Prove that the ring of integers
\[\Z[\sqrt{2}]=\{a+b\sqrt{2} \mid a, b \in \Z\}\] of the field $\Q(\sqrt{2})$ is a Euclidean Domain.

Proof.

First of all, it is clear that $\Z[\sqrt{2}]$ is an integral domain since it is contained in $\R$.

We use the norm given by the absolute value of field norm.
Namely, for each element $a+\sqrt{2}b\in \Z[\sqrt{2}]$, define
\[N(a+\sqrt{2}b)=|a^2-2b^2|.\] Then the map $N:\Z[\sqrt{2}] \to \Z_{\geq 0}$ is a norm on $\Z[\sqrt{2}]$.
Also, it is multiplicative:
\[N(xy)=N(x)N(y).\] Remark that since this norm comes from the field norm of $\Q(\sqrt{2})$, the multiplicativity of $N$ holds for $x, y \in \Q(\sqrt{2})$ as well.

We show the existence of a Division Algorithm as follows.
Let
\[x=a+b\sqrt{2} \text{ and } y=c+d\sqrt{2}\] be arbitrary elements in $\Z[\sqrt{2}]$, where $a,b,c,d\in \Z$.

We have
\begin{align*}
\frac{x}{y}=\frac{a+b\sqrt{2}}{c+d\sqrt{2}}=\frac{(ac-2bd)+(bc-ad)\sqrt{2}}{c^2-2d^2}=r+s\sqrt{2},
\end{align*}
where we put
\[r=\frac{ac-2bd}{c^2-2d^2} \text{ and } s=\frac{bc-ad}{c^2-2d^2}.\]

Let $n$ be an integer closest to the rational number $r$ and let $m$ be an integer closest to the rational number $s$, so that
\[|r-n| \leq \frac{1}{2} \text{ and } |s-m| \leq \frac{1}{2}.\]

Let
\[t:=r-n+(s-m)\sqrt{2}.\]

Then we have
\begin{align*}
t&=r+s\sqrt{2}-(n+m\sqrt{2})\\
&=\frac{x}{y}-(n+m\sqrt{2}).
\end{align*}

It follows that
\begin{align*}
yt=x-(n+m\sqrt{2})y \in \Z[\sqrt{2}].
\end{align*}

Thus we have
\begin{align*}
x=(n+m\sqrt{2})y+yt \tag{*}
\end{align*}
with $n+m\sqrt{2}, yt\in \Z[\sqrt{2}]$.

We have
\begin{align*}
N(t)&= |(r-n)^2-2(s-m)^2|\\
&\leq |r-n|^2+2|s-m|^2\\
& \leq \frac{1}{4}+2\cdot\frac{1}{4}=\frac{3}{4}.
\end{align*}

It follows from the multiplicativity of the norm $N$ that
\begin{align*}
N(yt)=N(y)N(t)\leq \frac{3}{4}N(y)< N(y). \end{align*} Thus the expression (*) gives a Division Algorithm with quotient $n+m\sqrt{2}$ and remainder $yt$.

Related Question.

For a proof of this problem, see the post “5 is prime but 7 is not prime in the ring $\Z[\sqrt{2}]$“.

The post The Ring $\Z[\sqrt{2}]$ is a Euclidean Domain first appeared on Problems in Mathematics.

For each positive integer $n$, prove that the polynomial
\[(x-1)(x-2)\cdots (x-n)-1\] is irreducible over the ring of integers $\Z$.

Proof.

Note that the given polynomial has degree $n$.
Suppose that the polynomial is reducible over $\Z$ and it decomposes as
\[(x-1)(x-2)\cdots (x-n)-1=f(x)g(x) \tag{*}\] for some polynomials $f(x)$ and $g(x)$ in $\Z[x]$ of degree less than $n$.

Evaluating at $x=1,2, \dots, n$, we obtain
\[f(k)g(k)=-1\] for $k=1, 2, \dots, n$.
Since $f(x)$ and $g(x)$ have integer coefficients, both $f(k)$ and $g(k)$ are integers for $k=1, 2, \dots, n$.
It follows that
\[f(k)=1=-g(k) \text{ or } f(k)=-1=-g(k). \tag{**}\]

Consider the polynomial $h(x):=f(x)+g(x)\in \Z[x]$.
Then we have
\begin{align*}
h(k)=f(k)+g(k)=0
\end{align*}
for $k=1, 2, \dots, n$ by (**).
Thus, the polynomial $h(x)$ has at least $n$ roots.

However, the degree of $h(x)$ is less than $n$ as so are the degrees of $f(x)$ and $g(x)$.
This forces that $h(x)=0$ for all $x$, and hence $f(x)=-g(x)$.

Then (*) becomes
\[(x-1)(x-2)\cdots (x-n)-1=-g(x)^2.\] The leading coefficient of the left hand side is $1$ but the leading coefficient of the right hand side is a negative number, hence it is a contradiction.

Therefore, the polynomial $(x-1)(x-2)\cdots (x-n)-1$ must be irreducible over $\Z$.

The post Polynomial $(x-1)(x-2)\cdots (x-n)-1$ is Irreducible Over the Ring of Integers $\Z$ first appeared on Problems in Mathematics.

Let $\Z$ be the ring of integers and let $R$ be a ring with unity.
Determine all the ring homomorphisms from $\Z$ to $R$.

Definition.

Recall that if $A, B$ are rings with unity then a ring homomorphism $f: A \to B$ is a map satisfying

for all $x, y \in A$ and $1_A, 1_B$ are unity elements of $A$ and $B$, respectively.

Proof.

We claim that there is one and only one ring homomorphism from $\Z$ to $R$.

Let us first remark that there is at least one ring homomorphism $\Z \to R$.
Define the map $f_0:\Z\to R$ by
\[f(n)=n.\] (To be more precise, this means $f(n) = n \cdot 1_R$, where $1_R$ is the unit element in $R$.)
Then it is clear that $f_0$ is a ring homomorphism from $\Z$ to $R$.
We want to prove that this is the only ring homomorphism.

Suppose that $f:\Z\to R$ is a ring homomorphism.
By definition, we must have
\[f(1)=1_R.\] Using property (1) with $x=y=0$, we see that
\[f(0)=f(0)+f(0).\] Thus, we have $f(0)=0$.
Next, we apply (1) with $x=1, y=-1$ and obtain
\[0=f(0)=f(1+(-1))=f(1)+f(-1).\] Thus we have
\[f(-1)=-f(1)=-1_R.\]

We want to determine the value $f(n)$ for any $n\in \Z$.
If $n$ is a positive integer, then we can write it as
\[n= \underbrace{1+\cdots+1}_{n\text{ times}} \] By property (a) applied repeatedly, we have
\begin{align*}
f(n) &= \underbrace{f(1)+\cdots+f(1)}_{n\text{ times}} \\
&=\underbrace{1_R+\cdots+1_R}_{n\text{ times}}=n.
\end{align*}

If $n$ is a negative integer, we express it as
\[n=\underbrace{(-1)+(-1)+\cdots+(-1)}_{n\text{ times}}\] and obtain
\begin{align*}
f(n)&=\underbrace{f(-1)+f(-1)+\cdots+f(-1)}_{n\text{ times}}\\
&=\underbrace{(-1_R)+(-1_R)+\cdots+(-1_R)}_{n\text{ times}}=n.
\end{align*}

Therefore, we have proved that
\[f(n)=n\] for any $n\in \Z$. Hence any ring homomorphism from $\Z$ to $R$ is the ring homomorphism $f_0$ that we saw at the beginning of the proof.

In conclusion, there is exactly one ring homomorphism from $\Z$ to $R$, which is given by
\[f_0(n)=n\] for any $n\in\Z$.

Comment.

In category theory, we say that the ring of integers $\Z$ is an initial object in the category of rings with unity.

The post There is Exactly One Ring Homomorphism From the Ring of Integers to Any Ring first appeared on Problems in Mathematics.