Math 380, Spring 2018, Assignment 4
I was at the mathematical school, where the master taught his pupils after a method scarce imaginable to us in Europe. The proposition and demonstration were fairly written on a thin wafer, with ink composed of a cephalic tincture. This the student was to swallow upon a fasting stomach, and for three days following eat nothing but bread and water. As the wafer digested the tincture mounted to the brain, bearing the proposition along with it.
- - Jonathan Swift, Gulliver's Travels
Read:
- Section 1.5.
- Section 2.1.
Carefully define the following terms, then give one example and one non-example of each:
- Monic polynomial (in k[x]).
- gcd(f,g) (where f,g∈k[x]).
Carefully state the following theorems (you do not need to prove them):
- Theorem concerning the division algorithm.
- Classification of ideals in k[x] ("Every ideal in k[x] is generated by...").
- Theorem concerning monic generators of ideals in k[x].
- Theorem relating gcd(f,g) to common divisors of f and g.
- Theorem relating ⟨f,g⟩ to ⟨g,r⟩ when f=gq+r.
Carefully describe the following algorithms:
- Division algorithm (to compute quotient and remainder when f is divided by g).
- Euclid's algorithm (to compute gcd(f,g)).
- Algorithm to replace any system of finitely many univariate equations by a single univariate equation.
- Algorithm to factor polynomials over C. (Note: this algorithm is conceptually simple but should NOT be used in engineering applications; it is numerically unstable. See books on numerical analysis for more practical algorithms.)
Solve the following problems:
- Section 1.5, problems 1, 3, 4, 11, and 12.
- Prove that for any f1,…,fs∈k[x], one has ⟨f1,f2,f3,…,fs⟩=⟨gcd(f1,f2),f3,…,fs⟩. (Hint: use the ideal containment criterion from the previous assignment.)
- Optional (but interesting): section 1.5, problem 17. (You may need to use the results of problems 13, 14, and 15.)
Questions:
Solutions:
1.5.1: It is known that every nonconstant polynomial in C[x] has a root in C. Let a1 be such a root. By the factor theorem, f is divisible by x−a1 with remainder zero, that is f=(x−a1)q where q is the quotient. If q is a nonconstant polynomial, repeat this argument with q instead of f, otherwise q is the constant c in the desired form. Thus every univariate polynomial of degree n>0 over the complex numbers may be written f=c(x−a1)⋯(x−an) where a1,…,an,c∈C and c≠0.
1.5.3: We wish to prove that I=⟨x,y⟩⊆k[x,y] is not a principal ideal, that is, that it is not generated by one element. Suppose it were. Let f be the single generator, i.e., ⟨x,y⟩=⟨f⟩. Obviously f cannot be a constant. Then every polynomial in ⟨x,y⟩ is equal to fg for some g∈k[x,y]. Consider x∈⟨x,y⟩. Then x=fg for some g. Taking the degree of both sides we have deg(x)=1=deg(fg)=deg(f)+deg(g). The only way this can be satisfied is if f or g is a constant and the other is degree 1. Since f cannot be a constant, g is. Say g=c, then x=cf⟹f=xc. Since f generates ⟨x,y⟩, it must also be that y=fh for some h∈k[x,y]. Then y=xch, but this clearly cannot be satisfied, so we have a contradiction. Thus ⟨x,y⟩ is not a principal ideal.
1.5.4: gcd(f,g) is the monic generator of the ideal ⟨f,g⟩, so, if h=gcd(f,g), then ⟨h⟩=⟨f,g⟩. By definition, this means that h∈⟨f,g⟩, so there exists A,B∈k[x] such that Af+Bg=h.
1.5.11: (a) As noted in class, every nonconstant polynomial has a root so V(f)≠∅ iff f is not a constant.
(b) If gcd(f1,…,fs)=1, we have ⟨f1,…,fs⟩=⟨1⟩ and so V(f1,…,fs)=V(1)=∅. If V(f1,…fs)=∅, then V(gcd(f1,…,fs))=∅. This means that gcd(f1,…,fs) is a polynomial in C[x] which doesn't have a root in C. Therefore it is a constant polynomial. By the definition of the gcd, it is monic, so gcd(f1,…,fs)=1.
(c) Determine gcd(f1,…,fs). If this is not equal to 1 then the affine variety V(f1,…,fs) is not empty.
1.5.12: (a) By inspection of the factored form of f, the polynomial vanishes at the points {a1,…al} and doesn't vanish at any others. Thus V(f)={a1,…,al}.
(b) We wish to show I(V(f))=⟨fred⟩. We proceed by mutual inclusion. I(V(f)) is the set of all polynomials which vanish at {a1,…,al}. We can clearly see that fred, as defined, vanishes at all of these points, so fred∈I(V(f))⟹⟨fred⟩⊆I(V(f)). Now let g∈I(V(f)). We want to show that g∈⟨fred⟩. Since g∈I(V(f)), g has, among its roots, {a1…,al}. As it is a univariate polynomial over C, we may write g in the fully factored form g=c(x−b1)n1⋯(x−bs)ns.
2: gcd(f1,f2) is the unique monic generator of ⟨f1,f2⟩. Consequently, ⟨gcd(f1,f2)⟩=⟨f1,f2⟩⟹gcd(f1,f2)∈⟨f1,f2⟩⟹gcd(f1,f2)∈⟨f1,…,fs⟩⟹⟨gcd(f1,f2),f3,…,fs⟩⊆⟨f1,…,fs⟩. Since gcd(f1,f2) is a divisor of f1, f1=h1gcd(f1,f2). Likewise f2=h2gcd(f1,f2). This shows that f1,f2∈⟨gcd(f1,f2)⟩⟹⟨f1,f2⟩⊆⟨gcd(f1,f2)⟩. We can add the generators f3,…,fs to both ideals without changing the inclusion, so we have ⟨f1,…,fs⟩⊆⟨gcd(f1,f2),f3,…,fs⟩.