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\begin{document}

\noindent MATH 720, Algebra I \\
Exercises 3\\
Due Fri 16 Sep

\begin{exerc}
Let $R$ be a ring, and fix an integer $n\geq 2$. Let $S_1,\ldots,S_n\subseteq R$ be subsets, and set $I_j=(S_j)R$ for $j=1,\ldots,n$.
\begin{enumerate}[\quad(a)]
\item
Prove that $\sum_jI_j=(\cup_jS_j)R$.
\item
Prove that if $R$ is commutative, then $\prod_jI_j=(\{a_1\cdots a_n\mid a_j\in S_j, j=1,\ldots,n\})R$.
\item
Prove that if $R$ is commutative, then any finite product of principal ideals is principal.
\end{enumerate}
\end{exerc}



\begin{exerc}
Let $\{R_{\alpha}\}_{\alpha\in A}$ be a set of rings,
and for each $\lambda\in\Lambda$ fix an ideal $I_{\lambda}\leq R_{\lambda}$.
\begin{enumerate}[\quad(a)]
\item
Prove that $\bigoplus_{\lambda}I_{\lambda}$ is an ideal of $\bigoplus_{\lambda}R_{\lambda}$
and that there is a ring isomorphism
$$\textstyle(\bigoplus_{\lambda}R_{\lambda})/(\bigoplus_{\lambda}I_{\lambda})\cong\bigoplus_{\lambda}(R_{\lambda}/I_{\lambda}).$$
\item
Prove that $\prod_{\lambda}I_{\lambda}$ is an ideal of $\prod_{\lambda}R_{\lambda}$.
and that there is a ring isomorphism
$$\textstyle(\prod_{\lambda}R_{\lambda})/(\prod_{\lambda}I_{\lambda})\cong\prod_{\lambda}(R_{\lambda}/I_{\lambda}).$$
\item
Prove that $\bigoplus_{\lambda}I_{\lambda}$ is an ideal of $\prod_{\lambda}R_{\lambda}$.
Deduce that $\bigoplus_{\lambda}R_{\lambda}$ is an ideal of $\prod_{\lambda}R_{\lambda}$.
\item
Given  elements $(r_{\lambda}),(s_{\lambda})\in\prod_{\lambda}R_{\lambda}$,
prove that the cosets $\ol{(r_{\lambda})}$ and $\ol{(s_{\lambda})}$ in $(\prod_{\lambda}R_{\lambda})/(\bigoplus_{\lambda}R_{\lambda})$ 
are equal  if and only if 
$r_{\lambda}=s_{\lambda}$ for all but finitely many $\lambda\in\Lambda$.
\end{enumerate}
\end{exerc}


\begin{exerc}
Prove or give a counterexample:
given rings $R_1$ and $R_2$ with identity, every ideal of $R_1\times R_2$ is of the form
$I_1\oplus I_2$ for some ideals $I_1\leq R_1$ and $I_2\leq R_2$.
\end{exerc}

\end{document}

\begin{exerc}
Prove that every finite integral domain is a field. 
(Note that $\bbz$ is an infinite integral domain that is not a field, so the finiteness assumption is crucial.)
\end{exerc}

\begin{exerc}
Let $\{R_{\alpha}\}_{\alpha\in A}$ be a set of rings with non-zero identity,
and for each $\lambda\in\Lambda$ fix an ideal $I_{\lambda}\leq R_{\lambda}$.
\begin{enumerate}[\quad(a)]
\item
Prove that $\prod_{\lambda}I_{\lambda}$ is a prime ideal of $\prod_{\lambda}R_{\lambda}$
if and only if there is an element $\mu\in\Lambda$ such that
$I_{\lambda}=R_{\lambda}$ for all $\lambda\neq\mu$ and $I_{\mu}$ is a prime ideal of $R_{\mu}$.
\item
Prove that $\prod_{\lambda}I_{\lambda}$ is a maximal ideal of $\prod_{\lambda}R_{\lambda}$
if and only if there is an element $\mu\in\Lambda$ such that
$I_{\lambda}=R_{\lambda}$ for all $\lambda\neq\mu$ and $I_{\mu}$ is a maximal ideal of $R_{\mu}$.
\item
Prove that $\bigoplus_{\lambda}I_{\lambda}$ is a prime ideal of $\prod_{\lambda}R_{\lambda}$
if and only if $\Lambda$ is finite and there is an element $\mu\in\Lambda$ such that
$I_{\lambda}=R_{\lambda}$ for all $\lambda\neq\mu$ and $I_{\mu}$ is a prime ideal of $R_{\mu}$.
Deduce that $\bigoplus_{\lambda}R_{\lambda}$ is a never a prime ideal of $\prod_{\lambda}R_{\lambda}$.
\item
Prove that $\bigoplus_{\lambda}I_{\lambda}$ is a maximal ideal of $\prod_{\lambda}R_{\lambda}$
if and only if $\Lambda$ is finite and there is an element $\mu\in\Lambda$ such that
$I_{\lambda}=R_{\lambda}$ for all $\lambda\neq\mu$ and $I_{\mu}$ is a maximal ideal of $R_{\mu}$.
Deduce that $\bigoplus_{\lambda}R_{\lambda}$ is a never a maximal ideal of $\prod_{\lambda}R_{\lambda}$.
\end{enumerate}
\end{exerc}

Chinese Remainder Theorem.

\begin{lem} \label{lem030401}
Let $R$ be a non-zero commutative ring with identity, and
fix ideals $I_1,\ldots,I_n\subseteq R$.
If $I_j+I_k=R$ for all $j\neq k$, then 
$I_j+\cap_{k\neq j}I_k=R$ for all $j$.
\end{lem} 

\begin{proof}
We prove the case $j=1$; the other cases are obtained by rearranging.
Assume without loss of generality that $n\geq 3$.

We prove that $I_1+\cap_{k= 2}^lI_k=R$ by induction on $l$.
(The case $l=n$ will give the desired result.)
The base case $l=2$ holds by assumption.
Assume $l\geq 3$ and $I_1+\cap_{k= 2}^{l-1}I_k=R$.
The induction hypothesis implies that there exist
$a\in I_1$ and $b\in\cap_{k= 2}^{l-1}I_k$ such that
$1=a+b$.  Our assumption implies
that there exist
$a'\in I_1$ and $b'\in I_l$ such that
$1=a'+b'$. 
Since $a,a'\in I_1$, we have $aa',ab',a'b\in I_1$ and so
$aa'+ab'+a'b\in I_1$.  Since $b\in\cap_{k= 2}^{l-1}I_k$
we have $bb'\in\cap_{k= 2}^{l-1}I_k$;
and since $b'\in I_l$, we have $bb'\in I_l$;
hence $bb'\in\cap_{k= 2}^{l}I_k$
Hence, we have
$$1=1\cdot 1=(a+b)(a'+b')=\underbrace{(aa'+ab'+a'b)}_{\in I_1}+
\underbrace{bb'}_{\in\cap_{k= 2}^lI_k}\in I_1+\cap_{k= 2}^lI_k$$
and it follows that 
$I_1+\cap_{k= 2}^lI_k=R$.
\end{proof}

\begin{ex} \label{ex030502}
For instance, the lemma says that
$3\bbz+(5\bbz\cap 7\bbz)=\bbz$ because
$3\bbz+5\bbz=\bbz$ and $3\bbz+ 7\bbz=\bbz$.
\end{ex}

\begin{thm}[Chinese Remainder Theorem, v.~1] \label{thm030401}
Let $R$ be a non-zero commutative ring with identity, and
let $I_1,\ldots,I_n$ be ideals of $R$.
\begin{enumerate}[\quad\rm(a)]
\item \label{thm030401a}
The function $\theta\colon R/(\cap_{j=1}^nI_j)\to\prod_{j=1}^nR/I_j$
given by $\ol r\mapsto (r+I_1,\ldots,r+I_n)$ is  well-defined.
Furthermore, it is a 1-1 homomorphism of rings with identity.
\item \label{thm030401b}
If $I_j+I_k=R$ for all $j\neq k$, then $\theta$ is an isomorphism
and $\cap_{j=1}^nI_j=\prod_{j=1}^nI_j$.
\end{enumerate}
\end{thm}

\begin{proof}
\eqref{thm030401a}
The map $\Phi\colon R\to\prod_{j=1}^nR/I_j$
given by $r\mapsto (r+I_1,\ldots,r+I_n)$ is a well-defined
homomorphism of rings with identity. 
From the definition, it follows that $\Ker(\Phi)=\cap_{j=1}^nI_j$.
Hence, the result follows from the First Isomorphism Theorem.

\eqref{thm030401b}
For the first conclusion, it suffices to show that $\Phi$ is surjective.  
Let $(r_1+I_1,\ldots,r_n+I_n)\in\prod_{j=1}^nR/I_j$.
For each $j=1,\ldots,n$, the previous lemma shows that there exist
$a_j\in I_j$ and $b_j\in \cap_{k\neq j}I_k$ such that 
$r_j=a_j+b_j$.  
Since $a_j\in I_j$, we have $r_j+I_j=b_j+I_j$.
For $l\neq j$, since $b_j\in \cap_{k\neq j}I_k\subseteq I_l$,
we have $b_j+I_l=0+I_l$.
With $b=b_1+\ldots+b_n$ it follows that
$b+I_j=r_j+I_j$ for each $j$, and so 
$\Phi(b)=(r_1+I_1,\ldots,r_n+I_n)$.

For the second conclusion, we argue by induction on $n$.  The case $n=1$ is
trivial, so assume that $n\geq 2$ and $\cap_{j=1}^{n-1}I_j=\prod_{j=1}^{n-1}I_j$.
From the previous lemma, we know 
$R=\cap_{j=1}^{n-1}I_j+I_n=\prod_{j=1}^{n-1}I_j+I_n$, and so
\begin{align*}
\cap_{j=1}^{n}I_j
&=\textstyle(\cap_{j=1}^{n-1}I_j)\cap I_n \\
&=\textstyle(\prod_{j=1}^{n-1}I_j)\cap I_n \\
&=\textstyle R[(\prod_{j=1}^{n-1}I_j)\cap I_n] \\
&=\textstyle [\prod_{j=1}^{n-1}I_j+I_n][(\prod_{j=1}^{n-1}I_j)\cap I_n] \\
&=\textstyle [\prod_{j=1}^{n-1}I_j][(\prod_{j=1}^{n-1}I_j)\cap I_n]+[I_n][(\prod_{j=1}^{n-1}I_j)\cap I_n] \\
&\subseteq\textstyle [\prod_{j=1}^{n-1}I_j][I_n]+[I_n][(\prod_{j=1}^{n-1}I_j)] \\
&\subseteq\textstyle [\prod_{j=1}^{n}I_j]+[(\prod_{j=1}^{n}I_j)] \\
&=\textstyle \prod_{j=1}^{n}I_j.
\end{align*}
From a homework exercise, we know $\prod_{j=1}^{n}I_j \subseteq\cap_{j=1}^{n}I_j$
so we have equality.
\end{proof}

%\section{Day 6} \label{sec0306}

\begin{defn} \label{defn030405}
If $I$ is an ideal in $R$, we write $a\equiv b\pmod{I}$ if
$a-b\in I$.  This is an equivalence relation on $R$.
\end{defn}

\begin{cor}[Chinese Remainder Theorem, v.~2] \label{cor030401}
Let $R$ be a non-zero commutative ring with identity, and
let $I_1,\ldots,I_n$ be ideals of $R$
such that $I_j+I_k=R$ for all $j\neq k$. For all $r_1,\ldots,r_n\in R$
there exists $b\in R$ such that $b\equiv r_j\pmod{I_j}$ for $j=1,\ldots,n$.
Furthermore, $b$ is uniquely determined up to congruence modulo the ideal
$\cap_{j=1}^nI_j=\prod_{j=1}^{n}I_j$.
\end{cor}

\begin{proof}
The first statement is a reformulation of the surjectivity of $\theta$ in
the previous theorem.  The second statement 
corresponds to the injectivity of $\theta$.
\end{proof}

The ``traditional'' Chinese Remainder Theorem is the special case where
$R=\bbz$.  Note that the proof of Theorem~\ref{thm030401}\eqref{thm030401b}
gives an algorithm for solving a system of congruences.

\begin{cor}[Chinese Remainder Theorem, v.~3] \label{cor030402}
Let $m_1,\ldots,m_n\in\bbz$ such that $\gcd(m_j,m_k)=1$ for each $j\neq k$.
For all $r_1,\ldots,r_n\in \bbz$
there exists $b\in \bbz$ such that $b\equiv r_j\pmod{m_j}$ for $j=1,\ldots,n$.
Furthermore, $b$ is uniquely determined up to congruence modulo 
$m=m_1\cdots m_n$. \qed
\end{cor}

Describe the units, nilpotents, and zero-divisors in $R[X]$.

Describe some ideals in in $R[X]$.

Develop $R[\![X]\!]$.

Describe the units, nilpotents, and zero-divisors in $R[\![X]\!]$.

Describe some ideals in in $R[\![X]\!]$.


Fermat's Little Theorem.

\begin{prop} \label{prop031102}
Let $k$ be a field.  Then any finite subgroup
$G\leq k^{\times}$ is cyclic.
\end{prop}

\begin{proof}
Assume $G\neq \{1\}$.
Then $G$ is a finite abelian group, so the fundamental theorem
of finitely generated abelian groups implies that 
there exist (not necessarily distinct) prime integers $p_1,\ldots,p_r\geq 2$ 
and positive integers $a_1,\ldots,a_r$ such that
$G\cong \oplus_{i=1}^r\bbz/p_i^{a_i}\bbz$.

Note that $G$ is written multiplicatively
and the $\bbz/p_i^{a_i}\bbz$ are written additively.
Identify each $\bbz/p_i^{a_i}\bbz$ with a subgroup
of $\oplus_{i=1}^r\bbz/p_i^{a_i}\bbz$ in the natural way.
Note that, under this identification, we have
$\bbz/p_i^{a_i}\bbz\cap \bbz/p_j^{a_j}\bbz=\{0\}$ when $i\neq j$.
There are $p_i$ distinct elements  $\ol n\in\bbz/p_i^{a_i}\bbz$
such that $p_i\ol n=0$, namely 
$\ol{p_i^{a_i-1}},2\ol{p_i^{a_i-1}}, \ldots,\ol{p_ip_i^{a_i-1}}=\ol 0$.
Thus, there are $p_i$ elements  $g\in G$ such that
$g^{p_i}=1$.

Suppose that $p_i=p_j$ for some $i\neq j$.
The  previous paragraph shows that
there are at least $2p_i-1>p_i$ elements  $g\in G$
such that $g^{p_i}=1$.  This shows that the polynomial
$x^{p_i}-1\in k[x]$ has more than $p_i$ roots,
contradicting Proposition~\ref{prop031101}\eqref{prop031101b}.

Thus, we have $\gcd(p_i, p_j)=1$ for all $i\neq j$, and it follows that
$$G\cong \oplus_{i=1}^r\bbz/p_i^{a_i}\bbz\cong
\bbz/(p_1^{a_1}\cdots p_r^{a_r})\bbz.$$
That is $G$ is cyclic.
\end{proof}

\begin{cor} \label{cor031101}
If $p$ is a positive prime integer, then
$(\bbz/p\bbz)^{\times}\cong\bbz/(p-1)\bbz$.
\end{cor}

\begin{proof}
The group $(\bbz/p\bbz)^{\times}$ is cyclic and has
exactly $p-1$ elements.
\end{proof}

\begin{cor}[Fermat's Little Theorem] \label{cor031102}
Let $p$ be a positive prime integer.
\begin{enumerate}[\quad\rm(a)]
\item \label{cor031102a}
For each $\ol n\in\bbz/p\bbz$, we have $\ol n^p=\ol n$.
\item \label{cor031102b}
In $\bbz/p\bbz[x]$ we have
$\prod_{\ol n\in \bbz/p\bbz}(x-\ol n)=x^p-x$.
\end{enumerate}
\end{cor}

\begin{proof}
\eqref{cor031102a}
The case where $\ol n=\ol 0$ is easy.  
For the remaining cases, use the isomorphism
$(\bbz/p\bbz)^{\times}\cong\bbz/(p-1)\bbz$:
Since $(p-1)(n+(p-1)\bbz)=0+(p-1)\bbz$,
this translates to $\ol n^{p-1}=\ol 1$ for all
$\ol n\neq \ol 0$ in $\bbz/p\bbz$,
and so $\ol n^p=\ol n$.

\eqref{cor031102b}
From part~\eqref{cor031102a}
we know that each element of $\bbz/p\bbz$
is a root of the polynomial $x^p-x\in \bbz/p\bbz[x]$.
Repeatedly apply the Remainder Theorem to write
$$\textstyle
x^p-x=q\prod_{\ol n\in \bbz/p\bbz}(x-\ol n)$$
for some $q\in \bbz/p\bbz[x]$.
Proposition~\ref{prop031101}\eqref{prop031101b}
shows that $\deg(q)=0$, and so $q$ is a constant.
Multiplying out, we see that the product is monic,
as $q$ times the product, and so we have $q=1$.
\end{proof}

UMP for $Q(R)$.

\begin{thm}[Rational Root Theorem] \label{thm031501}
Let $R$ be a UFD with $K=\Q(R)$.
Let $f=a_0+a_1x+\cdots+a_dx^d\in R[x]$ and assume that
$r=a/b\in K$ is a root of $f$ with $\gcd(a,b)=[1]$.
Then $a\divs a_0$ and $b\divs a_d$.
In particular, if $f$ is monic, then $r\in R$.
\end{thm}

\begin{proof}
We have
\begin{align*}
0
&=f(a/b) \\
&=a_0+a_1(a/b)+\cdots+a_d(a/b)^d\\
0
&=a_0b^d+a_1ab^{d-1}+\cdots+a_{d-1}a^{d-1}b+a_da^d\\
a_da^d
&=-(a_0b^d+a_1ab^{d-1}+\cdots+a_{d-1}a^{d-1}b)\\
&=-b(a_0b^{d-1}+a_1ab^{d-2}+\cdots+a_{d-1}a^{d-1})
\end{align*}
and so $b\divs a_da^d$.  Because $\gcd(a,b)=[1]$,
Exercise~\ref{exer031501} implies $b\divs a_d$.

Similarly, we have
\begin{align*}
a_0b^d
&=-(a_1ab^{d-1}+\cdots+a_{d-1}a^{d-1}b+a_da^d) \\
&=-a(a_1b^{d-1}+\cdots+a_{d-1}a^{d-2}b+a_da^{d-1}) 
\end{align*}
and so $a\divs a_0b^d$ and $a\divs a_0$.

If $f$ is monic, then $a_d=1$ and so $b\divs 1$.
This implies that $b$ is a unit, so $r=ab^{-1}\in R$.
\end{proof}

\begin{ex} \label{ex031601}
The only possible roots of $3x^7-7x+2$ in $\bbq$
are $\pm 1,\pm 2, \pm 1/3$ or $\pm 2/3$.
\end{ex}


\begin{lem} \label{lem031601}
Let $R$ be an integral domain,
and let $P\subset R$ be a prime ideal.
Let $f\in R[x]$ be monic,
and let $\ol f$ be the polynomial in $R/P[x]$ 
obtained by reducing the coefficients of $f$ modulo $P$.
If $\ol f$ is irreducible in $R/P[x]$,
then $f$ is irreducible in $R[x]$.
\end{lem}

\begin{proof}
Observe first that $R/P[x]$ is an integral domain.
Since $f$ is monic, so is $\ol f$, and
$\deg(\ol f)=\deg(f)$. Since $\ol f$ is irreducible and monic,
we have $\deg(\ol f)>0$.  
Suppose that $f=gh$ for some 
$g,h\in R[x]$. Since $R$ is an integral domain,
we know that the leading coefficient of $f$
is the product of the leading coefficients of $g$ and $h$.
Since $f$ is monic, 
it follows that the the leading coefficients of $g$ and $h$
are units.
%After dividing $g$ and $h$ by their leading coefficients,
%we  assume without loss of generality that
%$g$ and $h$ are monic.

It follows that the leading coefficients of $\ol g$ and $\ol h$
are units,  and $\deg(\ol g)=\deg(g)$ and $\deg(\ol h)=\deg(h)$.
Since $\ol f$ is irreducible, we conclude that either $\ol g$ or
$\ol h$ is a unit in $R/P[x]$.  Assume by symmetry that $\ol g$ is a unit.
Since $R/P$ is an integral domain, this implies that $\ol g$
is a constant, and so $\deg(g)=\deg(\ol g)=0$.  That is, $g$ is a constant,
necessarily equal to its leading coefficient, which is a unit.
It follows that $g$ is a unit in $R[x]$, so that $f$ is irreducible.
\end{proof}


\begin{prop} \label{prop031601}
Let $R$ be a UFD with quotient field $K=\Q(R)$,
and let $P\subset R$ be a prime ideal.
Let $f=a_nx^n+\cdots+a_0\in R[x]$ with $n\geq 1$, and 
assume that $a_0,\ldots,a_{n-1}\in P$ and $a_0\not\in P^2$ and $a_n\not\in P$.
Then $f$ is irreducible in $K[x]$.
If $f$ is primitive, then $f$ is irreducible in $R[x]$.
\end{prop}

\begin{proof}
Case 1:  $f$ is primitive. By Gauss' Lemma, it suffices to show that 
$f$ is irreducible in $R[x]$.
Write $f=gh$ for some $g,h\in R[x]$.
We need to show that either $g$ or $h$ is a unit in $R[x]$.
Set $k=\deg(g)$ and $l=\deg(h)$.

Case 1a: $k=0$.  Then $g=b_0$ is constant.
Since $f=gh=b_0h$  is primitive,
Lemma~\ref{lem031102}\eqref{lem031102b}
implies that $b_0$ is a unit in $R$, and so $g$ is a unit in $R[x]$. 

Case 1b: $l=0$.  Similar to Case 1a.

Case 1c:  $k,l\geq 1$.
Write
$g=b_kx^k+\cdots+b_0$ and 
$h=c_lx^l+\cdots+c_0$.
Since $R$ is an integral domain, we have $n=k+l$.
Our assumptions on the $a_i$
imply 
$$\ol {b_k}\ol{c_l}x^n+\cdots+\ol{b_0}\ol{c_0}=\ol g\ol h=\ol{a_n}x^n\neq 0.$$
in $R/P[x]$.
It follows that
$0\neq\ol{a_n}=\ol{b_k}\ol{c_l}=\ol{b_kc_l}$ in $R/P$.
Hence $b_kc_l\not\in P$ and so
$b_k,c_l\not\in P$. 
Let $r,s$ be the smallest integers such that
$r\leq k$ and $s\leq l$ and $b_r\not\in P$ and $c_s\not\in P$.


Suppose that $r+s=n$.  Since $r\leq k$ and $s\leq l$, this implies $r=k$
and $s=l$, and so $b_0,\ldots,b_{k-1},c_0,\ldots,c_{l-1}\in P$.
It follows that $\ol g=\ol{b_k}x^k$ and $\ol h=\ol{c_l}x^l$.  
It follows that $b_0,c_0\in P$ and so $a_0=b_0c_0\in P^2$, a contradiction.

Thus, we have $r+s<n$.  It follows that
$$\textstyle
0=\ol{a_{r+s}}=\sum_{i+j=r+s}\ol{b_i}\ol{c_j}.$$
For $i<r$, we have $\ol{b_i}=0$ by the choice of $r$ and so $\ol{b_i}\ol{c_j}=0$.
Similarly, for $j<s$, we have $\ol{b_i}\ol{c_j}=0$.
It follows that
$$\ol{b_r}\ol{c_s}=-\sum_{i+j=r+s, i\neq r}\ol{b_i}\ol{c_j}=0$$
in $R/P$.  Since $P$ is a prime ideal, it follows that
either $b_r\in P$ or $c_s\in P$, a contradiction.

%Since $n\geq 1$, we have 
%$\ol{b_0}\ol{c_0}=0$.  Since $R/P$ is an integral domain,
%we have $\ol{b_0}=0$ or $\ol{c_0}=0$.
%Furthermore, 


Case 2:  $f$ is not primitive.
By assumption, we have $a_n\not\in P$ and so $a_n\neq 0$
and $\deg(f)=n\geq 1$.  
Set $C(f)=[c]$. For each $i$, set $a_i^*=a_i/c\in R$, 
and set $f^*=a_n^*x^n+\cdots+a_0^*\in R[x]$.
It follows that $f^*$ is primitive, $ca_i^*=a_i$ for each $i$ and $cf^*=f$.
In particular, since $ca_n^*=a_n\not\in P$, we have $c,a_n^*\not\in P$.
For $i<n$, we have $ca_i^*=a_i\in P$; since $P$ is prime and $c\not\in P$,
this implies $a_i^*\in P$.
Furthermore, if $a_0^*\in P^2$, then $a_0=ca_0^*\in P^2$, a contradiction.
In particular, it follows that $f^*$ satisfies the hypotheses of the result.
By Case 1, it follows that $f^*$ is irreducible in $K[x]$. The fact that
$f=cf^*$ is a unit multiple of $f^*$ in $K[x]$, implies that $f$ is irreducible in $K[x]$.
\end{proof}

\begin{cor}[Eisenstein's Criterion] \label{cor031601}
Let $R$ be a UFD with quotient field $K=\Q(R)$,
and let $p\in R$ be a prime element.
Let $f=a_nx^n+\cdots+a_0\in R[x]$ with $n\geq 1$, and assume that
$p\divs a_i$ for $i=0,\ldots,n-1$ and $p^2\nmid a_0$ and $p\nmid a_n$.
Then $f$ is irreducible in $K[x]$.
If  $f$ is primitive, then $f$ is irreducible in $R[x]$.
\end{cor}

\begin{proof}
Use the prime ideal $P=(p)$ in Proposition~\ref{prop031601}.
\end{proof}

\begin{ex} \label{ex031602}
The polynomial $x^{42}-37$ is irreducible in $\bbq[x]$ and in $\bbz[x]$,
using $p=37$.
\end{ex}

\begin{ex} \label{ex031603}
Let $p\geq 2$ be a prime number.
Then the \emph{cyclotomic polynomial}
$f=x^{p-1}+x^{p-2}+\cdots+1$ is irreducible in $\bbq[x]$ and in $\bbz[x]$,
as follows. It is straightforward to show that
$(x-1)f=x^p-1$
and so $f=(x^p-1)/(x-1)$.
It follows that
\begin{align*}
f(x+1)
&=\frac{(x+1)^p-1}{(x+1)-1}
=\frac{x^p+\binom{p}{1}x^{p-1}+\cdots+\binom{p}{i}x^{p-i}+\cdots+\binom{p}{p-1}x}{x} \\
&=x^{p-1}+\binom{p}{1}x^{p-2}+\cdots+\binom{p}{i}x^{p-i-1}+\cdots+\binom{p}{p-1}.
\end{align*}
Now, recall that $p\divs\binom{p}{i}$ whenever $1<i<p$,
and furthermore that $\binom{p}{p-1}=p$.  It follows from
Eisenstein's criterion that $f(x+1)$ is irreducible over $\bbq$ and over $\bbz$, and it is straightforward 
to show that this implies that $f$ is irreducible over $\bbq$ and over $\bbz$ as well.
\end{ex}


\end{document}


\begin{exerc}
\end{exerc}















