Math 113, Second Midterm Exam
SOLUTIONS
(1) Part (a) is a special case of Exercise 20 in Section 26. We will check the two conditions
in Definition 18.9 on page 171. First, since Z3[x] is a commutative ring, we have
φ(f(x)g(x)) = (f(x)g(x))3=f(x)3·g(x)3=φ(f(x)) ·φ(g(x)).
Second, since Z3[x] is a ring of characteristic 3, we have
φ(f(x) + g(x)) = (f(x) + g(x))3=f(x)3+g(x)3=φ(f(x)) + φ(g(x)).
For part (b) we note that φ(f(x)) = f(x)3=f(x3) can only be the zero polynomial when
f(x) itself is the zero polynomial. Hence the kernel of φis the zero ideal h0iin Z3[x].
(2) (a) We shall prove that the ring R=Z3[x]hf(x)iis a field. First, the polynomial
f(x) = x3+x2+ 2 has no zeros in Z3because f(0) = f(2) = 2 and f(1) = 1. Theorem
23.10 on page 214 ensures that f(x) is irreducible in Z3[x]. By Theorem 27.25 on page
251, the ideal hf(x)iis maximal in Z3[x]. Hence Ris a field, by Theorem 27.9 on page 247.
(b) The field Ris an extension of degree 3 over Z3. Thus Ris a 3-dimensional vector
space over Z3. By Theorem 33.1 on page 200, the field Rhas precisely 33=27 elements.
(3) (a) The ideal hyiis non-zero and it is prime because R[x, y]/hyi ≃ R[x] is an integral
domain. Moreover, hyiis not maximal because it is strictly contained in the ideal hx, yi.
(b) The ideal hx, yiis not principal: it cannot be generated by one element because xand
yhave no common non-constant factor. It is maximal since R[x, y ]/hx, yi ≃ Ris a field.
(4) We fix the lexicographic term order with x > y. The two given generators of the ideal
Ihave the underlined leading terms:
f(x, y) = x+yand g(x, y) = x2+y2−1.
Their S-polynomial S(f, g) = xf (x, y )−g(x, y) = xy −y2+ 1 reduces to h(x, y) =
S(f, g )−yf(x, y) = −2y2+1by one step of the Division Algorithm. By Theorem 26.6,
the set G=f(x, y ), h(x, y)also generates I. We find that Gis a Gr¨obner basis of I
because S(f, h ) = 2yf(x, y) + h(x, y) = 2xy −1 reduces to zero by the Division Algorithm.
(b) The variety V(I) consists of the two points −1
√2,+1
√2and +1
√2,−1
√2.
(5) The assertion is the converse to Theorem 31.3 on page 283, and it is false, as seen in
Exercise 19 (c) in Section 31. For a counterexample, let F=Qbe the field of rational
numbers and let E=Qbe its algebraic closure in the field of complex numbers C. Then
Eis algebraic over F, by definition, but it is not a finite-dimensional vector space over F.