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Solution to Linear Algebra Hoffman & Kunze Chapter 4.4


Exercise 4.4.1

Let Q be the field of rational numbers. Determine which of the following subsets of Q[x] are ideals. When the set is an ideal, find its monic generator.

(a) all f of even degree;
(b) all f of degree 5;
(c) all f such that f(0)=0;
(d) all f such that f(2)=f(4)=0;
(e) all f in the range of the linear operator T defined by
T(i=0ncixi)=i=0ncii+1xi+1.Solution:

(a) This is not an ideal. Let f(x) be any polynomial of even degree. Let g(x)=x. Then deg(fg)=deg(f)+1 which is odd and is therefore the set of polynomials of even degree does not satisfy the necessary property that fg is in the set whenever f is in the set.

(b) This is not an ideal. Let f(x)=x5 and g(x)=x5+x4. Then f and g are in the set and therefore f+g must be in the set if it is an ideal. But f(x)+g(x)=x4 has degree equal to four and therefore is not in the set. In other words the set is not closed with respect to addition and therefore is not even a subspace.

(c) This is an ideal. Let I={fF[x]f(0)=0}. If f(0)=0 and g(0)=0 then (cf+g)(0)=cf(0)+g(0)=0 thus I is a subspace of F[x]. Now suppose fI and gF[x]. Then (fg)(0)=f(0)g(0)=0g(0)=0. Thus fgI. Thus I is an ideal. Let f(x)=a0+a1x++anxn. Then f(0)=a0. Thus if fI then necessarily a0=0. Let g(x)=x. Then gI and f(x)=a1x+a2x2+anxn=x(a1+a2x+anxn1)I. Thus g generates I.

(d) This is an ideal. Let I={fF[x]f(2)=f(4)=0}. If f(2)=f(4)=0 and g(2)=g(4)=0 then (cf+g)(2)=cf(2)+g(2)=0 and (cf+g)(4)=cf(4)+g(4)=0 thus I is a subspace of F[x]. Now suppose fI and gF[x]. Then (fg)(2)=f(2)g(2)=0g(2)=0 and (fg)(4)=f(4)g(4)=0g(4)=0. Thus fgI. Thus I is an ideal. Let g(x)=(x2)(x4). We claim g generates I. Let f(x)I. Then since f(2)=0, by Corollary 1 page 128 it follows that f(x)=(x2)q(x) for some q(x)F[x]. Now since f(4)=2q(4)=0 it follows that q(4)=0 Thus there is a p(x) such that q(x)=(x4)p(x). Thus f(x)=(x2)(x4)p(x)=g(x)p(x) and therefore f(x) is in the ideal generated by g(x).

(e) This is an ideal. In fact it is the same ideal as in part (c). Note that T(f) has no constant term, thus T(f)(0)=0 fF[x]. Now let f(x)=a1x+a2x2++anxn. Let g(x)=a1+2a2x+3a3x2++nanxn1. Then f=T(g). Thus f is in the image of T. Thus any polynomial with zero constant term is in the image of T. Thus it is exactly the same ideal is in part (c).


Exercise 4.4.2

Find the g.c.d. of each of the following paris of polynomials

(a) 2x5x33x26x+4, x4+x3x22x2;
(b) 3x4+8x23, x3+2x2+3x+6;
(c) x42x32x22x3, x3+6x2+7x+1.

Solution:

This question on its face seems to depend on what the base field is. I’m going to assume it is C.

(a) Let f(x)=2x5x33x26x+4 and g(x)=x4+x3x22x2. I have a feeling there’s a mistake and they really meant 2x5x33x26x+6 because then both polynomials are divisible by x22. But as it is, the g.c.d. of these two polynomials equals 1 – which does not seem that easy to prove using only what we know up to this point. If we knew about unique factorization in C[x] we could simply factor g(x) into linear factors and check none of the roots of g are roots of f. In fact the roots of g(x) are ±2 and 1±2i2.

But we can’t use that argument yet. Instead we are probably expected to argue using the comments on page 132 right after the proof of Theorem 7, commonly known as the “euclidean algorithm”.

Let I=f(x),g(x) be the ideal in C[x] generated by f and g. Then
2x5x33x26x+4=(x4+x3x22x2)(2x2)+(3x3x26x).Therefore 3x3x26x=f(x)(2x2)g(x)I.
x4+x3x22x2=(3x3x26x)(13x+49)+(139x2+23x2).Therefore 139x2+23x2I.
3x3x26x=(139x2+23x2)(2713x931132)+(2327132x23231132).Therefore 2327132x23231132I.139x2+23x2=(2327132x23231132)(1332347x+13219223472)+(132613272).Therefore 132613272I. And 327213261132613272=1 therefore 1I. Therefore I=C[x]. Therefore the g.c.d. of f(x) and g(x) equals 1: gcd(f,g)=1.

(b) Let f(x)=3x4+8x23 and g(x)=x3+2x2+3x+6. Let I=f(x),g(x). We have
3x4+8x23=(x3+2x2+3x+6)(3x6)+(11x2+33).Thus 11x2+33I.
Thus x2+3I.Now
x3+2x2+3x+6=(x2+3)(x+2).Thus x2+3 divides both f(x) and g(x). In particular
f(x)=(3x21)(x2+3)g(x)=(x+2)(x2+3)
and therefore it follows thatf(x),g(x)=x2+3.(c) As in part (a) I have a feeling there’s a typo and they really meant x3+6x2+7x+2 because then both are divisible by x+1. But as it is, x3+6x2+7x+1 is irreducible and the calculations are even worse than part (a). As I did in part (a), I’ll solve this in both the way they actually stated it and the was I think they intended it. First the way they stated it.

Let f(x)=x42x32x22x3 and g(x)=x3+6x2+7x+1 and I=f(x),g(x).
x42x32x22x3=(x3+6x2+7x+1)(x8)+(33x2+53x+5)Thus 33x2+53x+5I.
x3+6x2+7x+1=(33x2+53x+5)(133x+1451089332)+(2271089x+3641089).Thus
2271089x+3641089IAnd finally
33x2+53x+5=(2271089x+3641089)(108933227x2618282751529)+(900929751529).Thus 900929751529I and it follows that 1I and that gcd(f(x),g(x))=1.


Exercise 4.4.3

Let A be an n×n matrix over a field F. Show that the set of all polynomials f in F[x] such that f(A)=0 is an ideal.

Solution: Let I be the set {fF[x]f(A)=0}. Let f,gI and cF. Then (cf+g)(A)=cf(A)+g(A)=c0+0=0. Thus I is a subspace of F[x] as a vector space over F. Now suppose fI and gF[x]. Then (gf)(A)=g(A)f(A)=g(A)0=0. Thus I has the required property to be an ideal.


Exercise 4.4.4

Let F be a subfield of the complex numbers, and let
A=[1203].Find the monic generator of the ideal of all polynomials f in F[x] such that f(A)=0.

Solution: Let I={f(x)F[x]f(A)=0}. We know from the previous exercise that I is an ideal. It’s clear that f(A)0 if deg(f)1. Thus if we find any f(x)I such that deg(f)=2 then f must be a generator. Let f(x)=x24x+3. Then
f(A)=A24A+3=[1203][1203]4[1203]+3[1001]=[1809][48012]+[3003]=[0000].


Exercise 4.4.5

Let F be a field. Show that the intersection of any number of ideals in F[x] is an ideal.

Solution: Let A be a set. Let Iα be an ideal in F[x] for each αA. Let I=αAIα. By Theorem 2, page 36 (sec. 2.2), I is a subspace of F[x]. We must just show I satisfies the extra condition required to be an ideal. Specifically, let g(x)F[x] and f(x)I. We must show g(x)f(x)I. Since f(x)I, it follows that f(x)Iα αA. Since Iα is an ideal αA, it follows that g(x)f(x)Iα αA. Thus f(x)g(x)Iα αA. Thus f(x)g(x)I.


Exercise 4.4.6

Let F be a field. Show that the ideal generated by a finite number of polynomials f1,,fn in F[x] is the intersection of all ideals containing f1,,fn.

Solution: Let I=f1,,fn be the ideal generated by {f1,,fn}. Let J be any ideal containing {f1,,fn}. Let gI. Then g=g1f1++gnfn for some g1,,gn in F[x]. Since J is an ideal containing {f1,,fn} it must also contain g. Thus gI gJ. Thus IJ. Now by the definition of intersection, J is contained in every ideal which contains {f1,,fn}. Since I is such an ideal, it follows that JI. Thus we’ve shown IJ and JI. Thus J=I.


Exercise 4.4.7

Let K be a subfield of a field F, and suppose f, g are polynomials in K[x]. Let MK be the ideal generated by f and g in K[x] and MF be the ideal they generate in F[x]. Show that MK and MF have the same monic generator.

Solution 1: We first note a general fact. Let E be any field. Let f,gE[x] such that gcd(f,g)=k. Let hE[x] be monic. We claim gcd(kf,kg)=kh in E[x]. Since gcd(f,g)=k, the ideal generated by f and g is the same as the ideal generated by k. In other words fE[x]+gE[x]=kE[x]. Thus hfE[x]+hgE[x]=hkE[x]. Since h is monic by assumption and k is monic by definition of gcd, it follows that hk is monic – and therefore hk satisifes the definition of gcd(f,g) in E[x].

Now let h=gcd(f,g) in K[x]. So hK[x] and h=fa+gb for some a,bK[x]. And u,vK[x] such that f=uh and g=vh. Once you can write the three equations h=fa+gb, f=uh and g=vh it follows that the ideal in K[x] generated by f and g is the same as the ideal generated by h. That’s what it means for h to be the g.c.d. of f and g. But these three equalities also hold in F[x] and consequently the ideal in F[x] generated by f and g is the same as the one generated by h.

Solution 2: This might be easier if they had formalized what is discussed informally on page 132 after the proof of Theorem 7. What they are describing is the so-called “eucidean algorithm”. If you have two polynomials f and g and say deg(f)deg(g). Then you divide g into f and take the remainder rK[x]. That remainder is also in the ideal generated f and g. Then replace f and g with g and r and do the same thing, divide r into g and the remainder has degree less than r. Eventually you must arrive at a remainder equal to zero or of degree equal to zero (i.e. a scalar). If the remainder is zero then the previous remainder generates the ideal f,g in K[x]. If the remainder is a non-zero scalar then the ideal contains 1 and therefore f,g=K[x]. The euclidean algorithm allows us to always find the gcd in a finite number of steps.

Once we have the euclidean algorithm, all we need to do to solve this problem is to note that all operations in the euclidean algorithm happen in K[x] and so if KF then in F[x] the same operations will lead to the same generator which will therefore still live in the smaller ring K[x].

From http://greggrant.org

Linearity

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