MATH32012 SOLUTIONS
SECTION A
A1. All rings in this question are understood to be commutative.
(a) Give the definition of a noetherian ring.
A noetherian ring is a ring where every ideal is finitely generated.
(b)
...
MATH32012 SOLUTIONS
SECTION A
A1. All rings in this question are understood to be commutative.
(a) Give the definition of a noetherian ring.
A noetherian ring is a ring where every ideal is finitely generated.
(b) Give the definition of an irreducible element of a ring.
An element r of a ring R is irreducible, if r 6= 0, r is not invertible and r = ab implies
that a or b is invertible.
(c) Does there exist a noetherian ring which does not contain any irreducible elements? Justify
your
Yes, for example a field K. It is noetherian because its only ideals are <0> and
K = <1>. Every element of K is either zero or invertible, hence not irreducible.
Now let R[X, Y ] denote the ring of polynomials in two variables X, Y with real coefficients.
(d) State a reason why every irreducible element of R[X, Y ] is a prime element of R[X, Y ].
Because R[X, Y ] is a unique factorisation domain.
(e) Give an example of a prime element of R[X, Y ] which is not irreducible. Justify your example.
0, which is not irreducible by definition, and is prime because 0|fg =⇒ 0 = fg
=⇒ 0 = f or 0 = g (R[X, Y ] is a domain) =⇒ 0|f or 0|g.
(f) Is X4 + Y 4 an irreducible element of R[X, Y ]? Justify your
No, X4 + Y 4 = (X2 + Y 2)2 - 2X2Y 2 = (X2 + Y 2 - √2 XY )(X2 + Y 2 + √2 XY ), a
product of two non-invertible elements.
(g) Write the polynomial X4 - X4Y 4 + 2X2Y 2 - 3XY 3 - √2 Y 5 ∈ R[X, Y ] in standard form with
respect to each of the monomial orderings:
(1) ≺Lex with X ≻ Y ;
(2) ≺Deglex with X ≻ Y .
(h) Reduce f = Y 4 + XY 3 with respect to the set F = {f1 = Y 2 - XY 2, f2 = Y 2 - Y 3} using
the monomial ordering ≺Deglex with X ≻ Y . Express f in the form h1f1 + h2f2 + r for some
h1, h2 ∈ R[X, Y ], where r is the remainder that you found.
With respect to the chosen monomial ordering, the leading term of f1 is -XY 2 and
the leading term of f2 is -Y 3. At each step, we will underline the term in the polynomial
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MATH32012 SOLUTIONS
which is divisible by the leading term of f1 or of f2. One has
The remainder is r = 2Y 2 and moreover f = -Y f1 - (Y + 2)f2 + r. [NB: the remainder, r, is
uniquely determined as {f1, f2} is a Gr¨obner basis.]
In parts (i)–(l), you DO NOT need to justify your answers.
(i) Give an example of a maximal ideal of R[X, Y ].
For example, .
(j) Let I be the ideal of R[X, Y ] generated by all polynomials of the form XmY n where m, n ∈ Z,
m > 0 and n ≥ (m - 3)2. Give an example of a finite set T ⊂ R[X, Y ] such that I = .
For example, T = {XY 4, X2Y, X3}. To arrive at this answer, recall that a monomial
ideal is generated by its minimal monomials, and the minimal monomials form a finite set by
Dickson’s Lemma. One can work out the minimal monomials of I with the aid of a diagram
(black dots represent monomials in I):
(k) Identify the variety V(I) where I is the ideal defined in part (j).
V(I) = V({XY 4, X2Y, X3}) is the same as V({X}), i.e., the line {(0, t) : t ∈ R} ⊂ R2.
(l) Let C = {(cos t, sin t) ∈ R2 : t ∈ R, 0 ≤ t < 2π}. Identify the ideal I(C) of R[X, Y ].
C is the circle of radius 1 centred at the origin, and I(C) = .
MATH32012 SOLUTIONS
SECTION B
B2. Let R = K[X1, X2, . . . , Xn] where K is a field, and let ≺ be a monomial ordering for R.
(a) Give the definition of a Gr¨obner basis with respect to ≺ of an ideal of R.
A Gr¨obner basis of an ideal I is a finite subset {f1, f2, . . . , fn} of I such that the
leading monomials of the fi with respect to ≺ generate the ideal LT≺(I) of leading terms of I.
(b) What is a reduced Gr¨obner basis?
A reduced Gr¨obner basis is a Gr¨obner basis {g1, . . . , gn} such that all the gi are
monic polynomials and for any i 6= j, the polynomial gi is reduced with respect to gj.
(c) Let f1 = XY - X2, f2 = X - Y 2 in Q[X, Y ], and let I be the ideal of Q[X, Y ] generated by f1
and f2. With respect to the monomial ordering ≺Lex with X ≻ Y ,
(i) compute a Gr¨obner basis for I;
(ii) find a reduced Gr¨obner basis for I.
Buchberger algorithm: start with G = {XY -X2, X-Y 2}. We will double-underline
leading monomials with respect to ≺Lex. Compute the S-polynomial S(XY - X2, X - Y 2) =
XY - X2 + X(X - Y 2) = XY - XY 2. Reduce it with respect to G (when reducing, we
single-underline a monomial divisible by a leading monomial of one of the polynomials in
Compute the S-polynomials and reduce them with respect to G:
All the S-polynomials reduce to 0, so G = {XY - X2, X - Y 2, Y 3 - Y 4} is a Gr¨obner basis.
A short way to find a reduced Gr¨obner basis is as follows: the Gr¨obner basis found above shows
that the ideal of leading terms is , and that X2 can be deleted from the generating
set as it is divisible by X. (Alternatively, XY - X2 can be explicitly reduced to 0 with respect
to the other two polynomials.) Therefore, {X -Y 2, Y 3 -Y 4} is a Gr¨obner basis. No reductions
are possible, so, multiplying by constants to make all the polynomials monic, we obtain the
reduced Gr¨obner basis G′ = {X - Y 2, Y 4 - Y 3}. [NB: this is the only possible answer as the
reduced Gr¨obner basis with respect to the given monomial ordering is unique.]
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https://www.coursehero.com/file/9237825/2011-model-solutions/
MATH32012 SOLUTIONS
(d) Prove that the ideal I, defined in part (c), does not contain any monomials in Q[X, Y ].
Any element of I is of the form f = h1f1 + h2f2 where h1, h2 ∈ Q[X, Y ]. Note that
the zero set V(I) = V({f1, f2}) ∋ (1, 1) as f1(1, 1) = f2(1, 1) = 0. But no monomial XmY n
vanishes at the point X = Y = 1, thus no monomial belongs to I.
B3.
(a) What is meant by saying that a commutative ring R is a principal ideal domain?
R is a domain and every ideal of R can be generated by one element.
(b) Denote by S the set of all rational numbers of the form p
q
where p is an integer and q is an odd
integer. Show that S, with the standard operations of addition and multiplication of numbers,
is a ring.
It is enough to check that S is a subring of Q. Closure under addition: if p, q, r, s ∈ Z,
q, s are odd, p
q
+
r s
=
ps + qr
qs
∈ S because qs is odd. Closure under multiplication: p
q
r s
=
pr
qs
∈
S because qs is odd. Furthermore, S contains 0 = 0
1
and 1 = 1
1
(the denominator, 1, is odd)
and is closed under negation: -p
q
=
-p
q
∈ S where q is odd.
(c) Find the set U(S) of units of the ring S defined in part (b).
p
q
∈ S is invertible in S if and only if q
p
∈ S, if and only if p is odd. Thus
U(S) = {p
q
: p, q are odd integers }.
(d) Find the set of all irreducible elements of the ring S.
The set of irreducible elements of S is {2p
q
: p, q are odd integers }. Indeed, each
element of this set is not zero, and is not invertible by (c). If 2p
q
=
r s
t u
where p, q, s, u are odd,
then 2psu = qrt. The left-hand side is not divisible by 4, so r, t cannot both be even; therefore,
one of r, t is odd, so one of r
s
,
t u
is a unit. This proves that 2p
q
is irreducible. Any element
not in this set is either a unit or of the form 4p
q
where p, q ∈ Z, q is odd; as 4p
q
= 2 • 2
p q
is a
product of two non-units, this is not irreducible.
(e) Prove that the ring S is a principal ideal domain.
First of all, S is a domain because it is a subring of a domain Q. Every non-zero
element of S can be written as 2k p
q
where p, q are odd integers and k ∈ Z, k ≥ 0. Thus, every
element is associated to 2k, k ∈ Z, k ≥ 0. Let I be an ideal of the ring S. If I = {0}, then I
is principal. Otherwise, with every non-zero element 2k p
q
, the ideal I contains its associate 2k.
Let 2k0 be the lowest power of 2 contained in I. Then <2k0> ⊆ I, and for every 2k p
q
∈ I one
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MATH32012 SOLUTIONS
has 2k p
q
= (2k-k0 p
q
)2k0 ∈ <2k0>, so I ⊆ <2k0>. Thus, I = <2k0> is principal. Remark. In
fact, this ring is euclidean with euclidean norm N(2k p
q
) = k where p, q are odd integers.
B4. Let R be a commutative ring.
(a) Define the notion of a nilpotent element of R.
An element a ∈ R is nilpotent if there is n ∈ N such that an = 0.
(b) Let Nil(R) denote the set of all nilpotent elements of R. Prove that Nil(R) is an ideal of R.
(Note for the 2012 exam: it is enough to say that Nil(R) = √<0> therefore it is an
ideal.)
(c) (i) Define the radical, √I, of an ideal I of R.
√I = {a ∈ R | ∃n ∈ N : an ∈ I}.
(ii) Show that pI + Nil(R) = √I.
As I + Nil(R) ⊇ I, one has pI + Nil(R) ⊇ √I. On the other hand, I ⊆ √I, and
Nil(R) = p{0} ⊆ √I as {0} ⊆ I. Therefore, I + Nil(R) ⊆ √I and pI + Nil(R) ⊆ p√I =
√I.
(d) Prove that if x is a nilpotent element of R, then 1 - x is a unit of R.
Let n be such that xn = 0. Then 1 = 1 - xn = (1 - x)(1 + x + x2 + . . . + xn-1) which
shows that 1 - x is invertible.
(e) Give an example of a finite ring R such that Nil(R) consists of 2011 elements. Justify your
example. (Hint: 2011 is a prime number.)
For example, R = Z20112. This ring consists of residues of {0, 1, . . . , 20112 - 1}
modulo 20112. The invertible elements of this ring are residues of a such that a is co-prime
with 20112, i.e., a is not divisible by 2011. Invertible elements are not nilpotent. The noninvertible elements of Z20112 are the residues of 2011k for 0 ≤ k < 2011. That is, there are 2011
non-invertible elements; and each of them squares to zero, as (2011k)2 is divisible by 20112.
Thus, there are 2011 nilpotent elements in Z20112.
END OF EXAMINATION PAPER
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