MATH 435/735 Final Exam 12/21/12 Name: (1) Let R be a domain ...

MATH 435/735 Final Exam 12/21/12 Name:
(1) Let R be a domain.
(a) Define what it means for R to be a unique factorization domain.
(b) Define what it means for R to be Euclidean.
(c) Sketch a proof that R[x] is Euclidean.
(2) Let R = F2[x]/〈x 3 + 1〉 and α = x + 〈x 3 + 1〉 ∈ R.
(a) Show that |R| = 8 and (α 2 + α + 1)(α + 1) = 0.
(b) Show that α 2 + α + 1, α + 1, α 2 + α, and α 2 + 1 are not units in R.
(c) Show that R ∗ is cyclic of order 3.
Solution:
(a) Let f(x) = x 3 + 1 = (x 2 + x + 1)(x + 1) ∈ F2[x]. Then we proved in class that
R = {a2α 2 + a1α + a0 : a0, a1, a2 ∈ F2}, whence |R| = 2 3 = 8, and we also proved
that f(α) = 0, which gives the statement.
(b) By part (a), α 2 + α + 1, α + 1, α 2 + α = α(α + 1), and α 2 + 1 = α(α 2 + α) are all
zero divisors and hence cannot be units.
(c) We just found four zero divisors. The other nonzero elements in R are 1, α, and α 2 ;
they are all units since α · α 2 = 1 in R. Hence R ∗ is a group of order three, which
has to be cyclic.
(3) Let K be a field and consider I = 〈x 2 − y, x 3 − z〉 ∈ K[x, y, z].
(a) Define what a Gröbner basis for I is.
(b) Compute a Gröbner basis for I using graded lexicographic order with x ≥ y ≥ z.
(c) Is the polynomial x 3 z 2 + x 2 y 2 − xyz 2 − x 2 y − y 3 + xz + z in the ideal I?
Solution: (b) We use Buchberger’s algorithm and compute
S x 2 − y, x 3 − z = x x 2 − y − x 3 − z = −xy + z .
Thus x 3 −z ∈ (x 2 − y, −xy + z), and so I = 〈x 2 − y, −xy + z〉. Now S(x 2 − y, −xy + z) =
xz − y 2 . Let
G = x 2 − y, −xy + z, xz − y 2 .
Then we just derived that S(x 2 − y, −xy + z) G = 0. One computes
and
S x 2 − y, xz − y 2 = −yz + xy 2 = −y (−xy + z)
S −xy + z, xz − y 2 = z 2 − y 3 = z (−xy + z) + y xz − y 2 ,

so that S(x 2 − y, xz − y 2 ) G = S(−xy + z, xz − y 2 ) G = 0. Hence G is a (in fact, reduced)
Gröbner basis for I.
(c) The division algorithm gives
x 3 z 2 + x 2 y 2 − xyz 2 − x 2 y − y 3 + xz + z = x 2 − y xz 2 + y 2 + (−xy + z) x + z ,
so that (x 3 z 2 + x 2 y 2 − xyz 2 − x 2 y − y 3 + xz + z) G = z is nonzero, whence
x 3 z 2 + x 2 y 2 − xyz 2 − x 2 y − y 3 + xz + z /∈ I .
(4) Let R, S, and T be rings, and let M and N be R-modules.
(a) Define what it means for φ : S → T to be a ring homomorphism.
(b) Define what it means for ψ : M → N to be an R-module homomorphism.
(c) Show that the factor rings F3[x]/〈x 2 + 1〉 and F3[x]/〈x 2 + x〉 are
• isomorphic as F3-modules,
• non-isomorphic as rings.
Solution: (c) Let f = x 2 + 1 and g = x 2 + x. Then any polynomial h ∈ F3[x] can be
written as h = a1x + a0 + 〈f〉 and h = b1x + b0 + 〈g〉 for some a1, a0, b1, b0 ∈ F3.
The map φ : F3[x]/〈f〉 → F3[x]/〈g〉 given by
φ (cx + d + 〈f〉) = cx + d + 〈g〉
is a module isomorphism: If c1x + d1 + 〈f〉 , c2x + d2 + 〈f〉 ∈ F3[x]/〈f〉, then for any
λ ∈ F3,
φ (λ (c1x + d1 + 〈f〉) + (c2x + d2 + 〈f〉)) = φ ((λc1 + c2) x + (λd1 + d2) + 〈f〉)
= (λc1 + c2) x + (λd1 + d2) + 〈g〉
= λ (c1x + d1 + 〈g〉) + (c2x + d2 + 〈g〉)
= λ φ (c1x + d1 + 〈f〉) + φ (c2x + d2 + 〈f〉)
and ker(φ) = 〈f〉. Since φ is apparently onto, the module isomorphism theorem gives
the statement.
The polynomial f is irreducible over F3 (because it has no roots), so F3[x]/〈f〉 is a field.
On the other hand, F3[x]/〈g〉 has zero divisors (e.g., (x + 〈g〉) (x + 1 + 〈g〉) = 〈g〉), and
so F3[x]/〈g〉 cannot be ring-isomorphic to F3[x]/〈f〉.