(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.

Let $R$ be an integral domain and let $S=R[t]$ be the polynomial ring in $t$ over $R$. Let $n$ be a positive integer.

Prove that the polynomial
\[f(x)=x^n-t\] in the ring $S[x]$ is irreducible in $S[x]$.

Proof.

Consider the principal ideal $(t)$ generated by $t$ in $S$.
Then the ideal $(t)$ is a prime ideal in $S$ since the quotient
\[S/(t)=R[t]/(t)\cong R\] is an integral domain.

The only non-leading coefficient of $f(x)=x^n-t$ is $-t$, and $-t$ is in the ideal $(t)$ but not in the ideal $(t)^2$.
Then by Eisenstein’s criterion, the polynomial $f(x)$ is irreducible in $S[x]$.

(Remark that $S=R[t]$ is an integral domain since $R$ is an integral domain.)

The post Irreducible Polynomial Over the Ring of Polynomials Over Integral Domain first appeared on Problems in Mathematics.

Let $R$ be a commutative ring with $1$. Show that if $R$ is an integral domain, then the characteristic of $R$ is either $0$ or a prime number $p$.

Definition of the characteristic of a ring.

The characteristic of a commutative ring $R$ with $1$ is defined as follows.
Let us define the map $\phi: \Z \to R$ by sending $n \in \Z$ to
\[\phi(n)= \begin{cases}
\underbrace{1+\cdots+1}_{n\text{ times}} \text{ if } n>0\\
0 \text{ if } n=0\\
– (\underbrace{1+\cdots+1}_{-n\text{ times}}) \text{ if } n<0. \end{cases} \] Then this map $\phi$ is a ring homomorphism and we define the characteristic $c$ of $R$ to be the integer $c$ such that
\[\ker(\phi)=(c).\] (Note that the kernel of $\phi$ is an ideal in $\Z$, and $Z$ is a principal ideal domain (PID), thus such an integer $c$ exists.)

Proof.

Let us now prove the problem.
Let $c$ be the characteristic of an integral domain $R$.

Then by the first isomorphism theorem with the ring homomorphism $\phi: \Z\to R$ as above, we have an injective homomorphism
\begin{align*}
\Zmod{c}=\Z/\ker(\phi) \to R.
\end{align*}
Since $R$ is an integral domain, $\Zmod{c}$ is also an integral domain.

This yields that $c\Z$ is a prime ideal of $\Z$.
Therefore $c=0$ or $c$ is a prime number.

The post Characteristic of an Integral Domain is 0 or a Prime Number first appeared on Problems in Mathematics.

In the ring
\[\Z[\sqrt{2}]=\{a+\sqrt{2}b \mid a, b \in \Z\},\] show that $5$ is a prime element but $7$ is not a prime element.

Hint.

An element $p$ in a ring $R$ is prime if $p$ is non zero, non unit element and whenever $p$ divide $ab$ for $a, b \in R$, then $p$ divides $a$ or $b$.

Equivalently, an element $p$ in the ring $R$ is prime if the principal ideal $(p)$ generated by $p$ is a nonzero prime ideal of $R$.

Proof.

5 is a prime element in the ring $\Z[\sqrt{2}]$.

We first show that $5$ is prime in the ring $\Z[\sqrt{2}]$.
Suppose that
\[5|(a+\sqrt{2}b)(c+\sqrt{2}d)\] for $a+\sqrt{2}b, c+\sqrt{2}d \in \Z[\sqrt{2}]$.
By taking the norm, we obtain
\[25| (a^2-2b^2)(c^2-2d^2)\] in $\Z$.
From this, we may assume that $5|a^2-2b^2$.
Now look at the following table.

\begin{array}{ |c|c|c|c| }
\hline
a, b & a^2, b^2 \pmod{5} & 2b^2 \pmod{5} \\
\hline
0 & 0 & 0 \\
1& 1 & 2 \\
2& 4 & 3 \\
3 & 4 & 3\\
4 & 1 & 2\\
\hline
\end{array}

From this table, we see that $a^2-2b^2=0 \pmod{5}$ if and only if $a, b$ are both divisible by $5$.
Therefore $5|a+\sqrt{2}b$, and $5$ is a prime element in $\Z[\sqrt{2}]$.

7 is not a prime element in the ring $\Z[\sqrt{2}]$.

Next, we show that $7$ is not a prime element in $\Z[\sqrt{2}]$.
To see this, note that we have
\[7=(3+\sqrt{2})(3-\sqrt{2})\] and $7$ does not divide $3+\sqrt{2}$ and $3-\sqrt{2}$.
Hence $7$ is not a prime element in the ring $\Z[\sqrt{2}]$.

Related Question.

For a proof of this fact, see that post “The Ring $\Z[\sqrt{2}]$ is a Euclidean Domain“.

The post 5 is Prime But 7 is Not Prime in the Ring $\Z[\sqrt{2}]$ first appeared on Problems in Mathematics.

Let $R$ be a commutative ring with $1$. Prove that the principal ideal $(x)$ generated by the element $x$ in the polynomial ring $R[x]$ is a prime ideal if and only if $R$ is an integral domain.

Prove also that the ideal $(x)$ is a maximal ideal if and only if $R$ is a field.

Proof.

We claim that we have a ring isomorphism
\[ R[x]/(x) \cong R.\] Let us for the moment assume that this claim is true and prove the problem.
If the ideal $(x)$ is a prime ideal of $R[x]$, then $R[x]/(x)$ is an integral domain.
Hence, by the claim $R\cong R[x]/(x)$ is also an integral domain.

On the other hand, if $R$ is an integral domain, then $R[x]/(x)$ is also an integral domain.
This yields that the ideal $(x)$ is a prime ideal.

Similarly, we see that the ideal $(x)$ is a maximal ideal if and only if $R\cong R[x]/(x)$ is a field.

Thus it remains to prove the claimed isomorphism of rings.

Proof of the claim

Define
\[\Psi: R[x] \to R\] by mapping $f(x) \in R[x]$ to $f(0)\in R$. (Evaluating the polynomial $f(x)$ at $x=0$.)
Then the map $\Psi$ is a ring homomorphism since we have for $f, g\in R[x]$ and $r\in R$
\begin{align*}
\Psi(f+g)&=(f+g)(0)=f(0)+g(0)=\Psi(f)+\Psi(g)\\
\Psi(rf)&=(rf)(0)=rf(0)=r\Psi(f).
\end{align*}

The homomorphism $\Psi$ is surjective since for any $r\in R$, letting $f(x)=r$ we see that $\Psi(f)=r$.
Therefore by the isomorphism theorem for rings, we have
\[R[x]/\ker(\Psi) \cong R. \tag{*}\] It remains to show that
\[\ker(\Psi)= (x).\]

$(\implies)$ Let $f\in \ker(\Psi)$. Then we have $\Psi(f)=f(0)=0$.
Let us write
\[f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1 x +a_0.\] Then since $f(0)=0$, we have $a_0=0$ and thus we have
\[f(x)=x(a_nx^{n-1}+a_{n-1}x^{n-2}+\cdots +a_1) \in (x).\] Thus $\ker(\Psi) \subset R$.

$(\impliedby)$ Let $f\in (x)$. Then we can write $f$ as $f=xg$ for some $g\in R[x]$.
It follows that we have
\[\Psi(f)=f(0)=0\cdot g(0)=0\] and $f\in \ker(\Psi)$.
Hence $R \subset \ker(\Psi)$.

Thus putting the two inclusions together gives $\ker(\Psi)= (x)$, and combining this with the isomorphism (*) we obtain
\[R[x]/(x)\cong R\] as claimed.

The post The Ideal $(x)$ is Prime in the Polynomial Ring $R[x]$ if and only if the Ring $R$ is an Integral Domain first appeared on Problems in Mathematics.