Math 380, Spring 2018, Assignment 4

From cartan.math.umb.edu
Revision as of 20:09, 18 February 2018 by John.DeBrota (talk | contribs) (Solutions:)

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:

  1. Section 1.5.
  2. Section 2.1.

Carefully define the following terms, then give one example and one non-example of each:

  1. Monic polynomial (in k[x]).
  2. gcd(f,g) (where f,gk[x]).

Carefully state the following theorems (you do not need to prove them):

  1. Theorem concerning the division algorithm.
  2. Classification of ideals in k[x] ("Every ideal in k[x] is generated by...").
  3. Theorem concerning monic generators of ideals in k[x].
  4. Theorem relating gcd(f,g) to common divisors of f and g.
  5. Theorem relating f,g to g,r when f=gq+r.

Carefully describe the following algorithms:

  1. Division algorithm (to compute quotient and remainder when f is divided by g).
  2. Euclid's algorithm (to compute gcd(f,g)).
  3. Algorithm to replace any system of finitely many univariate equations by a single univariate equation.
  4. 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:

  1. Section 1.5, problems 1, 3, 4, 11, and 12.
  2. Prove that for any f1,,fsk[x], one has f1,f2,f3,,fs=gcd(f1,f2),f3,,fs. (Hint: use the ideal containment criterion from the previous assignment.)
  3. Optional (but interesting): section 1.5, problem 17. (You may need to use the results of problems 13, 14, and 15.)
--------------------End of assignment--------------------

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 xa1 with remainder zero, that is f=(xa1)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(xa1)(xan) where a1,,an,cC and c0.

1.5.3: We wish to prove that I=x,yk[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 gk[x,y]. Consider xx,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=cff=xc. Since f generates x,y, it must also be that y=fh for some hk[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 hf,g, so there exists A,Bk[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 fredI(V(f))fredI(V(f)). Now let gI(V(f)). We want to show that gfred. Since gI(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(xb1)n1(xbs)ns.

But it must be that l of the roots are the ai, so we can arrange to write g as g=c(xa1)n1(xal)nl(xbsl)nsl(xbs)ns.
By inspection, then, we can see that g=hfred for an appropriately choosen hC[x], and so gfredI(V(f))fred.

2: gcd(f1,f2) is the unique monic generator of f1,f2. Consequently, gcd(f1,f2)=f1,f2gcd(f1,f2)f1,f2gcd(f1,f2)f1,,fsgcd(f1,f2),f3,,fsf1,,fs. Since gcd(f1,f2) is a divisor of f1, f1=h1gcd(f1,f2). Likewise f2=h2gcd(f1,f2). This shows that f1,f2gcd(f1,f2)f1,f2gcd(f1,f2). We can add the generators f3,,fs to both ideals without changing the inclusion, so we have f1,,fsgcd(f1,f2),f3,,fs.