本次美国代写主要为抽象代数的takehome assignment
在本作业中，除非另有说明，否则所有环均为一体环（尽管它们不会
必然是可交换的），并且所有同态都是单位同态。
1.让我们开始探索单元组。回想一下，如果R是一个（单元）环，则R是
单位，具有通过R中的乘法给出的组结构（请参阅HW10问题2）。
（a）让’：R！ S是环的（单位）同构。证明如果r 2 R则
‘（r）2 S。举一个反例，其中’不是单位。
（b）证明’对R的限制是基团同态’：R！ which
如果’是，则是单射的。
（c）类似的陈述并不适用于“形容词”。举一个形容词的例子
（单位）同态’：R！ S，但是使得单位组上的诱导图
‘：R！ S不是排斥的。
（d）让’：R！ S是环的一个射影（单位）同构，并假设
ker’J（R）（其中J是TH3问题4中的Jacobson部首）。证明
诱导图’：R！ S是客体的。
2.在基本演算中，人们经常使用这样的事实，即实数上的次数为n的多项式
数字最多有n个根。事实证明，这在任何领域都是正确的！对于这个问题，我们
X场F.
（a）设f（x）2 F [x]，并假设f（a）= 0约a 2F。证明（x a）除
f（x）。 （提示：记得F [x]是欧几里得域）。
（b）令f（x）2 F [x]，并假设f（a1）= f（a2）= = f（ar）= 0，对于ai 2 F都是不同的。
通过归纳证明（x a1）（x a2）（x ar）除以f（x）。
（c）从（b）部分推论出，如果f（x）的阶数为n，则f（x）最多具有n个根。
（d）作为推论，令f（x）2 F [x]为2或3的多项式。证明F [x] =（f（x））
当且仅当f（x）在F中没有根时为。
阶数为4的多项式
3.本学期我们使用了很多次（例如，在对HW9中的组进行分类时），
单元组（Z = pZ）为p 1阶的循环，更普遍的是，如果p为奇数，则
（Z = pnZ）是循环的。但是，如果您一直在密切注意，您应该注意到我们
尚未真正证明这一事实！因此，让我们来个完整的圈子，并将这一事实推论为
问题1和2的后果
（a）考虑不变因子形式的nite阿贝尔群G = Zn1 Zn2 Znk
nkjnk 1j jn2jn1）。证明如果k = 1则G中有超过nk个元素
其顺序除以nk。
（b）令F为域，令F为F单元组的nite子组。
G是循环的。推论（Z = pZ）= Zp1。（提示：您可以表达条件
（a）中关于F [x]的多项式的解？
现在，我们得出奇数素数p的（Z = pnZ）的类似结果。
（c）令G为一个尼特阿贝尔群，并假设其所有Sylow子群都是循环的。表演
G是循环的。
In this assignment unless otherwise indicated, all rings are unital rings (although they will not
necessarily be commutative), and all homomorphisms are unital homomorphisms.
1. Let’s begin by exploring unit groups. Recall that if R is a (unital) ring, then R is the set of
units, endowed with a group structure given by multiplication in R (cf. HW10 Problem 2).
(a) Let ‘ : R ! S be a (unital) homomorphism of rings. Show that if r 2 R then
‘(r) 2 S. Give a counterexample where ‘ is not unital.
(b) Show that the restriction of ‘ to R is a group homomorphism ‘ : R ! S, which
is injective if ‘ is.
(c) The analogous statement does not hold for ‘ surjective. Give an example of a surjective
(unital) homomorphism ‘ : R ! S, but such that the induced map on unit groups
‘ : R ! S is not surjective.
(d) Let ‘ : R ! S be a surjective (unital) homomorphism of rings, and suppose that
ker ‘ J(R) (where J is the Jacobson radical from TH3 Problem 4). Prove that the
induced map ‘ : R ! S is surjective.
2. In elementary calculus one often uses the fact that a polynomial of degree n over the real
numbers has at most n roots. This turns out to be true over any eld! For this problem we
x a eld F.
(a) Let f(x) 2 F[x], and suppose that f(a) = 0 for some a 2 F. Show that (x a) divides
f(x). (Hint: recall that F[x] is Euclidean domain).
(b) Let f(x) 2 F[x], and suppose f(a1) = f(a2) = = f(ar) = 0, for ai 2 F all distinct.
Prove by induction that (x a1)(x a2) (x ar) divides f(x).
(c) Deduce from part (b) that if the degree of f(x) is n, then f(x) has at most n-roots.
(d) As a corollary, let f(x) 2 F[x] be a polynomial of degree 2 or 3. Prove that F[x]=(f(x))
is a eld if and only if f(x) has no roots in F. Give an example to show this is not true
for polynomials of degree 4.
3. We used many times this semester, (for example when classifying groups like in HW9) that
the unit group (Z=pZ) is cyclic of order p 1, and more generally that if p is odd then
(Z=pnZ) is cyclic. But if you’ve been paying close attention you should notice that we
haven’t actually proved that fact yet! So let’s come full circle and deduce this fact as a
consequence of Problems 1 and 2.
(a) Consider a nite abelian group G = Zn1 Zn2 Znk in invariant factor form (so
that nkjnk 1j jn2jn1). Prove that if k = 1 then there are more than nk elements in G
whose order divides nk.
(b) Let F be a eld, and let G F be a nite subgroup of the unit group of F. Prove
that G is cyclic. Deduce that (Z=pZ) = Zp 1. (Hint: Can you express the condition
in (a) in terms of solutions to a polynomial in F[x]?)
Let’s now deduce the analogous result of (Z=pnZ) for an odd prime p.
(c) Let G be a nite abelian group and suppose all it’s Sylow subgroups are cyclic. Show
that G is cyclic.