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\centerline{ \bigbf Notes on Problem 1, August 28, 1989 }
\bs
\centerline { \bigbf Math 401 -- Second Course in Abstract Algebra, I}
\ms
\centerline { \bigbf Prof. Daniel Grayson }
\bs

All students should learn the proofs of the theorems in the book referred to
here, and be able to reproduce them on demand.  I expect the average student to
spend about three hours a week reading from the book, and about three hours a
week doing exercises, including those assigned for discussion in class.

All students should already know the definition of {\sl group}, {\sl abelian
group}, and {\sl field}.  The notion of polynomial is discussed in III.5.  The
division algorithm for polynomials is discussed in III.6.  The correspondence
between roots of polynomials and linear factors of polynomials is explained in
III.6.6.  A polynomial of degree $n$ over a field has at most $n$ roots (see
III.6.7).

Some people wondered what I meant by {\sl irreducible} in the problem.
Everyone should have wondered about that; an undefined word in a sentence
should be noticed immediately by alert students, and even those who already
knew the definition of irreducible should have been bothered by the fact that I
didn't mention the ground field under consideration.  A polynomial is {\sl
irreducible} if it can't be factored into two polynomials of positive degree,
and one must mention a field $F$ from which the coefficients of the factors are
allowed to be drawn.  Such a polynomial is said to be irreducible over $F$, and
we speak of factorizations over $F$.  Alternatively, one may speak of
factorizations in the ring $F[t]$ of polynomials in $t$ with coefficients in $F$.

One might try to guess which ground field was under consideration in problem 1 as
follows.

The Fundamental Theorem of Algebra (V.3.19) says that every polynomial with
complex coefficients has a complex root, i.e., $\C$ is an algebraically closed
field.  This is a hard theorem which we hope to prove by the end of the course.
It follows that every irreducible polynomial over the field $\C$ of complex
numbers is linear, i.e., has degree $1$, and that every polynomial of degree
$n$ factors as a product of linear factors.

If $f(t)$ is a polynomial with real coefficients, and if $\alpha$ is a root of
$f$, then there are two possibilities.  Either $\alpha$ is real, and $t-\alpha$
is a real factor of $f$, or $\alpha$ is complex, in which case
$(t-\alpha)(t-\bar\alpha)$ is a real factor of $f$.  Thus it follows from the
fundamental theorem that every irreducible polynomial over $\R$ has degree $2$
or $1$.

In problem 1 some of the factors exhibited have degrees greater than $2$, thus
couldn't be irreducible over $\R$ over $\C$.  Thus it must have been some other
field we had in mind, and in fact, the field the computer worked with is the
field $\Q$ of rational numbers.

In fact, all the polynomials exhibited have integers as coefficients, and no
fractions actually appear.  Read about Gauss' lemma (in III.6.11) to see why
this is true.  Gauss' lemma applies not only to the ring $\Z$ of integers and
its fraction field $\Q$, but also to any unique factorization domain and its
fraction field.  Thus you must learn about unique factorization domains (in
III.3.5) and why the rings $\Z$ and $F[T]$ are unique factorization domains
(see Introduction.6.7 and III.6.14).  For these two rings, the proof of unique
factorization depends on the Euclidean algorithm, which in turn depends on the
division algorithm.

\bs

\noindent
{\bf Homework:} Do III.3, Exercises 3, 4, 6, 7, 8, 13.  Turn in Monday, Sept 4.

\bs

Some people noticed that the number of irreducible factors of $t^n-1$ seems to
be the same as the number of divisors of $n$.  This is a good conjecture which
we will try to prove.

Others noticed that all factors of $t^n-1$ but one appear earlier in the chart
as a factor of $t^m-1$ for some integer $m$ dividing $n$.  Lets us call that
one factor the {\sl new} factor.

Some groups noticed that the sharing of factors can be explained as follows.
In fact, if $n = m k$, then one can prove that $t^m-1$ divides $t^n-1$ as
follows.  Write $t^n-1 = (t^m)^k-1 = u^k-1$, where $u = t^m$.  Then $u-1$
divides $u^k-1$ because $1$ is a root of $u^k-1$ (temporarily think of $u$ as a
variable).  Alternatively, one might consider the roots of the two polynomials;
every root of $t^m-1$ is also a root of $t^n-1$, and the roots occur with
multiplicity one in the complete factorizations, so the statement follows (but
there is something to check here; do you see what it is?).

Some people decided to examine the roots of the polynomial $t^n-1$.  (This sort
of thing is a good idea, and should become instinctive.)  There are $n$
distinct roots, and they are the complex numbers of the form $\exp(2 \pi i
j/n)$ with $j = 0,\dots,n-1$.  I will assume you all know about exponentials of
complex numbers, i.e. that $\exp(x + i y) = e^x \( \cos y + i \sin y \)$, and
that $\exp(w+z) = \exp(w) \exp(z)$.  You may review this in any basic analysis
book.

Some students observed that the Eisenstein criterion (III.6.15) can be used to
prove that the polynomial $t^{p-1} + \dots + t + 1$ is irreducible over $\Q$
when $p$ is a prime number (see III.6, Exercise 10(b)).  The idea is to replace
$t$ by $t+1$, arriving at a polynomial that has many coefficients divisible by
$p$, and then to exploit that divisibility appropriately.  Everyone should
learn this technique, but do try to figure it out yourself before reading the proof.

\bs

\noindent
{\bf Points for further consideration.}

\ms

\noindent
1.1.  What common algebraic structure do the complex roots of $t^n-1$ constitute?

\ms

\noindent
1.2.
Which complex numbers are the roots of the new factor?  

\ms

\noindent
1.3.
What is the degree of the new factor?  Try to write down a concise formula for
the degree.

\ms

\noindent
1.4.
Can you write down an explicit formula for the new factor in
terms of polynomials of the form $t^n-1$?  For example, the new factor
appearing in $t^8-1$ is $(t^8-1)/(t^4-1)$.

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