpotent element of $R$ for each positive integer $n$.
We claim that $b^n(1-ba)\neq b^m(1-ba)$ for each pair of distinct integers $m, n$.
Without loss of generality, we may assume that $m > n$.

We state simple facts which are needed below.
We have
\begin{align*}
a^nb^n&=1\\
a^nb^m&=b^{m-n}.
\end{align*}
Note that $a^nb^n$ and $a^nb^m$ look like
\[aa\cdots a\cdot bb\cdots b.\] Then we use the relation $ab=1$ from the middle successively, and we obtain the right-hand sides.

Now we prove that $b^n(1-ba)\neq b^m(1-ba)$ for each pair of distinct integers $m, n$.
Assume on the contrary $b^n(1-ba)= b^m(1-ba)$ for $m > n$.
Then we multiply by $a^n$ on the left and get
\begin{align*}
a^n b^n(1-ba)= a^n b^m(1-ba).
\end{align*}

Using the facts stated above, we obtain
\[1-ba=b^{m-n}(1-ba).\] Note that the left-hand side is a nonzero idempotent element by part (a).
On the other hand, the right-hand side is nilpotent by part (b).
Since a nonzero idempotent element can never be nilpotent, this is a contradiction.

Therefore, $b^n(1-ba)\neq b^m(1-ba)$ for each pair of distinct integers $m, n$.
Hence there are infinitely many nilpotent elements in $R$.

The post A Ring Has Infinitely Many Nilpotent Elements if $ab=1$ and $ba \neq 1$ first appeared on Problems in Mathematics.

Let $R$ be a ring with $1\neq 0$. Let $a, b\in R$ such that $ab=1$.

(a) Prove that if $a$ is not a zero divisor, then $ba=1$.

(b) Prove that if $b$ is not a zero divisor, then $ba=1$.

Definition.

An element $x\in R$ is called a zero divisor if there exists a nonzero element $y\in R$ such that $xy=0$ or $yx=0$.

So if $x$ is not a zero dividor, then $xy=0$ implies that $y=0$. Similarly, $yx=0$ implies that $y=0$.

Proof.

(a) Prove that if $a$ is not a zero divisor, then $ba=1$.

Suppose that $a$ is not a zero divisor. We compute
\begin{align*}
a(ba-1)&=aba-a && \text{by distributivity}\\
&=1\cdot a -a &&\text{by $ab=1$}\\
&=a-a=0.
\end{align*}

Since $a$ is not a zero divisor, this yields that $ba-1=0$, and hence $ba=1$.

(b) Prove that if $b$ is not a zero divisor, then $ba=1$.

Suppose that $b$ is not a zero divisor. We calculate
\begin{align*}
(ba-1)b&=bab-b && \text{by distributivity}\\
&=b\cdot 1 -b &&\text{by $ab=1$}\\
&=b-b=0.
\end{align*}

As $b$ is not a zero divisor, the equality $(ba-1)b=0$ implies that $ba-1=0$.
Hence we have $ba=1$.

The post If $ab=1$ in a Ring, then $ba=1$ when $a$ or $b$ is Not a Zero Divisor first appeared on Problems in Mathematics.