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.

(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.

(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.

Prove that any field automorphism of the field of real numbers $\R$ must be the identity automorphism.

Proof.

We prove the problem by proving the following sequence of claims.

Let $\phi:\R \to \R$ be an automorphism of the field of real numbers $\R$.

1. Claim 1. For any positive real number $x$, we have $\phi(x)>0$.
2. Claim 2. For any $x, y\in \R$ such that $x>y$, we have $\phi(x) > \phi(y)$.
3. Claim 3. The automorphism $\phi$ is the identity on positive integers.
4. Claim 4. The automorphism $\phi$ is the identity on rational numbers.
5. Claim 5. The automorphism $\phi$ is the identity on real numbers.

Let us now start proving the claims.
Let $\phi:\R \to \R$ be an automorphism of the field of real numbers $\R$.

Claim 1. For any positive real number $x$, we have $\phi(x)>0$.

Since $x$ is a positive real number, we have $\sqrt{x}\in \R$ and
\[\phi(x)=\phi\left(\sqrt{x}^2\right)=\phi(\sqrt{x})^2 \geq 0.\]

Note that since $\phi(0)=0$ and $\phi$ is bijective, $\phi(x)\neq 0$ for any $x\neq 0$.
Thus, it follows that $\phi(x) > 0$ for each positive real number $x$.
Claim 1 is proved.

Claim 2. For any $x, y\in \R$ such that $x>y$, we have $\phi(x) > \phi(y)$.

Since $x > y$, we have $x-y > 0$ and it follows from Claim 1 that
\[0<\phi(x-y)=\phi(x)-\phi(y).\] Hence, $\phi(x)> \phi(y)$.

Claim 3. The automorphism $\phi$ is the identity on positive integers.

Let $n$ be a positive integer. Then we have
\begin{align*}
\phi(n)=\phi(\underbrace{1+1+\cdots+1}_{\text{$n$ times}})=\underbrace{\phi(1)+\phi(1)+\cdots+\phi(1)}_{\text{$n$ times}}=n
\end{align*}
since $\phi(1)=1$.

Claim 4. The automorphism $\phi$ is the identity on rational numbers.

Any rational number $q$ can be written as $q=\pm m/n$, where $m, n$ are positive integers.
Then we have
\begin{align*}
\phi(q)=\phi\left(\, \pm \frac{m}{n} \,\right)=\pm \frac{\phi(m)}{\phi(n)}=\pm \frac{m}{n}=q,
\end{align*}
where the third equality follows from Claim 3.

Claim 5. The automorphism $\phi$ is the identity on real numbers.

In this claim, we finish the proof of the problem.

Let $x$ be any real number.
Seeking a contradiction, assume that $\phi(x)\neq x$.

There are two cases to consider:
\[x < \phi(x) \text{ or } x > \phi(x).\]

First, suppose that $x < \phi(x)$. Then there exists a rational number $q$ such that \[x< q < \phi(x).\] Then we have \begin{align*} \phi(x) &< \phi(q) && \text{by Claim 2 since $x < q$}\\ &=q && \text{by Claim 4 since $q$ is rational}\\ &<\phi(x) && \text{by the choice of $q$}, \end{align*} and this is a contradiction. Next, consider the case when $x > \phi(x)$.
There exists a rational number $q$ such that
\[\phi(x) < q < x.\] Then by the same argument as above, we have \[\phi(x) < q =\phi(q) < \phi(x),\] which is a contradiction. Thus, in either case we reached a contradiction, and hence we must have $\phi(x)=x$ for all real numbers $x$. This proves that the automorphism $\phi: \R \to \R$ is the identity map.

The post Any Automorphism of the Field of Real Numbers Must be the Identity Map 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.

Let $R$ be a ring with $1$. Prove that the following three statements are equivalent.

1. The ring $R$ is a field.
2. The only ideals of $R$ are $(0)$ and $R$.
3. Let $S$ be any ring with $1$. Then any ring homomorphism $f:R \to S$ is injective.

Proof.

We prove the equivalences $(1) \Leftrightarrow (2)$ and $(2) \Leftrightarrow (3)$.

$(1) \implies (2)$

Suppose that $R$ is a field. Let $I$ be an ideal of $R$.
If $I=(0)$, then there is nothing to prove.
So assume that $I\neq (0)$.

Then there is a nonzero element $x$ in $I$.
Since $R$ is a field, we have $x^{-1}\in R$.

Since $I$ is an ideal, we have
\[1=x^{-1}\cdot x\in I.\] This yields that $I=R$.

$(2) \implies (1)$

Suppose now that the only ideals of $R$ are $(0)$ and $R$.
Let $x$ be a nonzero element of $R$. We show the existence of the inverse of $x$.
Consider the ideal $(x)=xR$ generated by $x$.

Since $x$ is nonzero, the ideal $(x)\neq 0$, and thus we have $(x)=R$ by assumption.
Thus, there exists $y\in R$ such that
\[xy=1.\]

So $y$ is the inverse element of $x$.
Hence $R$ is a field.

$(2)\implies (3)$

Suppose that the only ideals of $R$ are $(0)$ and $R$.
Let $S$ be any ring with $1$ and $f:R\to S$ be any ring homomorphism.
Consider the kernel $\ker(f)$. The kernel $\ker(f)$ is an ideal of $R$, and thus $\ker(f)$ is either $(0)$ or $R$ by assumption.

If $\ker(f)=R$, then the homomorphism $f$ sends $1\in R$ to $0\in S$, which is a contradiction since any ring homomorphism between rings with $1$ sends $1$ to $1$.
Thus, we must have $\ker(f)=0$, and this yields that the homomorphism $f$ is injective.

$(3) \implies (2)$

Suppose that statement 3 is true. That is, any ring homomorphism $f:R\to S$, where $S$ is any ring with $1$, is injective.
Let $I$ be a proper ideal of $R$: an ideal $I\neq R$.
Then the quotient $R/I$ is a ring with $1$ and the natural projection
\[f:R\to R/I\] is a ring homomorphism.

By assumption, the ring homomorphism $f$ is injective, and hence we have
\[(0)=\ker(f)=I.\] This proves that the only ideals of $R$ are $(0)$ and $R$.

The post Three Equivalent Conditions for a Ring to be a Field first appeared on Problems in Mathematics.

Prove that any algebraic closed field is infinite.

Definition.

A field $F$ is said to be algebraically closed if each non-constant polynomial in $F[x]$ has a root in $F$.

Proof.

Let $F$ be a finite field and consider the polynomial
\[f(x)=1+\prod_{a\in F}(x-a).\] The coefficients of $f(x)$ lie in the field $F$, and thus $f(x)\in F[x]$. Of course, $f(x)$ is a non-constant polynomial.

Note that for each $a \in F$, we have
\[f(a)=1\neq 0.\] So the polynomial $f(x)$ has no root in $F$.
Hence the finite field $F$ is not algebraic closed.

It follows that every algebraically closed field must be infinite.

The post Prove that any Algebraic Closed Field is Infinite first appeared on Problems in Mathematics.