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\begin{document}

\begin{center}
{\bf \Large \underline{Honors Algebra 4, MATH 371 Winter 2010}}\\ 
	{\large Assignment 4}\\Due Wednesday, February 17 at 08:35
\end{center}

\begin{enumerate}
	\item Let $R$ be a commutative ring with $1\neq 0$. 
	\begin{enumerate} 
		\item Prove that the nilradical of $R$ is equal to the intersection of
		the prime ideals of $R$.  Hint: it's easy to show using the definition
		of prime that the nilradical is contained in every prime ideal.  Conversely,
		suppose that $f$ is not nilpotent and consider the set $S$ of ideals
		$I$ of $R$ with the property that ``$n>0\implies f^n\not\in I$."
		Show that $S$ has maximal elements and that any such maximal element must 
		be a prime ideal. 

\medskip\noindent{\bf Solution:} Suppose that $f\in R$ is not nilpotent and let $S$
 be the set $S$ of ideals $I$ of $R$ with the property that ``$n>0\implies f^n\not\in I$."
Ordering $S$ by inclusion, note that every chain is bounded above: if $I_1\subseteq I_2\subseteq \cdots$
is a chain, then $I=\cup I_i$ is an upper bound which clearly lies in $S$.
By Zorn's lemma, $S$ has a maximal element, say $M$, which we claim is prime.
Indeed, suppose that $uv\in M$ but that $u\not\in M$ and $v\not\in M$.  Then the ideals
$M+(u)$ and $M+(v)$ strictly contain $M$ so do not belong to $S$ by maximality of $M$.
Thus, there exist $m$ and $n$ such that $f^n\in M+(u)$ and $f^m\in M+(v)$.
It follows that $f^{m+n}\in M+(uv)=M$ and hence that $M$ is not in $S$, a contradiction.
Thus, either $u$ or $v$ lies in $M$ and $M$ is prime.  We deduce that $f$ is not contained
in the prime ideal $M$, and hence that $f$ is not contained in the intersection of all prime
ideals.  

Conversely, if $\p$ is any prime ideal and $f$ is nilpotent then $0=f^n\in \p$ for some $n$, 
and an easy induction argument using that $\p$ is prime shows that $f$ must be in $\p$.

		\item 	Suppose that $R$ is {\em reduced}, {\em i.e.} that the nilradical of $R$ is the zero ideal.
		If $\p$ is a minimal prime ideal of $R$, show that the localization $R_{\p}$
		 has a unique prime ideal and conclude that $R_{\p}$ is a field.
		 
\medskip\noindent{\bf Solution:} By a previous exercise, the prime ideals of the localization
$R_{\p}$ are those prime ideals of $R$ not meeting $R\setminus \p$, or in other words, the prime
ideals of $R$ contained in $\p$.  As $\p$ is minimal, there is a unique such prime ideal: namely 
$\p$ itself.  We claim that the image of $\p$ in $R_{\p}$ is the zero ideal.  
Indeed, by part (a), any $f\in \p$ is nilpotent in $R_{\p}$ so there exists $s\in R\setminus \p$
such that $sf^n=0$ for some $n>0$.  We deduce by commutativity that $sf\in R$ is nilpotent, 
whence it must be zero since $R$ is reduced, and we conclude that $f$ is zero in $R_{\p}$
as desired.  Thus, $R_{\p}$ is a ring whose only prime ideal is $0$ and therefore must be a field.
(If $T$ is any such ring and $x\in T$ is nonzero, then $(x)$ can not be contained in
any maximal ideal, since maximal ideals are prime and hence $(x)$ is the unit ideal so $x$ is a unit.)		 
	
		 \item Again supposing $R$ to be reduced, prove that $R$ is isomorphic to a subring
		 of a direct product of fields.

\medskip\noindent{\bf Solution:} We have a canonical ring homomorphism
$$R\rightarrow \prod_{\p\ \text{minimal}} R_{\p}$$
whose kernel is the intersection of all minimal primes.  By part (a), this kernel 
is the nilradical of $R$, so since $R$ is reduced the above map is injective.  
By part (b), the right hand side is a product of fields, so we conclude that $R$ is isomorphic
to a subring of a direct product of fields.		 
		 
	\end{enumerate}
	
\item Let $R$ be a commutative ring with $1\neq 0$ and let $\varphi:R\rightarrow R$
be a ring homomorphism.  If $R$ is noetherian and $\varphi$ is surjective, 
show that $\varphi$ must be injective too, and hence an isomorphism.
(Hint: Consider the iterates of $\varphi$ and their kernels.)
Can you give a counter-example to this when $R$ is not noetherian?

\medskip\noindent{\bf Solution:} Let $\varphi^n$ be the composition of $\varphi$
with itself $n$-times and denote by $I_n$ the kernel of $\varphi^n$.
Then we have a chain of ideals 
$$I_1\subseteq I_2 \subseteq I_3\subseteq \cdots$$
in the noetherian ring $R$, so we conclude that this chain stabilizes,
hence $I_n=I_{n+1}$ for some $n\ge 1$.  Suppose $x\in \ker\varphi$.
Since $\varphi^n$ is surjective, we can write $x=\varphi^{n}(y)$
whence $0=\varphi(x)=\varphi^{n+1}(y)$ and we deduce that $y\in I_{n+1}=I_n$
and hence that $x=\varphi^n(y)=0$.  Thus, $\varphi$ is injective.

As a counterexample in the case of non-noetherian $R$, consider the 
ring $R$ of infinitely differentiable real-valued functions on the interval $[0,1]$
and the map $\varphi:R\rightarrow R$ given by differentiation.  
This map is surjective, since for any $f$, the function $F(x):=\int_0^x f(u) du$
is well-defined and infinitely differentiable.  However, $\varphi$ is not injective
as it kills the constant functions.
 


	
\item As usual, for a prime $p$ we write $\F_p=\Z/p\Z$ for the field with $p$ elements.
\begin{enumerate}
	\item Find all monic irreducible polynomials in $\F_p[X]$ of degree $\le 3$ for $p=2,3,5$.\label{calc}

\medskip\noindent{\bf Solution:} For $p=2$ the monic irreducibles are 
\begin{align*}
	& x^3 + x^2 + 1,\ x^3 + x + 1,\ x^2 + x + 1,\ x + 1,\ x
\end{align*}
for $p=3$ they are
\begin{align*}
	&x^3 + x^2 + x + 2,\ x^3 + x^2 + 2x + 1,\ x^3 + x^2 + 2,\ x^3 + 2x^2 + x + 1\\
    &x^3 + 2x^2 + 2x + 2,\ x^3 + 2x^2 + 1,\ x^3 + 2x + 1,\ x^3 + 2x + 2\\
	&x^2 + x + 2,\ x^2 + 2x + 2,\ x^2 + 1,\ x + 1,\ x + 2,\ x
\end{align*}
for $p=5$ they are
\begin{align*}
	x^3 + x^2 + x + 3,\ x^3 + x^2 + x + 4,\ x^3 + x^2 + 3x + 1,\ x^3 + x^2 + 3x + 4\\
	x^3 + x^2 + 4x + 1,\ x^3 + x^2 + 4x + 3,\ x^3 + x^2 + 1,\ x^3 + x^2 + 2,\ x^3 + 2x^2 + x + 3\\
	x^3 + 2x^2 + x + 4,\ x^3 + 2x^2 + 2x + 2,\ x^3 + 2x^2 + 2x + 3,\ x^3 + 2x^2 + 4x + 2\\
	x^3 + 2x^2 + 4x + 4,\ x^3 + 2x^2 + 1,\ x^3 + 2x^2 + 3,\ x^3 + 3x^2 + x + 1,\ x^3 + 3x^2 + x + 2\\
	x^3 + 3x^2 + 2x + 2,\ x^3 + 3x^2 + 2x + 3,\ x^3 + 3x^2 + 4x + 1,\ x^3 + 3x^2 + 4x + 3\\
	x^3 + 3x^2 + 2,\ x^3 + 3x^2 + 4,\ x^3 + 4x^2 + x + 1,\ x^3 + 4x^2 + x + 2,\ x^3 + 4x^2 + 3x + 1\\
	x^3 + 4x^2 + 3x + 4,\ x^3 + 4x^2 + 4x + 2,\ x^3 + 4x^2 + 4x + 4,\ x^3 + 4x^2 + 3,\ x^3 + 4x^2 + 4\\
	x^3 + x + 1,\ x^3 + x + 4,\ x^3 + 2x + 1,\ x^3 + 2x + 4,\ x^3 + 3x + 2,\ x^3 + 3x + 3,\ x^3 + 4x + 2\\
	x^3 + 4x + 3,\ x^2 + x + 1,\ x^2 + x + 2,\ x^2 + 2x + 3,\ x^2 + 2x + 4,\ x^2 + 3x + 3,\ x^2 + 3x + 4\\
	x^2 + 4x + 1,\ x^2 + 4x + 2,\ x^2 + 2,\ x^2 + 3,\ x + 1,\ x + 2,\ x + 3,\ x + 4,\ x
\end{align*}

	\item Prove that for $f\in \F_p[X]$ monic and irreducible, the ideal $(f(X))$ is maximal
	and hence that $\F_p[X]/(f(X))$ is a field.  Show that $\F_p[X]/(f(X))$ has finite 
	cardinality $p^{\deg{f}}$ and use part (\ref{calc}) to explicitly construct finite fields of 
	orders $8,9,25,125$.

\medskip\noindent{\bf Solution:} We showed that $\F_p[X]$ is Euclidean and hence a PID and hence a UFD.
In particular, irreducible implies prime (using UFD) and prime implies maximal (using PID).
We conclude that for a monic irreducible $f$, the ring $\F_p[X]/(f(X))$ is a field.
As an $\F_p$-vector space, $\F_p[X]/(f(X))$ has basis $1,X,X^2,\ldots,X^{\deg{f} -1}$
and hence this field has cardinality $p^{\deg{f}}$.  Choosing specific examples
of monic irreducibles as found in part (a) yields specific examples of finite fields
of size $2^3$, $3^2$, $5^2$, and $5^3$.	

	\item Prove that $\F_{7}[X]/(X^2+2)$ and $\F_7[X]/(X^2+X+3)$ are both finite fields of size
	49.  Show that these fields are isomorphic by exhibiting an explicit isomorphism between them.

\medskip\noindent{\bf Solution:} Both $X^2+X+3$ and $X^2+2$ are monic irreducibles in $\F_7[X]$.
Any ring map $\varphi:\F_7[X]\rightarrow \F_7[Y]/(Y^2+Y+3)$ has the form 
$$X\mapsto aY + b,$$ 
so $X^2 + 2$ maps to 
\begin{equation*}
	(aY+b)^2+2 = a^2Y^2 +2abY + b^2 + 2 = a^2(-Y-3) + 2abY + b^2 +2 = -(a^2-2ab)Y + b^2+2-3a^2
\end{equation*}
so since $1,Y$ is an $\F_7$-basis of the target, if $X^2+2$ is to map to zero we must have 
$a^2=2ab$ and $b^2=3a^2+2=0$.  If $a=0$ then $\varphi$ would not be surjective, 
so we must have $a\neq 0$.  Then $a=2b$ and $b^2 = 4$ so $b=\pm 2$.  
Taking $b=2$ and $a=4$ gives the map $X\mapsto 4Y+2$ which by our calculation induces
a nonzero map of fields $\F_7[X]/(X^2+2)\rightarrow \F_7[Y]/(Y^2+Y+3)$ which must therefore
be an isomorphism.
	
\end{enumerate}

\item   Let $R$ ba a ring with $1\neq 0$ and $M$ an $R$-module.  
Show that if $N_1\subseteq N_2\subseteq \cdots $
is an ascending chain of submodules of $M$ then $\cup_{i\ge 1} N_i$ is a submodule of $N$.
Show by way of counterexample that modules over a ring need not have maximal proper
submodules (in contrast to the special case of ideals in a ring with $1$).

\medskip\noindent{\bf Solution:} The argument is identical to that for the special case of ideals.
For a counterexample, consider $\Q$ as a $\Z$-module.



\item Let $R$ be any commutative ring with $1\neq 0$ and $M$ and $R$-module.  Show that the 
canonical map
$$\Hom_R(R,M)\rightarrow M$$
sending $\varphi$ to $\varphi(1)$ is an isomorphism of $R$-modules.

\medskip\noindent{\bf Solution:} One must first check that the given map really is a map of
$R$-modules; as this is straightforward and tedious, we omit it.
For $m\in M$ let $\varphi_m:R\rightarrow M$
be the map defined by $\varphi_m(r):=rm$.  It is easy to see that this is an $R$-module
homomorphism and is inverse to the canonical map $\Hom_R(R,M)\rightarrow M$.

\item Let $F=\R$ and let $V=\R^3$.  Consider the linear map $\varphi:V\rightarrow V$
given by rotation through an angle of $\pi/2$ about the $z$-axis.  Consider $V$
as an $F[X]$-module by defining
$$(a_nX^n+a_{n-1}X^{n-1}+\cdots + a_1X + a_0)v :=(a_n\varphi^n+a_{n-1}\varphi^{n-1}+\cdots + a_1\varphi + a_0)v,$$
where $\varphi^i$  is the composition of $\varphi$ with itself $i$-times.
\begin{enumerate}
	\item What are the $F[X]$-submodules of $V$?
	
	\medskip\noindent{\bf Solution:} The $F[X]$-submodules are precisely that $F$-subspaces
	of the vector space $\R^3$ which are stable under multiplication by $X$, i.e.
	the subspaces preserved by $\varphi$.  Thinking geometrically, these are 
	the $x-y$ plane and the $z$-axis.
	
	\item Show that $V$ is naturally a module over the quotient ring $F[X]/(X^3-X^2+X-1)$.

	\medskip\noindent{\bf Solution:} Thinking geometrically, $\varphi$
	has eigenvalues $1$ (the $z$-axis is an eigenvector) and $\pm i$
	(the restriction of $\varphi$ to the $x-y$ plane is rotation through $\pi/2$,
	whose characteristic polynomial is clearly $X^2+1$).  
	We conclude that the characteristic polynomial of $\varphi$ is 
	$$(X-1)(X^2+1)=(X^3-X^2+X-1)$$
	and hence that this element of $F[X]$ acts trivially on $V$ by 
	the Cayley-Hamilton theorem.  Thus, $V$ is a module over the quotient
	ring $F[X]/(X^3-X^2+X-1)$.
	
	
	
	
	
\end{enumerate}

\item Let $R$ be a ring with $1\neq 0$.  
\begin{enumerate}
	\item For a left ideal $I$ of $R$ and an $R$-module $M$, define
	$$IM:=\left\{  r_1m_1 + r_2m_2 + \cdots + r_k m_k\ :\ r_i\in R,\ m_i\in M,\ k\in \Z_{\ge 0}\right\}.$$
	Show that $IM$ is an $R$-submodule of $M$.

	\medskip\noindent{\bf Solution:} Obvious.

	\item Prove that for any ideal $I$ of $R$ and any positive integer $n$, 
	there is a canonical isomorphism of $R$-modules
	$$R^n/IR^n \simeq R/IR \times R/IR \times \cdots \times R/IR$$
	with $n$-factors in the product on the right.\label{fieldtrick}
	
	\medskip\noindent{\bf Solution:} The map 
	$$R^n \simeq R/IR \times R/IR \times \cdots \times R/IR$$
	defined by $(r_1,\ldots,r_n)\mapsto (r_1+I,\ldots, r_n+I)$
	is a well-defined and surjective $R$-module homomorphism.
	The kernel consists of exactly those $(r_1,\ldots,r_n)$
	with $r_i\in I$ for all $I$, which is easily seen to be the ideal $IR^n$.
	
	\item Suppose now that $R$ is commutative and that $R^n\simeq R^m$ as $R$-modules.
	Show that $m=n$.  Hint: reduce to the case of finite dimensional vector spaces
	over a field by applying (\ref{fieldtrick}) with $I$ a maximal ideal of $R$. 
	
	\medskip\noindent{\bf Solution:} Let $I$ be a maximal ideal of $R$ so $F:=R/IR$
	is a field.  By (\ref{fieldtrick}), we deduce that 
	$$F^m\simeq R^m/IR^m \simeq R^n/IR^n simeq F^n$$
	which forces $m=n$ since all bases of a finite dimensional vector space have the same
	cardinality (i.e. dimension is well-defined).
	
	 	
	\item If $R$ is commutative and $A$ is any finite set of cardinality $n$, show that $F(A)\simeq R^n$
	as $R$-modules (Hint: Show that $R^n$ satisfies the same universal mapping property as $F(A)$ 
	and deduce from this that one has maps in both directions whose composition in either order
	must be the identity).  Conclude that the rank of a free module over a commutative ring
	is well-definied if it is finite.
	
	\medskip\noindent{\bf Solution:} 
	Let $E:=\{e_i\}_{i=1}^n$ be the standard basis of $R^n$ and suppose given
	a map of sets $\psi:E\rightarrow M$ for an $R$-module $M$.  We extend $\psi$
	to an $R$-module homomorphism $R^n\rightarrow M$ by the rule
	$$\sum r_ie_i\mapsto \sum r_i\psi(e_i).$$
	This is well-defined because $\{e_i\}$ is a basis of $R^n$, so every element of $R^n$
	has a unique representation as a sum $\sum r_ie_i$.  Moreover, this map is
	obviously a homomorphism of $R$-modules, and is uniquely determined by $\psi$
	(because $\{e_i\}$ spans $R^n$).  Thus, $R^n$ with the set $E$ satisfies the same universal
	property as $F(E)$ so the two must be isomorphic as $R$-modules.  As $F(E)\simeq F(A)$
	(because $A$ and $E$ are in bijection as sets) we conclude as desired.
\end{enumerate}

\item Let $R$ be a ring with $1\neq 0$ and $M$ an $R$-module.  We say that $M$ is {\em irreducible}
if $M\neq 0$ and the only submodules of $M$ are $0$ and $M$.  
\begin{enumerate}
	\item Show that $M$ is irreducible if and only if $M$ is a nonzero cyclic $R$-module.
	
\medskip\noindent{\bf Solution:} Let $m\in M$ be any nonzero element.  Then $Rm$ is a nonzero cyclic submodule
of $M$ (since it contains $m$) and by irreducibility of $M$ we must have $Rm=M$.
The converse is false in general (example $2\Z\subseteq \Z$), and we must require the additional
phrase ``with any nonzero element as a generator" to get the desired equivalence.
Suppose that $M$ is a nonzero cyclic $R$-module with any nonzero element as a generator.
If $N\subseteq M$ is a submodule which is nonzero, then any nonzero $n\in N$ generates
$M$ as an $R$-module so $N=M$ and $M$ is irreducible.

	\item If $R$ is commutative, show that $M$ is irreducible if and only if $M\simeq R/I$
	as $R$-modules for some maximal ideal $I$ of $R$.

\medskip\noindent{\bf Solution:} By part (a), if $M$ is irreducible then there is a natural surjective
map of $R$-modules $\varphi:R\rightarrow M$ given by $r\mapsto rm$ for any (fixed) nonzero $m\in M$.
The kernel of this map is a submodule of $R$, i.e. an ideal $I$ of $R$.
Since the submodules of $M\simeq R/\ker(\varphi)$ are those ideals of $R$ containing $\ker(\varphi)$,
by irreducibility of $M$ we conclude that $I$ must be maximal.


	\item Prove Schur's lemma: if $M_1$ and $M_2$ are irreducible $R$-modules then any nonzero
	$R$-module homomorphism $\varphi:M_1\rightarrow M_2$ is an isomorphism.
	
\medskip\noindent{\bf Solution:} The kernel of $\varphi$ is a submodule
of $M_1$ so by irreducibility of $M_1$ must be zero (as $\varphi$ is not the zero map).
Since $M_1$ is nonzero (definition of irreducible)
we conclude that $M_1$ is isomorphic to a nonzero submodule of $M_2$ and hence $\varphi$ is an isomorphism
by irreducibility of $M_2$.  	
	
\end{enumerate}



\end{enumerate}
\end{document}
