这一节先说了一些多项式整除的结论（Theorem 4）和多项式空间中的Taylor公式（Theorem 5），Theorem 6揭示了多项式的根的multiplicity和求导算子D之间的关系。
下一部分引入的是ideal，Theorem 7证明了任何一个ideal都由唯一一个monic polynomial生成，通过这一定理的推论可以得到最大公约数（greatest common divisor）的概念，用ideal的generator引入g.c.d.是一个很有意思的做法，看起来不好理解，但实际计算时也会方便，由于所考虑的ideal是形如
p
1
F
[
x
]
+
⋯
+
p
n
F
[
x
]
p_1F[x]+\cdots+p_nF[x]
p1F[x]+⋯+pnF[x]的组合，g.c.d.是这个ideal的monic generator，那么只要找到在此ideal中的任何一个多项式，g.c.d.一定比此多项式低阶。
Exercises
1.Let
Q
Q
Q be the field of rational numbers. Determing which of the following subsets of
Q
[
x
]
Q[x]
Q[x] are ideals. When the set is an ideal, find its monic generator.
( a ) all
f
f
f of even degree;
( b ) all
f
f
f of degree
≥
5
\geq 5
≥5;
( c ) all
f
f
f such that
f
(
0
)
=
0
f(0)=0
f(0)=0;
( d ) all
f
f
f such that
f
(
2
)
=
f
(
4
)
=
0
f(2)=f(4)=0
f(2)=f(4)=0;
( e ) all
f
f
f in the range of the linear operator
T
T
T defined by
T
(
∑
i
=
0
n
c
i
x
i
)
=
∑
i
=
0
n
c
i
i
+
1
x
i
+
1
T\left(\sum_{i=0}^nc_ix^i\right)=\sum_{i=0}^n\frac{c_i}{i+1}x^{i+1}
T(i=0∑ncixi)=i=0∑ni+1cixi+1
Solution:
( a ) No. As
g
=
x
2
g=x^2
g=x2 is in this set, but
x
g
=
x
3
xg=x^3
xg=x3 is not.
( b ) No. As
f
=
x
5
+
x
2
f=x^5+x^2
f=x5+x2 and
g
=
x
5
g=x^5
g=x5 are in this set, but
f
−
g
=
x
2
f-g=x^2
f−g=x2 is not.
( c ) Yes, its monic generator is
d
=
x
d=x
d=x.
( d ) Yes, its monic generator is
d
=
(
x
−
2
)
(
x
−
4
)
d=(x-2)(x-4)
d=(x−2)(x−4).
( e ) Yes, its monic generator is
d
=
x
d=x
d=x.
2.Find the g.c.d. of each of the following pairs of polynomials
( a )
2
x
5
−
x
3
−
3
x
2
−
6
x
+
4
,
x
4
+
x
3
−
x
2
−
2
x
−
2
2x^5-x^3-3x^2-6x+4,x^4+x^3-x^2-2x-2
2x5−x3−3x2−6x+4,x4+x3−x2−2x−2;
( b )
3
x
4
+
8
x
2
−
3
,
x
3
+
2
x
2
+
3
x
+
6
3x^4+8x^2-3,x^3+2x^2+3x+6
3x4+8x2−3,x3+2x2+3x+6;
( c )
x
4
−
2
x
3
−
2
x
2
−
2
x
−
3
,
x
3
+
6
x
2
+
7
x
+
1
x^4-2x^3-2x^2-2x-3,x^3+6x^2+7x+1
x4−2x3−2x2−2x−3,x3+6x2+7x+1.
Solution:
( a ) If we let
f
=
2
x
5
−
x
3
−
3
x
2
−
6
x
+
4
,
g
=
x
4
+
x
3
−
x
2
−
2
x
−
2
f=2x^5-x^3-3x^2-6x+4,g=x^4+x^3-x^2-2x-2
f=2x5−x3−3x2−6x+4,g=x4+x3−x2−2x−2, and denote
M
=
f
F
[
x
]
+
g
F
[
x
]
M=fF[x]+gF[x]
M=fF[x]+gF[x], then
f
−
(
2
x
−
2
)
g
=
3
x
3
−
x
2
−
6
x
∈
M
x
(
3
x
3
−
x
2
−
6
x
)
−
3
g
=
−
4
x
3
−
3
x
2
+
6
x
+
6
∈
M
f-(2x-2)g=3x^3-x^2-6x\in M \\ x(3x^3-x^2-6x)-3g=-4x^3-3x^2+6x+6\in M
f−(2x−2)g=3x3−x2−6x∈Mx(3x3−x2−6x)−3g=−4x3−3x2+6x+6∈M
combined we get
−
13
x
2
−
6
x
+
18
∈
M
-13x^2-6x+18\in M
−13x2−6x+18∈M, similarly we can see
31
x
2
+
14
x
∈
M
31x^2+14x\in M
31x2+14x∈M, continue these steps we shall find
f
f
f and
g
g
g are relatively prime.
( b ) Since
3
x
4
+
8
x
2
−
3
=
(
3
x
2
−
1
)
(
x
2
+
3
)
3x^4+8x^2-3=(3x^2-1)(x^2+3)
3x4+8x2−3=(3x2−1)(x2+3) and
x
3
+
2
x
2
+
3
x
+
6
=
(
x
+
2
)
(
x
2
+
3
)
x^3+2x^2+3x+6=(x+2)(x^2+3)
x3+2x2+3x+6=(x+2)(x2+3), we see
x
2
+
3
x^2+3
x2+3 divides both polynomials and further
1
11
(
3
x
4
+
8
x
2
−
3
)
+
6
−
3
x
11
(
x
3
+
2
x
2
+
3
x
+
6
)
=
x
2
+
3
\frac{1}{11}(3x^4+8x^2-3)+\frac{6-3x}{11}(x^3+2x^2+3x+6)=x^2+3
111(3x4+8x2−3)+116−3x(x3+2x2+3x+6)=x2+3
the desired g.c.d is
x
2
+
3
x^2+3
x2+3.
( c ) A similar procedure can show they are relatively prime.
3.Let
A
A
A be an
n
×
n
n\times n
n×n matrix over a field
F
F
F. Show that the set of all polynomials
f
∈
F
[
x
]
f\in F[x]
f∈F[x] such that
f
(
A
)
=
0
f(A)=0
f(A)=0 is an ideal.
Solution: Let
M
=
{
f
∈
F
[
x
]
:
f
(
A
)
=
0
}
M=\{f\in F[x]:f(A)=0\}
M={f∈F[x]:f(A)=0}, then if
f
,
g
∈
M
f,g\in M
f,g∈M, then for any
c
∈
F
c\in F
c∈F we have
(
c
f
+
g
)
(
A
)
=
c
f
(
A
)
+
g
(
A
)
=
0
⇒
c
f
+
g
∈
M
(cf+g)(A)=cf(A)+g(A)=0\Rightarrow cf+g\in M
(cf+g)(A)=cf(A)+g(A)=0⇒cf+g∈M
also if
g
∈
M
g\in M
g∈M, then
g
(
A
)
=
0
g(A)=0
g(A)=0, and for any
f
∈
F
[
x
]
f\in F[x]
f∈F[x], we have
(
f
g
)
(
A
)
=
f
(
A
)
g
(
A
)
=
f
(
A
)
0
=
0
(fg)(A)=f(A)g(A)=f(A)0=0
(fg)(A)=f(A)g(A)=f(A)0=0, so
f
g
∈
M
fg\in M
fg∈M.
4.Let
F
F
F be a subfield of complex numbers, and let
A
=
[
1
−
2
0
3
]
A=\begin{bmatrix}1&-2\\0&3\end{bmatrix}
A=[10−23]. Find the monic generator of the ideal of all polynomials
f
f
f in
F
[
x
]
F[x]
F[x] such that
f
(
A
)
=
0
f(A)=0
f(A)=0.
Solution: Since
A
2
=
[
1
−
8
0
9
]
A^2=\begin{bmatrix}1&-8\\0&9\end{bmatrix}
A2=[10−89], we see
A
2
−
4
A
+
3
I
=
[
1
−
8
0
9
]
−
4
[
1
−
2
0
3
]
+
3
[
1
0
0
1
]
=
[
0
0
0
0
]
A^2-4A+3I=\begin{bmatrix}1&-8\\0&9\end{bmatrix}-4\begin{bmatrix}1&-2\\0&3\end{bmatrix}+3\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}
A2−4A+3I=[10−89]−4[10−23]+3[1001]=[0000]
thus the polynomial
f
=
x
2
−
4
x
+
3
f=x^2-4x+3
f=x2−4x+3 is in the ideal. As no polynomial
g
g
g of degree 1 can make
g
(
A
)
=
0
g(A)=0
g(A)=0 and
f
f
f is monic,
f
f
f is the result.
5.Let
F
F
F be a field. Show that the intersection of any number of ideals in
F
[
x
]
F[x]
F[x] is an ideal.
Solution: Let
{
M
α
}
\{M_{\alpha}\}
{Mα} be any number of ideals in
F
[
x
]
F[x]
F[x], and
M
=
∩
α
M
α
M=\cap_{\alpha}M_{\alpha}
M=∩αMα.
First let
f
,
g
∈
M
,
c
∈
F
f,g\in M,c\in F
f,g∈M,c∈F, then
f
,
g
∈
M
α
f,g\in M_{\alpha}
f,g∈Mα for all
α
\alpha
α, thus
c
f
+
g
∈
M
α
cf+g\in M_{\alpha}
cf+g∈Mα for all
α
\alpha
α, and
c
f
+
g
∈
M
cf+g\in M
cf+g∈M, and
M
M
M is a subspace of
F
[
x
]
F[x]
F[x].
Next let
f
∈
F
[
x
]
f\in F[x]
f∈F[x] and
g
∈
M
g\in M
g∈M, then
g
∈
M
α
g\in M_{\alpha}
g∈Mα for all
α
\alpha
α, as
M
α
M_{\alpha}
Mα is an ideal, we see
f
g
∈
M
α
fg\in M_{\alpha}
fg∈Mα for all
α
\alpha
α, which means
f
g
∈
M
fg\in M
fg∈M. This shows
M
M
M is an ideal.
6.Let
F
F
F be a field. Show that the ideal generated by a finite number of polynomials
f
1
,
…
,
f
n
f_1,\dots,f_n
f1,…,fn in
F
[
x
]
F[x]
F[x] is the intersection of all ideals containing
f
1
,
…
,
f
n
f_1,\dots,f_n
f1,…,fn.
Solution: Let
M
M
M be the intersection of all ideals containing
f
1
,
…
,
f
n
f_1,\dots,f_n
f1,…,fn. From Exercise 5 we know
M
M
M is an ideal which contains
f
1
,
…
,
f
n
f_1,\dots,f_n
f1,…,fn, thus contains
f
1
F
[
x
]
+
⋯
+
f
n
F
[
x
]
f_1F[x]+\cdots+f_nF[x]
f1F[x]+⋯+fnF[x]. On the other hand,
f
1
F
[
x
]
+
⋯
+
f
n
F
[
x
]
f_1F[x]+\cdots+f_nF[x]
f1F[x]+⋯+fnF[x] is an ideal containing
f
1
,
…
,
f
n
f_1,\dots,f_n
f1,…,fn, thus shall contain
M
M
M.
7.Let
K
K
K be a subfield of a field
F
F
F, and suppose
f
,
g
f,g
f,g are polynomials in
K
[
x
]
K[x]
K[x]. Let
M
K
M_K
MK be the ideal generated by
f
f
f and
g
g
g in
K
[
x
]
K[x]
K[x] and
M
F
M_F
MF be the ideal they generate in
F
[
x
]
F[x]
F[x]. Show that
M
K
M_K
MK and
M
F
M_F
MF have the same monic generator.
Solution: Let
d
K
,
d
F
d_K,d_F
dK,dF be the monic generator of
M
K
M_K
MK and
M
F
M_F
MF, then there is
p
,
q
∈
K
[
x
]
p,q\in K[x]
p,q∈K[x] such that
d
K
=
f
p
+
g
q
d_K=fp+gq
dK=fp+gq, since
K
K
K is a subfield of
F
F
F,
p
,
q
∈
F
[
x
]
p,q\in F[x]
p,q∈F[x] as well, thus
d
K
∈
M
F
d_K\in M_F
dK∈MF,so
d
K
=
h
d
F
d_K=hd_F
dK=hdF for some
h
∈
F
[
x
]
h\in F[x]
h∈F[x]. Since
d
K
d_K
dK divides
f
f
f and
g
g
g in
F
F
F, by Corollary of Theorem 7 we know
d
F
d_F
dF is divisible by
d
K
d_K
dK, so
d
F
=
h
′
d
K
d_F=h'd_K
dF=h′dK for some
h
′
∈
F
[
x
]
h'\in F[x]
h′∈F[x]. Now we have
(
d
K
=
h
d
F
=
h
h
′
d
K
)
⇒
(
h
h
′
=
1
)
⇒
(
h
=
h
′
=
1
)
⇒
(
d
K
=
d
F
)
(d_K=hd_F=hh'd_K)\Rightarrow(hh'=1)\Rightarrow(h=h'=1)\Rightarrow(d_K=d_F)
(dK=hdF=hh′dK)⇒(hh′=1)⇒(h=h′=1)⇒(dK=dF)
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4.4 Polynomial Ideals
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