Abstract Algebra 作业
Abstract Algebra 作业
HW7
题目
Recall in our course, all rings have identity elements.
A ring R such that a^2=a for all a\in R is called a Boolean ring. Prove that every Boolean ring R is commutative and a+a=0 for all a\in R.
Prove: if A is the abelian group \mathbb Z\oplus \mathbb Z, then \operatorname{End} A is a non-commutative ring.
A finite ring with more than one element and no zero divisors is a division ring.
An element of a ring is nilpotent if a^n=0 for some n. Prove that in a commutative ring a+b is nilpotent if a and b are nilpotent. Show that this result may be false if R is not commutative. (Give counterexamples in the non-commutative case.)
In a ring R the following conditions are equivalent.
(a) R has no nonzero nilpotent elements.
(b) If a\in R and a^2=0, then a=0.
答案
Since a^2=a, applying this to a+a gives
a+a=(a+a)^2=a^2+a^2+a^2+a^2=4a.
Hence 2a=0 for every a\in R.Applying the Boolean condition to a+b,
a+b=(a+b)^2=a^2+ab+ba+b^2=a+ab+ba+b.
Thus ab+ba=0. Since every element has additive inverse equal to itself, this means ab=ba. Therefore R is commutative and a+a=0 for all a\in R.Identify \operatorname{End}(\mathbb Z\oplus\mathbb Z) with 2\times 2 integer matrices, where addition is matrix addition and multiplication is composition, i.e. matrix multiplication.
Let
A=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad B=\begin{pmatrix}0&0\\1&0\end{pmatrix}.
ThenAB=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad BA=\begin{pmatrix}0&0\\0&1\end{pmatrix}.
Hence AB\ne BA, so \operatorname{End}A is non-commutative.Let R be finite, |R|>1, and suppose R has no zero divisors. For any nonzero a\in R, the map
L_a:R\to R,\qquad x\mapsto ax
is injective, since ax=ay implies a(x-y)=0, hence x-y=0. Since R is finite, L_a is surjective. Therefore there exists b\in R such that ab=1.Similarly the map R_a:x\mapsto xa is injective and hence surjective, so there exists c\in R such that ca=1. Then
c=c(ab)=(ca)b=b.
Thus b is a two-sided inverse of a. Every nonzero element is invertible, so R is a division ring.Suppose R is commutative, a^m=0, and b^n=0. By the binomial theorem,
(a+b)^{m+n-1}=\sum_{i=0}^{m+n-1}\binom{m+n-1}{i}a^i b^{m+n-1-i}.
For each term, either i\ge m or m+n-1-i\ge n. Hence every term is 0, so (a+b)^{m+n-1}=0. Therefore a+b is nilpotent.In the non-commutative ring M_2(F), take
A=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad B=\begin{pmatrix}0&0\\1&0\end{pmatrix}.
Then A^2=B^2=0, so both are nilpotent. ButA+B=\begin{pmatrix}0&1\\1&0\end{pmatrix}, \qquad (A+B)^2=I.
Thus A+B is not nilpotent.(a)\Rightarrow(b) is immediate: if a^2=0, then a is nilpotent, so a=0.
For (b)\Rightarrow(a), suppose a is nilpotent and a\ne0. Let k be minimal such that a^k=0. Since a\ne0, we have k>1. Put m=\lceil k/2\rceil. Then m<k, so a^m\ne0, but
(a^m)^2=a^{2m}=0.
This contradicts (b). Hence there are no nonzero nilpotent elements.
HW8
题目
Let I be an ideal in a commutative ring R. Let
\operatorname{Rad} I=\{r\in R\mid r^n\in I\text{ for some }n\}.
Show that \operatorname{Rad} I is also an ideal.Let F be a field. Let S=\operatorname{Mat}_{2\times 2}(F) be the ring of 2 by 2 matrices over F. Recall the center of a ring consists of elements commuting with all elements. Prove the center of S consists of all matrices of the form
\begin{pmatrix}a&0\\0&a\end{pmatrix}.Let F be a field. Let S=\operatorname{Mat}_{2\times 2}(F) be the ring of 2 by 2 matrices over F. Show that the center of S is not an ideal in S.
Let F be a field. Let S=\operatorname{Mat}_{n\times n}(F) be the ring of n\times n matrices over F. What is its center? Prove your answer.
Let R be a ring and S the ring of all n\times n matrices over R. Prove that J is an ideal of S if and only if J is the set of all n\times n matrices over I for some ideal I in R.
答案
Let a,b\in\operatorname{Rad}I. Then a^m\in I and b^n\in I for some m,n. Since R is commutative,
(a-b)^{m+n-1}=\sum_i \binom{m+n-1}{i}a^i(-b)^{m+n-1-i}.
Each term has either i\ge m or m+n-1-i\ge n, so each term lies in I. Hence a-b\in\operatorname{Rad}I.If r\in R, then
(ra)^m=r^m a^m\in I,
so ra\in\operatorname{Rad}I. Therefore \operatorname{Rad}I is an ideal.Let
A=\begin{pmatrix}a&b\\c&d\end{pmatrix}
be central. Commuting withE_{11}=\begin{pmatrix}1&0\\0&0\end{pmatrix}
givesAE_{11}=\begin{pmatrix}a&0\\c&0\end{pmatrix},\qquad E_{11}A=\begin{pmatrix}a&b\\0&0\end{pmatrix}.
Thus b=c=0. So A=\begin{pmatrix}a&0\0&d\end{pmatrix}.Now commute with
E_{12}=\begin{pmatrix}0&1\\0&0\end{pmatrix}.
ThenAE_{12}=\begin{pmatrix}0&a\\0&0\end{pmatrix},\qquad E_{12}A=\begin{pmatrix}0&d\\0&0\end{pmatrix}.
Hence a=d. Therefore the center is exactly {aI_2:a\in F}.The center contains I_2. If it were an ideal of S, then for every A\in S we would have AI_2=A in the center. That would force every matrix to be central, which is false. For example E_{12} is not central. Hence the center is not an ideal.
The center of \operatorname{Mat}_{n\times n}(F) is
Z(S)=\{aI_n:a\in F\}.
Indeed, if A=(a_{ij}) commutes with every matrix unit E_{rs}, then AE{rs}=E{rs}A for all r,s. Comparing entries shows all off-diagonal entries of A are 0, and all diagonal entries are equal. Thus A=aI_n. Conversely, every scalar matrix aI_n commutes with all matrices.Suppose I is an ideal of R. Let \operatorname{Mat}_n(I) be the set of matrices whose entries lie in I. It is closed under addition and additive inverses. If A\in \operatorname{Mat}_n(I) and B\in \operatorname{Mat}_n(R), then each entry of AB and BA is a finite sum of elements of the form r i or i r, so lies in I. Hence \operatorname{Mat}_n(I) is an ideal of S.
Conversely, let J be an ideal of S. Define
I=\{r\in R:\ r\text{ is the }(1,1)\text{-entry of some matrix in }J\}.
Then I is an ideal of R: addition and negatives are clear, and if r\in I occurs as the (1,1)-entry of A\in J, then s r and r s occur as (1,1)-entries of sE_{11}A and AE{11}sE{11}, which lie in J.Now take any A=(a_{ij})\in J. For each i,j,
E_{1i}AE_{j1}
has a_{ij} in the (1,1)-entry and 0 elsewhere, so a_{ij}\in I. Thus J\subseteq\operatorname{Mat}_n(I).Conversely, if r\in I, choose A\in J whose (1,1)-entry is r. Then
E_{i1}AE_{1j}
is the matrix with r in the (i,j)-entry and 0 elsewhere, so every matrix unit multiple rE_{ij} lies in J. Hence every matrix over I lies in J. Therefore J=\operatorname{Mat}_n(I).
HW9
题目
All rings in this HW are commutative rings with identity.
Let R be a commutative ring with identity and suppose that the ideal A of R is contained in a finite union of prime ideals
P_1\cup\cdots\cup P_n.
Show that A\subset P_i for some i.Let f:R\to S be an epimorphism of rings with kernel K.
(a) If P is a prime ideal in R that contains K, then f(P) is a prime ideal in S.
(b) If Q is a prime ideal in S, then f^{-1}(Q) is a prime ideal in R that contains K.
(c) There is a one-to-one correspondence between the set of all prime ideals in R that contain K and the set of all prime ideals in S, given by P\mapsto f(P).
(d) If I is an ideal in a ring R, then every prime ideal in R/I is of the form P/I, where P is a prime ideal in R that contains I.
An ideal M\ne R in a commutative ring R with identity is maximal if and only if for every r\in R-M, there exists x\in R such that 1_R-rx\in M.
Consider the infinite direct product ring
R=\prod_{i=1}^{\infty}\mathbb Z.
Construct an ideal I such that I is not of the form\prod_{i=1}^{\infty} I_i,
where each I_i is an ideal of \mathbb Z.
答案
We prove by prime avoidance. Suppose no P_i contains A. Removing redundant primes if necessary, assume for each j that
A\cap P_j\not\subseteq \bigcup_{i\ne j}P_i.
Choosea_j\in (A\cap P_j)-\bigcup_{i\ne j}P_i.
Thenx=a_1+a_2a_3\cdots a_n
lies in A. But x\notin P_1, because a_1\in P_1 while a_2\cdots a_n\notin P_1 by primality. For j\ge2, we have a_2\cdots a_n\in P_j but a_1\notin P_j, so x\notin P_j. Hence x\notin P_1\cup\cdots\cup P_n, contradicting A\subseteq\bigcup_iP_i. Therefore A\subseteq P_i for some i.(a) Since f is surjective, f(P) is an ideal of S. Also f(P)\ne S, because if 1_S\in f(P), then some p\in P has f(p)=1_S, so p-1_R\in K\subseteq P, hence 1_R\in P, impossible.
If ab\in f(P), choose r,s\in R with f(r)=a, f(s)=b. Then f(rs)\in f(P), so rs-p\in K for some p\in P. Since K\subseteq P, we get rs\in P. Since P is prime, r\in P or s\in P, so a\in f(P) or b\in f(P). Thus f(P) is prime.
(b) The preimage f^{-1}(Q) is an ideal and contains K=f^{-1}(0). It is proper because 1_S\notin Q. If ab\in f^{-1}(Q), then f(a)f(b)=f(ab)\in Q. Since Q is prime, f(a)\in Q or f(b)\in Q, so a\in f^{-1}(Q) or b\in f^{-1}(Q). Thus f^{-1}(Q) is prime.
(c) The two maps are inverse. If P\supseteq K, then
f^{-1}(f(P))=P.
Indeed, if x\in f^{-1}(f(P)), then f(x)=f(p) for some p\in P, so x-p\in K\subseteq P, hence x\in P. The reverse inclusion is clear. Also, because f is surjective,f(f^{-1}(Q))=Q
for every ideal Q of S. Hence prime ideals of S correspond exactly to prime ideals of R containing K.(d) Apply (c) to the quotient map \pi:R\to R/I. The kernel is I. Thus every prime ideal of R/I is \pi(P)=P/I for some prime ideal P\supseteq I.
If M is maximal, then R/M is a field. For r\notin M, the coset r+M is nonzero, so it has an inverse x+M. Hence
(r+M)(x+M)=1+M,
meaning 1-rx\in M.Conversely, suppose the stated condition holds. Let J be an ideal with M\subsetneq J\subseteq R. Choose r\in J-M. By assumption, there exists x\in R such that 1-rx\in M. Since r\in J, we have rx\in J, and since M\subseteq J, we get 1=(1-rx)+rx\in J. Thus J=R. Hence M is maximal.
Let
I=\bigoplus_{i=1}^{\infty}\mathbb Z
be the set of integer sequences with finite support. This is an ideal of R=\prod_{i=1}^{\infty}\mathbb Z, because multiplying a finite-support sequence coordinatewise by any sequence still gives finite support.Suppose I=\prod_{i=1}^{\infty} I_i. For each coordinate i, the sequence with 1 in the i-th position and 0 elsewhere lies in I, so I_i=\mathbb Z for every i. Then \prod_i I_i=\prod_i\mathbb Z=R, but I\ne R, since (1,1,1,\ldots)\notin I. Contradiction. Therefore I is not of product form.
HW10
题目
Feel free to use the following easy fact:
Let d\in\mathbb Z. Let
(a) We have
but 1+i\notin(2) and 1-i\notin(2). Hence (2) is not prime.
(b) Suppose (a+bi)(c+di)\in(3). Then 3 divides
Since 3 is prime in \mathbb Z, 3\mid a^2+b^2 or 3\mid c^2+d^2. Modulo 3, the only squares are 0 and 1, so a^2+b^2\equiv0\pmod3 implies a\equiv b\equiv0\pmod3. Hence a+bi\in(3); similarly for c+di. Therefore (3) is prime.
(c) Since
we have (2+i)(2-i)\in(5), but neither 2+i nor 2-i is in (5). Hence (5) is not prime.
(d) Suppose (a+bi)(c+di)\in(7). Taking norms gives
If 7\mid a^2+b^2, then modulo 7 the squares are 0,1,2,4. The congruence a^2+b^2\equiv0\pmod7 forces a\equiv b\equiv0\pmod7, since -1 is not a square modulo 7. Thus a+bi\in(7). Similarly for the other factor. Hence (7) is prime.
HW11
题目
Let F be a field. Prove F[x] is a Euclidean domain.
(a) If F is a field, then every nonzero element of Fx is of the form x^k u with u\in Fx a unit.
(b) Fx is a principal ideal domain whose only ideals are 0, Fx=(1_F)=(x^0), and (x^k) for each k\ge1.
3.
(a) The polynomial x+1 is a unit in the power series ring \mathbb Zx, but is not a unit in \mathbb Z[x].
(b) x^2+3x+2 is irreducible in \mathbb Zx, but not in \mathbb Z[x].
- If c_0,c_1,\ldots,c_n are distinct elements of an integral domain D and d_0,\ldots,d_n are any elements of D, then there is at most one polynomial f of degree \le n in D[x] such that f(c_i)=d_i for i=0,1,\ldots,n.
答案
Let f,g\in F[x] with g\ne0. The polynomial division algorithm over a field gives
f=qg+r,\qquad r=0\text{ or }\deg r<\deg g.
Therefore F[x] is Euclidean with Euclidean function \delta(f)=\deg f for nonzero f.(a) Let
f=a_kx^k+a_{k+1}x^{k+1}+\cdots
where a_k\ne0 and k is minimal. Thenf=x^k(a_k+a_{k+1}x+\cdots)=x^ku.
Since the constant term of u is a_k\ne0, u is a unit in Fx.(b) Let I be a nonzero ideal of Fx. Choose f\in I with minimal order k, meaning f=x^ku where u is a unit. Then x^k=fu^{-1}\in I, so (x^k)\subseteq I. Conversely, every element of I has order at least k, so it is divisible by x^k. Hence I\subseteq(x^k). Therefore I=(x^k). The zero ideal is the remaining case.
(a) In \mathbb Zx,
(1+x)(1-x+x^2-x^3+\cdots)=1,
so 1+x is a unit. In \mathbb Z[x], the only units are \pm1, so 1+x is not a unit.(b) In \mathbb Z[x],
x^2+3x+2=(x+1)(x+2),
so it is reducible.In \mathbb Zx, x+1 is a unit, so x^2+3x+2 is associated to x+2. If
x+2=fg
in \mathbb Zx, then the constant terms of f and g multiply to 2. Thus one constant term is \pm1, so one factor is a unit. Hence x+2 is irreducible, and therefore x^2+3x+2 is irreducible in \mathbb Zx.Suppose f,g\in D[x] both have degree \le n and satisfy f(c_i)=g(c_i)=d_i. Then h=f-g has degree \le n and has roots c_0,\ldots,c_n. Since D is an integral domain, a nonzero polynomial of degree \le n has at most n roots. But h has n+1 distinct roots, so h=0. Hence f=g.
HW12
题目
Let R be a commutative ring with identity and c,b\in R with c a unit.
(a) Show that the assignment x\mapsto cx+b induces a unique automorphism of R[x] that is the identity on R. What is its inverse?
(b) If D is an integral domain, then show that every automorphism of D[x] that is the identity on D is of the type described in (a).
Prove that x^5+48x+24 is irreducible in \mathbb Q[x].
Let
f=\sum_{i=0}^n a_ix^i\in\mathbb Z[x]
have degree n. Suppose that for some k with 0<k<n and some prime p, we have:- p\nmid a_n and p\nmid a_k;
- p\mid a_i for all 0\le i\le k-1;
- p^2\nmid a_0.
Show that f has a factor g of degree at least k that is irreducible in \mathbb Z[x].
Let H be a subgroup of G. The centralizer of H is
C_G(H)=\{g\in G\mid hg=gh\text{ for all }h\in H\}.
Show that C_G(H) is a normal subgroup of N_G(H).Use the setup in the above problem. Prove:
N_G(H)/C_G(H)
is isomorphic to a subgroup of \operatorname{Aut}H.Find all the Sylow 2-subgroups and 3-subgroups of S_5.
If |G|=200, then G contains a normal Sylow subgroup.
答案
(a) By the universal property of polynomial rings, any choice of an element of R[x] determines a unique R-algebra homomorphism R[x]\to R[x]. Thus x\mapsto cx+b induces a unique homomorphism \varphi fixing R.
Since c is a unit, define
\psi(x)=c^{-1}x-c^{-1}b.Then\psi(\varphi(x))=\psi(cx+b)=c^{-1}(cx+b)-c^{-1}b=x,and
\varphi(\psi(x))=\varphi(c^{-1}x-c^{-1}b)=c^{-1}(cx+b)-c^{-1}b=x.
Both maps fix R, so \psi=\varphi^{-1}. Hence \varphi is an automorphism.(b) Let \varphi be an automorphism of D[x] fixing D. Put f=\varphi(x). Since \varphi is surjective, there exists g\in D[x] such that g(f)=x. Taking degrees over the integral domain D gives
\deg(g(f))=(\deg g)(\deg f)=1.
Hence \deg f=1. Thereforef=cx+b
with c\ne0. Applying the same argument to \varphi^{-1} shows that c must be a unit. Hence every such automorphism has the form in (a).The polynomial
f(x)=x^5+48x+24
is Eisenstein at p=3: the leading coefficient is not divisible by 3, every other coefficient is divisible by 3, and the constant term 24 is not divisible by 9. Hence f is irreducible in \mathbb Z[x], and by Gauss’s lemma it is irreducible in \mathbb Q[x].Factor f into irreducibles in \mathbb Z[x]:
f=g_1g_2\cdots g_m.
Reducing modulo p, the hypotheses say\overline f=x^k\overline h
in \mathbb F_p[x], where \overline h(0)=\overline{a_k}\ne0 and the leading coefficient is nonzero.Since p\mid a_0 but p^2\nmid a_0, exactly one irreducible factor g_j has constant term divisible by p. All other factors have nonzero constant term modulo p. Therefore all of the factor x^k in \overline f must come from \overline{g_j}. Hence x^k\mid\overline{g_j}, so
\deg g_j\ge k.
This g_j is irreducible in \mathbb Z[x] and has degree at least k.First C_G(H)\le N_G(H), because if c commutes with every element of H, then cHc^{-1}=H.
Let n\in N_G(H) and c\in C_G(H). For any h\in H, since n^{-1}hn\in H, we have
c(n^{-1}hn)=(n^{-1}hn)c.
Multiplying by n on the left and n^{-1} on the right gives(ncn^{-1})h=h(ncn^{-1}).
Thus ncn^{-1}\in C_G(H). Therefore C_G(H)\trianglelefteq N_G(H).Define
\Phi:N_G(H)\to\operatorname{Aut}(H)
by\Phi(n)(h)=nhn^{-1}.
Since n\in N_G(H), conjugation by n maps H to itself, so \Phi(n) is an automorphism of H. Also \Phi is a group homomorphism.Its kernel consists of those n\in N_G(H) such that nhn^{-1}=h for every h\in H, i.e.
\ker\Phi=C_G(H).
By the first isomorphism theorem,N_G(H)/C_G(H)\cong \operatorname{Im}\Phi,
and \operatorname{Im}\Phi is a subgroup of \operatorname{Aut}H.Since
|S_5|=120=2^3\cdot3\cdot5,
a Sylow 2-subgroup has order 8, and a Sylow 3-subgroup has order 3.The Sylow 3-subgroups are exactly
\langle(abc)\rangle
where {a,b,c} runs over all 3-element subsets of {1,2,3,4,5}. There are \binom53=10 of them.The Sylow 2-subgroups are obtained as follows. Choose one element a to be fixed, and partition the remaining four elements into two unordered pairs:
\{b,c\}\sqcup\{d,e\}.
ThenP_{a;\{b,c\},\{d,e\}} = \langle (bc),\ (de),\ (bd)(ce)\rangle
is a subgroup of order 8, isomorphic to D_8. These are all Sylow 2-subgroups. There are5\cdot3=15
such subgroups.Since
|G|=200=2^3\cdot5^2,
the number n_5 of Sylow 5-subgroups satisfiesn_5\equiv1\pmod5,\qquad n_5\mid 8.
The divisors of 8 are 1,2,4,8, and the only one congruent to 1 modulo 5 is 1. Hence n_5=1. Therefore the Sylow 5-subgroup is unique, and hence normal.
