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.

Prove that the rings $\Z[x]$ and $\Q[x]$ are not isomoprhic.

Proof.

We give three proofs.
The first two proofs use only the properties of ring homomorphism.

The third proof resort to the units of rings.

If you are familiar with units of $\Z[x]$, then the third proof might be concise and easy to follow.

The First Proof

Assume on the contrary that the rings $\Z[x]$ and $\Q[x]$ are isomorphic.
Let
\[\phi:\Q[x] \to \Z[x]\] be an isomorphism.

The polynomial $x$ in $\Q[x]$ is mapped to the polynomial $\phi(x)\in \Z[x]$.

Note that $\frac{x}{2^n}$ is an element in $\Q[x]$ for any positive integer $n$.
Thus we have
\begin{align*}
\phi(x)&=\phi(2^n\cdot \frac{x}{2^n})\\
&=2^n\phi\left(\frac{x}{2^n}\right)
\end{align*}
since $\phi$ is a homomorphism.

As $\phi$ is injective, the polynomial $\phi(\frac{x}{2^n})\neq 0$.
Since $\phi(\frac{x}{2^n})$ is a nonzero polynomial with integer coefficients, the absolute values of the nonzero coefficients of $2^n\phi(\frac{x}{2^n})$ is at least $2^n$.

However, since this is true for any positive integer, the coefficients of the polynomial $\phi(x)=2^n\phi(\frac{x}{2^n})$ is arbitrarily large, which is impossible.
Thus, there is no isomorphism between $\Q[x]$ and $\Z[x]$.

The Second Proof

Seeking a contradiction, assume that we have an isomorphism
\[\phi:\Q[x] \to \Z[x].\]

Since $\phi$ is a ring homomorphism, we have $\phi(1)=1$.
Then we have
\begin{align*}
1&=\phi(1)=\phi \left(2\cdot \frac{1}{2}\right)\\
&=2\phi\left( \frac{1}{2} \right)
\end{align*}
since $\phi$ is a homomorphism.

Since $\phi\left( \frac{1}{2} \right)\in \Z[x]$, we write
\[\phi\left( \frac{1}{2} \right)=a_nx^n+a_{n-1}x^{n-1}+\cdots a_1x+a_0,\] for some integers $a_0, \dots, a_n$.

Since $2\phi\left( \frac{1}{2} \right)=1$, it follows that
\[2a_n=0, \dots, 2a_1=0, 2a_0=1.\] Since $a_0$ is an integer, this is a contradiction.
Thus, such an isomorphism does not exists.
Hence $\Q[x]$ and $\Z[x]$ are not isomorphic.

The Third Proof

Note that in general the units of the polynomial ring $R[x]$ over an integral domain $R$ is the units $R^{\times}$ of $R$.

Since $\Z$ and $\Q$ are both integral domain, the units are
\[\Z[x]^{\times}=\Z^{\times}=\{\pm 1\} \text{ and } \Q[x]^{\times}=\Q^{\times}=\Q\setminus \{0\}.\] Since every ring isomorphism maps units to units, if two rings are isomorphic then the number of units must be the same.

As seen above, $\Z[x]$ contains only two units although $\Q[x]$ contains infinitely many units.
Thus, they cannot be isomorphic.

The post The Polynomial Rings $\Z[x]$ and $\Q[x]$ are Not Isomorphic first appeared on Problems in Mathematics.

Is there a (not necessarily commutative) ring $R$ with $1$ such that the equation
\[x+x=1 \] has more than one solutions $x\in R$?

Solution.

We claim that there is at most one solution $x$ in the ring $R$.
Suppose that we have two solutions $r, s \in R$. That is, we have
\[r+r=1 \text{ and } s+s=1.\]

Then we have $r+r=s+s$. Putting $a=r-s\in R$, we have
\[a+a=0.\] Now we compute
\begin{align*}
0&=1\cdot 0 =(r+r)(a+a)\\
&=ra+ra+ra+ra\\
&=(r+r)a+r(a+a)\\
&=1\cdot a+r\cdot 0\\
&=a.
\end{align*}

Therefore we obtain $a=0$ and thus $r=s$.
It follows that the equation $x+x=1$ has only one solution (at most).

The post How Many Solutions for $x+x=1$ in a Ring? first appeared on Problems in Mathematics.

Show that any finite integral domain $R$ is a field.
 

Definition.

A commutative ring $R$ with $1\neq 0$ is called an integral domain if it has no zero divisors.
That is, if $ab=0$ for $a, b \in R$, then either $a=0$ or $b=0$.

Proof.

We give two proofs.

Proof 1.

Let $r \in R$ be a nonzero element in $R$.
We show that $r$ is a unit.

Consider the map $f: R\to R$ sending $x\in R$ to $f(x)=rx$.
We claim that the map $f$ is injective.
Suppose that we have $f(x)=f(y)$ for $x, y \in R$. Then we have
\[rx=ry\] or equivalently, we have
\[r(x-y)=0.\]

Since $R$ is an integral domain and $r\neq 0$, we must have $x-y=0$, and thus $x=y$.
Hence $f$ is injective. Since $R$ is a finite set, the map is also surjective.

Then it follows that there exists $s\in R$ such that $rs=f(s)=1$, and thus $r$ is a unit.
Since any nonzero element of a commutative ring $R$ is a unit, $R$ is a field.

Proof 2.

Let $r\in R$ be a nonzero element.
We show that the inverse element of $r$ exists in $R$ as follows.
Consider the powers of $r$:
\[r, r^2, r^3,\dots.\] Since $R$ is a finite ring, not all of the powers cannot be distinct.
Thus, there exist positive integers $m > n$ such that
\[r^m=r^n.\]

Equivalently we have
\[r^n(r^{m-n}-1)=0.\] Since $R$ is an integral domain, this yields either $r^n=0$ or $r^{m-n}-1=0$.
But the former gives $r=0$, and this is a contradiction since $r\neq 0$.

Hence we have $r^{m-n}=1$, and thus
\[r\cdot r^{m-n-1}=1.\] Since $r-m-1 \geq 0$, we have $r^{m-n-1}\in R$ and it is the inverse element of $r$.

Therefore, any nonzero element of $R$ has the inverse element in $R$, hence $R$ is a field.

Related Question.

For a solution, check out the post “Every Prime Ideal of a Finite Commutative Ring is Maximal“.

The post Finite Integral Domain is a Field first appeared on Problems in Mathematics.

Denote by $i$ the square root of $-1$.
Let
\[R=\Z[i]=\{a+ib \mid a, b \in \Z \}\] be the ring of Gaussian integers.
We define the norm $N:\Z[i] \to \Z$ by sending $\alpha=a+ib$ to
\[N(\alpha)=\alpha \bar{\alpha}=a^2+b^2.\]

Here $\bar{\alpha}$ is the complex conjugate of $\alpha$.
Then show that an element $\alpha \in R$ is a unit if and only if the norm $N(\alpha)=\pm 1$.
Also, determine all the units of the ring $R=\Z[i]$ of Gaussian integers.

Proof.

Suppose that an element $\alpha$ is a unit of $R$.
Then there exists $\beta \in R$ such that $\alpha \beta=1$.

Then the norm of $\alpha \beta$ is
\begin{align*}
N(\alpha \beta)&=(\alpha \beta)(\overline{\alpha \beta})\\
&=\alpha \bar{\alpha} \beta \bar{\beta}\\
&=N(\alpha)N(\beta).
\end{align*}

Since the norm $N(1)=1$, we obtain
\[1=N(\alpha)N(\beta)\] in the ring $\Z$. Since $N(\alpha)$ and $N(\beta)$ are both integers, it follows that we have
\[N(\alpha)=\pm 1 \text{ and } N(\beta)=\pm 1.\]

On the other hand, suppose that $N(\alpha)=\pm 1$ for an element $\alpha\in R$.
Then let $\beta:=N(\alpha)^{-1}\bar{\alpha}$.

Since $N(\alpha)^{-1}=\pm 1$, the element $\beta \in R$. We have
\begin{align*}
\beta \alpha&=N(\alpha)^{-1}\bar{\alpha}\cdot \alpha\\
&=N(\alpha)^{-1}N(\alpha)=1.
\end{align*}
Thus the element $\alpha$ is a unit in $R$.

Using this result, let us determine all units of the ring $R$ of Gaussian integers.
An element $\alpha=a+ib \in R$ is a unit if and only if
\[N(\alpha)=a^2+b^2=1,\] where $a, b \in \Z$. Thus only solutions are
\[(a,b)=(\pm 1, 0), (0, \pm 1).\] Therefore the units of $R=\Z[i]$ are
\[\pm 1, \pm i.\] Click here if solved 31

The post Ring of Gaussian Integers and Determine its Unit Elements first appeared on Problems in Mathematics.

Let $R$ be a commutative ring.

Then prove that $R$ is a field if and only if $\{0\}$ is a maximal ideal of $R$.
 

Proof.

$(\implies)$: If $R$ is a field, then $\{0\}$ is a maximal ideal

Suppose that $R$ is a field and let $I$ be a non zero ideal:
\[ \{0\} \subsetneq I \subset R.\]

Then the ideal $I$ contains a nonzero element $x \neq 0$. Since $R$ is a field, we have the inverse $x^{-1}\in R$.
Then it follows that $1=x^{-1}x \in I$ since $x$ is in the ideal $I$.

Since $1\in I$, any element $r \in R$ is in $I$ as $r=r\cdot 1 \in I$.
Thus we have $I=R$ and this proves that $\{0\}$ is a maximal ideal of $R$.

$(\impliedby)$: If $\{0\}$ is a maximal ideal, then $R$ is a field

Let us now suppose that $\{0\}$ is a maximal ideal of $R$.
Let $x$ be any nonzero element in $R$.

Then the ideal $(x)$ generated by the element $x$ properly contains the ideal $\{0\}$.
Since $\{0\}$ is a maximal ideal, we must have $(x)=R$.

Since $1\in R=(x)$, there exists $y\in R$ such that $1=xy$.
This implies that the element $x$ is invertible. Therefore any nonzero element of $R$ is invertible, and hence $R$ is a field.

The post Ring is a Filed if and only if the Zero Ideal is a Maximal Ideal first appeared on Problems in Mathematics.

Let $R$ be a commutative ring with $1 \neq 0$.
An element $a\in R$ is called nilpotent if $a^n=0$ for some positive integer $n$.

Then prove that if $a$ is a nilpotent element of $R$, then $1-ab$ is a unit for all $b \in R$.
 

We give two proofs.

Proof 1.

Since $a$ is nilpotent, we have $a^n=0$ for some positive integer $n$.
Then for any $b \in R$, we have $(ab)^n=a^nb^n=0$ since $R$ is commutative.

Then we have the following equality:
\[(1-ab)(1+(ab)+(ab)^2+\cdots+(ab)^{n-1})=1.\] Therefore $1-ab$ is a unit in $R$.

Proof 2.

There exists $n \in \N$ such that $a^n=0$ since $a$ is nilpotent.
Assume that $1-ab$ is not a unit for some $b \in R$.
Then there exists a prime ideal $\frakp$ of $R$ such that $\frakp \ni 1-ab$.

Since $a^n=0\in \frakp$, we have $a\in \frakp$ since $\frakp$ is an prime ideal.
Then $ab \in \frakp$ and we have
\[1=(1-ab)+ab \in \frakp.\]

However, this implies that $\frakp=R$ and this is a contradiction. Thus $1-ab$ is a unit of the ring $R$ for all $b \in R$.

The post Nilpotent Element a in a Ring and Unit Element -ab$ first appeared on Problems in Mathematics.