Download Solutions — Ark 1

Solutions — Ark 1
From the book: Number 8, 9, 10 and 12 on page 11.
Number 8 : Show that the set of prime ideals of A has a minimal element with respect
to inclusion.
Solution: By Zorn’s lemma it sufficies to show that any descending chain of prime
ideals contains a prime ideal. Let {pi }i∈I be such a chain, i.e., the set I is linearly
ordered and pi ⊆ pj whenever i ≤ j. If the chain is finite, there is a smallest ideal in the
chain, which is prime,�
so we may assume the chain
� to be infinit.
We shall see that i pi is prime. Let xy ∈ i pi . Then for a given index i, either
x ∈ pi or y ∈ pi . It follows that either x or y must be contained in infinitly many
of the pi ’s, and because the chain is descending, the one that is, is contained in the
intersection.
�
√
Number 9 : Let a be a proper ideal in a ring A. Show that a = a is equivalent to a
being an intersection of prime ideals.
√
Solution: Since the radical a is
a,
�
√ the intersection of all the primes containing
clearly a is such an intersection if a = a. Conversly, assume that a = i pi . Then
all the pi are among the prime ideals containing a, of course, and hence we have the
inclusions
�
√
a⊆ a⊆
pi .
As a =
�
i
pi it follows that a =
√
i
a.
�
Number 10 : Let A be a ring with nilradical N. Then the following are equivalent:
1. A has exactly one prime ideal;
2. every element of A is either a unit or nilpotent;
3. A/N is a field.
Solution: .
The point is that all three are equivalent to N being a maxiamal ideal.Clearly this
is equivalent to iii).
Now N is the interection of all prime ideals in A, so if A has only one prime ideal,
that ideal must be N and, being the only one, it is of course maximal. On the other
hand, if N is maximal, it is the only prime ideal, since it is contained in any other.
Solutions Ark 1
MAT4200 — autumn 2011
If N is maximal, and hence the only prime ideal, A is a local ring. Elements in A not
in N are therefore invertible, and those in N are nilpotent, so elements in A are either
invertible or nilpotent. On the other hand, if every non-nilpotent element is invertible,
N is a maximal ideal. This shows that ii) is equivalent to N being maximal.
�
Number 12 : A local ring contains no idempotents other than 0 or 1.
Solution: Let e ∈ A be an idempotent, i.e., e2 = e. Then e(e − 1) = 0. Now the
ring being local, either e or e − 1 is invertible (both can not lie in the maximal ideal).
It follows that either e = 0 or e − 1 = 0.
�
Oppgave 1. Show that the principal ideal (P (X1 , . . . , Xn )) in the polynomial ring
k[X1 , . . . , Xn ] over the field k is prime if and only if P (X1 , . . . , Xn ) is irreducible.
(Hint: Use that k[X1 , . . . , Xn ] is UFD.)
Solution: In fact, we are going to show that in any ring A being a UFD an element
f is irreducible if and only if the principal ideal (f ) is prime.
The easy implication — which does not use that A is UFD, and hence is true in
general — is that (f ) prime implies f irreducible. To see that, assume that f = xy.
Then xy ∈ (f ). Hence, (f ) being prime by assumption, one, of them, say x, is in
(f ), and consequently x = af . Substituting back into f = xy we get f = af y, and
cancelling f (which we safely can do since A is a domain) we obtain 1 = ay. This shows
that y is a unit.
To prove the other way around, assume f is irreducible and pick elements x and y
with xy ∈ (f ). Then xy = af . As our ring A is a UFD, we may factor x, y and a into
irreducibles: x = x1 · · · · · xr , y = y1 · · · · · ys and a = a1 · · · · · at .
This gives the two factorisations of xy into irreducibles:
xy = x1 · · · · · xr y1 · · · · · ys = a1 · · · · · at f
(1)
Such factorisations being essentially unique, f has to be a unit times either one of the
xi ’s or one of the yj ’s. In the first case x ∈ (f ) and in the second y ∈ (f ).
�
Oppgave 2. Let k be a field . Show that the ideal (XY − ZW ) ⊂ k[X, Y, Z, W ] is
prime. Is it maximal? (Hint: Use the previous exercise and take a look at the degrees
of the polynomials involved).
Let x, y, z and w cosets of the variables X, Y, Z and W in the quotientring A =
k[X, Y, Z, W ]/(XY − ZW ). Show that x is irreducible inA, but that (x)A is not prime.
Is k[X, Y, Z, W ]/(XY − ZW ) UFD?
—2—
Solutions Ark 1
MAT4200 — autumn 2011
Solution: We shall show that (XY − ZW ) is irreducible, and then appeal to the
previous exercise. Assume therefore that P Q = (XY − ZW ). We may write P =
P0 + P1 + · · · + Pn and Q = Q0 + Q1 + · · · + Qm where the Pi ’s and the Qi ’s are
homogenous polynomials of degree i. If Pn and Qm both are not constant, it follows
that n + m = 2 since (XY − ZW ) is homogenous of degree 2. Hence either m = 0
or n = 0 and we are through, or n = m = 1 and P and Q are linear forms. Putting
Z = 0, one sees that W is not occuring in a nonzero term of either P or Q, putting
W = 0 gives that Z does not occure, so either P = X and Q = Y or vise versa. Both
cases are absurd.
The ideal (XY − ZW ) is not maximal e.g., since (XY − ZW ) ⊂ (X, Y ) which is a
proper ideal.
Let us see that x is irreducible. Assume x = f g and
� pick representatives
� F and G
in k[X, Y, Z, W ] for f and g respectively. Then F = si=0 Fi and G = ti=1 Gi where
Fi and Gi are homogeneus polynomials of degree i. We are allowed to freely change
both F and G with elements of the ideal (XY − ZW ), so we may assume that no term
Fi or Gi is in (XY − ZW ). We get:
� s
�� t
�
�
�
X = F G + C(XY − ZW ) =
Fi
Gi + C(XY − ZW )
i=0
=
�
i<s and j<t
i=1
Fi Gj + Fs Gt + C(XY − ZW ).
Now if s + t > 1, Fs Gt + Cs+t−2 (XY − ZW ) being the only term of degree s + t in
the equation must vanish: Fs Gt + Cs+t−2 (XY − ZW ) = 0. Hence either Fs or Gt has
(XY − ZW ) as a factor (the polynomial ring is UFD!) contrary to the hypothesis that
no Fi or Gi is in (XY − ZW ). It follows that either F or G is a constant.
The ring A is not UFD — in fact in some sense it is the “generic” or minimal example
of a non-UDF — since x is irreducible, but not prime. ( We have that zw ∈ (x) but
neither z nor w lies in (x).)
�
√
√
√
Oppgave 3. Let Z[i 5] = {a + ib 5 | a, b ∈ Z} ⊂ C. Show that Z[i 5] is a subring
of C and that there is an isomorphism
√
Z[X]/(X 2 + 5) � Z[i 5]
(2)
√
given by X �→ i 5.
√
√
Show that 2 · 3 �
= (1 + i 5)(1 − i 5) and that the four numbers involved all�
are
irreducible in Z[i 5]. (Hint: Let N (z) = z z̄ and check that an element w ∈ Z[i 5]
is a unit if and only if N (w) = 1.)
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Solutions Ark 1
Is Z[i
MAT4200 — autumn 2011
�
5] a UFD?
√
√
√
Show
that
a
=
(2,
1+i
5)
⊂
Z[i
5]
is
a
maximal
ideal.
(Hint:
Check
that
Z[i
5]/(2, 1+
√
i 5) � Z[X]/(2, 1 + x) � F2
�
Vis at a2 = (2). Hence (2) = a2 .
√
Solution: It is easy to check that Z[i 5] is a subring og the complex field C — it is
closed under multiplication and addition.
√
√
There is a ring homomorphism
Z[X]
→
Z[i
5]
sending
P
(X)
to
P
(i
5). It is
√
obviously surjective. If P (i 5) = 0 it follows from the fundamental theorem of algebra
that P (X) = (X 2 + 5)(a0 X n + a1 X n−1 + · · · + an ) where ai ∈ R. Multiplying out, on
gets ai+2 + 5ai ∈ Z which gives a√
on i.
i ∈ Z by induction
√
It is trivial that 2 · 3 = √
(1 + i 5)(1 − i 5)
We check that w ∈ Z[i 5] is a unit if and only if N (w) = 1. In that case ww̄=1,
so w̄ is an invers for w. If ww� = 1 it follows that N (w)N (w� ) = 1, and since the norm
always is a non-negative integer, it follows that N (w) = 1.
Now, if wz√
= 2, we get N (w)N (z) = 4 and as a2 + 5b2 = 2 has no solution in Z, no
element in Z[i 5] has norm 2. Hence either w or z is of norm 1 and thus a unit.
√
In a similar way, a2 + 5b2 = 3 has no integral solution, so no element i Z[i 5] has
norm 3. This means
that if zw = 3, either z or w is of norm 1 and hence a unit.√
√
As N (1 ± i 5) = 6 it follows in the same way as for 2 and 3 that 1 ± i 5 is
irreducible.
We have √
just produced a factorisation into irreducibles of an element in two different
ways, so Z[i 5] is not UFD.
√
The invers image of the ideal a under the map Z[X] → Z[i 5] above is (2,
√X +
1, X 2 + 5) = (2, X + 1) since X 2 + 5 = (X + 1)(X − 1) + 3 · 2. Hence Z[i 5] ≈
Z[X]/(2, X + 1, X 2 + 5) =√Z[X]/(2, X + 1)√≈ Z/(2)Z √
= F2 which is a field.
2
2
√We have a =
√ (2, 1 + i 5) = (4, 2 + 2i 5, −4 + 2i 5) = (2) since 2 = −4 − (−4 +
2i 5) + (2 + 2i 5).
�
—4—