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{\Huge 189-235A: Basic Algebra I} \\ \vskip 0.1in
{\Huge Assignment 6} \\ \vskip 0.1in
{\Large Due: Wednesday, November 9, 2005 }
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1. For which odd primes $p\le 23$ is the polynomial $x^2+1$ irreducible
in $\Z_p[x]$? Can you detect a pattern?
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2. Find a polynomial of degree $2$ in $\Z_6[x]$ that has four roots in
$\Z_6$. Why does this not contradict the theorem shown in class that
a polynomial in $F[x]$ of degree $d$ has at most $d$ roots?
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3. Find the inverse of $[x^2+x+1] $ in the ring
$\Z_2[x]/(x^3+x+1)$.
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4. Write down all the powers of $[x]$ in
the finite ring $\Z_2[x]/(x^3+x+1)$.
What is the smallest $j>1$ such that $[x]^j=1$?
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5. If $p$ is an odd prime of the form $3+4m$, show that the polynomial
$x^2+1$ is irreducible in $\Z_p[x]$, so that
$\Z_p[x]/(x^2+1)$ is a field.
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6.
Which of the following subsets $I$ of a commutative
ring $R$ are ideals of $R$?
Justify your answer.
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6a. $R=F[X]$, where $F$ is a field, and $I=F$ is the set of constant polynomials.
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6b. $R=\Z\times\Z$, and $I=\{ (m,0) \quad \mbox{ where } m\in \Z \}$.
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6c. The set of {\em nilpotent elements} of a ring $R$, i.e., those
$a\in R$ such that $a^n=0$ for some $n$.
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6d. $R$ is the ring of functions from $\Z$ to the real numbers ${\bf R}$,
and $I$ the subset of those
functions $f$ satisfying $f(0)=f(1)$.
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6e. $R$ is the ring of functions from $\Z$ to ${\bf R}$,
and $I$ the subset of those
functions $f$ satisfying $f(0)=f(1)=0$.
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7. Let $R$ be the polynomial ring $F[x]$
with coefficients in a field. Adapt the argument given in class for
$R=\Z$ to show that
every ideal of $R$ is principal.
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{\bf Extra credit problems}
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Let $\Q(\sqrt{-5})= \{a+b\sqrt{-5}, a,b\in \Q\}$, and
$\Z[\sqrt{-5}]= \{a+b\sqrt{-5}, a,b\in \Z \}$.
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8. Show that $\Q(\sqrt{-5})$ is a field, and that $\Z[\sqrt{-5}]$ is a subring.
It is called the {\em ring of integers} of $\Q(\sqrt{-5})$ and
plays the role of the usual integers in the arithmetic of
$\Q(\sqrt{-5})$.
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9. Show that the invertible elements in $\Z[\sqrt{-5}]$ are exactly
$1$ and $-1$.
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10. Show that the elements $2$, $3$, $1+\sqrt{-5}$ and $1-\sqrt{-5}$ are
irreducible. (I.e., they cannot be written in the form $ab$ where
$a,b\ne \pm 1$.)
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11. Using 9, show that the ring $\Z[\sqrt{-5}]$ is not a unique factorization
ring. (I.e., the ``integers" in $\Z[\sqrt{-5}]$ cannot be written uniquely
as a product of irreducible elements.)
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12. Show that the ideals $(2,1+\sqrt{-5})$, $(3,1+\sqrt{-5})$, and
$(3,1-\sqrt{-5})$ are not principal, and that
they are
{\em irreducible}, i.e.,they cannot be factored further into products of
non-trivial ideals.
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13. If $I$ and $J$ are ideals, define the product
$IJ$ to be the ideal generated
by the elements of the form $ij$ with $i\in I$ and $j\in J$.
Show that
$ (2, 1+\sqrt{5})^2 = (2), (3,1+\sqrt{-5})(3,1-\sqrt{-5}) = (3),$
and conclude that the ideal $(6)$ factorizes as a product of $4$
(non-principal) ideals:
$ (6) = (2,1+\sqrt{-5})^2 (3,1+\sqrt{-5})(3,1-\sqrt{-5}).$
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{\em Remark}: It can be shown that this factorization of the principal ideal
$(6)$ into a product of
irreducible ideals is {\em unique}, up to the order of the factors.
This is a general phenomenon: although the ring $\Z[\sqrt{-5}]$ fails to
satisfy unique factorization, its {\em ideals} can be expressed
uniquely as products of irreducible ideals.
The introduction of ideals in the late 19-$th$ century by Dedekind was
an attempt to salvage unique factorization in such rings, by showing
it was true on the level of ideals which were viewed as a kind of
``ideal number". This is where the terminology comes from...
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