MTH 461 University of Miami Modern Algebra Questions

MTH 461: Survey of Modern Algebra, Spring 2021Homework 9
Homework 9
1. Determine which of the following subsets are ideals. For example, Q Ă R means: is Q
an ideal in the ring R? Justify your answer in each case.
(a) 2Z Ă Z
(b) Q Ă R
?
?
(c) ta ` b ´5 ∶ a ` b ” 0 pmod 6qu Ă Zr ´5s
(d) R Ă Rrxs
(e) tf px, yq P Rrx, ys ∶ f p4, 5q “ 0u
?
2. Consider the ideal in the ring Zr ´5s given by
?
?
?
I “ p3, 1 ` ´5q “ t3x ` p1 ` ´5qy ∶ x, y P Zr ´5su
?
?
?
Suppose I is principal and generated by some a`b ´5 P Zr ´5s, i.e. I “ pa`b ´5q.
?
?
(a) Since 3 P I, we have
3
“
pa
`
b
´5qpc
`
d
´5q for some c, d P Z. Find all possi?
bilities for a ` b ´5 based on this relation.
(b) Do the same as in (a) but using 1 `
?
´5 P I instead of 3 P I.
(c) Using your results in (a) and (b) show that I cannot be a principal ideal.
3. Let R be a commutative ring. Consider the ring of polynomials Rrxs. For an element
a P R define the evaluation at a to be the map
φa ∶ Rrxs ÐÑ R,
φa pf pxqq “ f paq
given by plugging a into a polynomial f pxq P Rrxs to get an element f paq P R.
(a) Show that φa is a homomorphism of rings.
(b) Apply the 1st Isomorphism Theorem of rings to this homomorphism.
4. (a) Given ideals I, J Ă Rrx, ys show that the varieties satisfy V pI X Jq “ V pIq Y V pJq.
(b) Let I “ px2 ´y ´ 21 q and J “ px2 `y 2 ´1q. Draw the varieties V pI `Jq and V pI XJq.
1
MTH 461: Survey of Modern Algebra, Spring 2021
Lecture 25
Principal ideal domains
Last lecture we introduced the notion of a principal ideal in a commutative ring. These are
ideals in a commutative ring R that take the form
paq “ aR “ tar ∶ r P Ru
A principal ideal domain (PID) is an integral domain R (a commutative ring such that ab “ 0
implies a “ 0 or b “ 0) such that every ideal in R is principal.
The ring of integers Z is the most basic example of a PID. The ideals in Z are
p0q,
p1q,
p2q,
p3q,
p4q,
…
PIDS are an important class of rings because they occur frequently and the theory of PIDS
is considerably simpler than that of general (commutative) rings.
§ Let R be a non-zero commutative ring.
The following are equivalent:
(i) R is a field.
(ii) The only ideals in R are p0q “ t0u and p1q “ R.
(iii) Every homomorphism φ ∶ R Ñ R1 where R1 ‰ t0u is 1-1.
Proof. We show (i) implies (ii). Assume (i), i.e. R is a division ring. Consider an ideal I Ă R
with I ‰ p0q. Choose a P I, a ‰ 0. Since R is a division ring, a´1 P R. Then 1 “ a´1 a P I
as a´1 P R and a P I. Now for any b P R, we have b “ b1 P I as b P R and 1 P I. Thus (ii) holds.
Next, we show (ii) implies (iii). Assume (ii): the only ideals in R are t0u and R. Consider a
homomorphism φ ∶ R Ñ R1 where R1 ‰ t0u. Then kerpφq Ă R is a proper ideal as 1 R kerpφq.
By our assumption it must be t0u. This is equivalent to φ being 1-1. Thus (iii) holds.
Finally, we show (iii) implies (i). Assume (iii), i.e. every homomorphism φ ∶ R Ñ R1
where R1 ‰ t0u is 1-1. Now take a P R with a ‰ 0. Consider the natural homomorphism
φ ∶ R Ñ R{paq. Suppose R{paq ‰ t0u, so that φ is 1-1. Then kerpφq “ p0q. On the other
hand, kerpφq “ paq. Then paq “ p0q implies a “ 0, a contradiction. Thus R{paq “ t0u,
impying paq “ R “ p1q. In particular, 1 “ ar for some r P R, so a has a multiplicative
inverse. We have shown that every non-zero a P R is invertible, and thus R is a field.
A corollary of this result is the following:
§ If R is a field, then R is a PID.
This holds simply because the only ideals in a field R are the ideals p0q “ t0u and p1q “ R
which are principal ideals.
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MTH 461: Survey of Modern Algebra, Spring 2021
Lecture 25
Thus the fields Q, R, C are all examples of PIDs.
A more exotic example of a PID is the ring of Gaussian integers
Zris “ ta ` bi ∶ a, b P Zu Ă C
We will show this is a PID later.
?
We saw last lecture
that
the
ring
Zr
´3s is not a PID, because we showed that the ideal
?
consisting of a ` b ´3 with a ” b (mod 2) is not principal.
Polynomial rings
Another important example involves the following construction. Let R be any ring. Define
Rrxs “ tpolynomials in x with coefficients in Ru
That is, a typical element in Rrxs is an expression of the form
f pxq “ an xn ` an´1 xn´1 ` ⋯a1 x ` a0
where n is a non-negative integer and a0 , . . . , an are elements of the ring R. The sum
f pxq ` gpxq and product f pxqgpxq of two such polynomials are defined in the usual fashion,
and this makes Rrxs into a ring. If R is commutative, so is Rrxs. Also, if R is an integral
domain, so too is the polynomial ring Rrxs.
§ If R is a field, then Rrxs is a PID.
Before explaining the proof, let’s explore the consequences of this result.
The rings Qrxs, Rrxs, Crxs are all PIDS. This means, for example, that for any ideal I Ă Qrxs
there exists a polynomial f pxq with rational coefficients such that I “ pf pxqq. The polynomial f pxq is unique up to multiplication by an element in Qˆ .
Let p be a prime. Then Zp is a field, so Zp rxs is a PID. For example, consider Z2 rxs,
polynomials with coefficients in Z2 . Some elements in this ring are
0,
1,
x,
1 ` x,
x2 ,
1 ` x2 ,
1 ` x ` x2 ,
…
For any ideal I Ă Z2 rxs we can find some such element f pxq such that I “ pf pxqq.
The condition that R be a field in the result is necessary. For example, consider Zrxs. Of
course Z is not a field, so the result does not apply here. In fact Zrxs is not a PID: the ideal
I “ t2f pxq ` xgpxq ∶ f pxq, gpxq P Zrxsu Ă Zrxs
is not a principal ideal, as you can verify.
2
MTH 461: Survey of Modern Algebra, Spring 2021
Lecture 25
Now let us prove the statement. We will need the following:
§ (Division algorithm for polynomials) Let R
be a field. Consider polynomials
f pxq and gpxq ‰ 0 in Rrxs. Then there exist unique qpxq, rpxq P Rrxs such that
f pxq “ gpxqqpxq ` rpxq
and where either degprpxqq ă degpgpxqq or rpxq “ 0.
ř
Here the degree of f pxq “ ai xi P Rrxs is the largest n such that an ‰ 0. We omit the proof
of the division algorithm and now show that if R is a field then Rrxs is a PID.
Let I Ă Rrxs be a PID. We must show I is principal. If I “ t0u then it is the principal ideal
p0q, so assume I ‰ p0q. Let gpxq P I be non-zero and of minimal possible degree among all
non-zero polynomials in I. Note pgpxqq Ă I. Now consider any other element f pxq P I. Then
the division algorithm gives us qpxq, rpxq P Rrxs satisyfing
f pxq “ gpxqqpxq ` rpxq
and degprpxqq ă degpgpxqq or rpxq “ 0. Suppose rpxq ‰ 0. Then
rpxq “ f pxq ´ gpxqqpxq
is in I, because f pxq, gpxq P I and qpxq P Rrxs. Furthermore, degprpxqq ă degpgpxqq. But
this contradicts our assumption that gpxq has the minimal possible degree among non-zero
polynomials in I. So we must have rpxq “ 0. Then
f pxq “ qpxqqpxq
This shows f pxq P pgpxqq. Thus I “ pgpxqq, and I is a principal ideal. This completes the
proof that all ideals in Rrxs are principal.
Finally, we remark that if you continue to add variables to your ring, and consider polynomials in several variables, you will not get a PID. For example, consider
Qrxsrys “ Qrx, ys “ tpolynomials in x, y with coefficients in Qu
Then the ideal generated by the polynomials x and y, which is given by I “ txf px, yq `
ygpx, yq ∶ f px, yq, gpx, yq P Qrxsu, is not a principal ideal in Qrx, ys.
3
MTH 461: Survey of Modern Algebra, Spring 2021
Lecture 26
Rings and geometry
In this lecture we continue studying ideals in rings, and also explain the connection of ring
theory to the geometry of algebraic varieties.
Let R be a commutative ring and consider the ring of polynomials Rrx, ys. In fact for the
time being we will take R “ R to be the real numbers, so that we may easily draw pictures.
An (algebraic) variety in R2 is a set V “ V pSq Ă R2 which consists of points px, yq in the
plane R2 that are the common roots of a given collection of polynomials S Ă RrXs:
V pSq “ tpx0 , y0 q P R2 ∶ f px0 , y0 q “ 0 for all f P Su Ă R2
For example, consider the following subsets of Rrx, ys:
S1 “ tx2 ` y 2 ´ 1u
S2 “ tpx ` y ´ 1qpx ´ y ´ 1qu
S3 “ ty 2 ´ x2 px ` 34 qu
S4 “ tx, y ` 1u
The corresponding varieties V pS1 q, V pS2 q, V pS3 q are shown below, while V pS4 q “ tp0, ´1qu:
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Given a subset S Ă R in any ring, we can also associate an ideal to S, called IpSq Ă R: it is
the smallest ideal containing S. Explicitly, we can define this ideal as follows:
n
ÿ
IpSq “ t ri fi ∶ ri P R, fi P Su
i“1
Each of the examples above yield principal ideals because the sets contain one element:
IpS1 q “ px2 ` y 2 ´ 1q
IpS2 q “ ppx ` y ´ 1qpx ´ y ´ 1qq
IpS3 q “ py 2 ´ x2 px ` 43 qq
IpS4 q “ px, y ` 1q “ pxq ` py ` 1q
(See below for the sum of ideals.) We have the following (note the reversal of inclusion!):
§ For subsets S Ă T
in Rrx, ys we have V pT q Ă V pSq.
1
MTH 461: Survey of Modern Algebra, Spring 2021
Lecture 26
Proof. Let px0 , y0 q P V pT q. This means f px0 , y0 q “ 0 for all polynomials f P T . Since S Ă T
it is then certainly true that f px0 , y0 q “ 0 for all f P S. Thus px0 , y0 q P V pSq.
The next observation is that the variety corresponding to a set S Ă Rrx, ys of polynomials
is the same as the variety associated to the ideal IpSq.
§ For S Ă Rrx, ys we have V pSq “ V pIpSqq.
Proof. As S Ă IpSq, the previous result gives us V pIpSqq Ă V pSq.
For the reverse inclusion,
řn suppose px0 , y0 q P V pSq. Then f px0 , y0 q “ 0 for all f P S. Let
f P IpSq. Then f “ i“1 ri fi for some ri P R and fi P Rrx, ys. Then
f px0 , y0 q “
n
ÿ
ri fi px0 , y0 q “
n
ÿ
ri 0 “ 0
i“1
i“1
Thus f px0 , y0 q “ 0 for all f P IpSq, and px0 , y0 q P V pIpSqq. Thus V pSq Ă V pIpSqq.
For example, for the ideals I “ px ` y ´ 1q and I2 “ IpS2 q “ ppx ` y ´ 1qpx ´ y ´ 1qq we
have I2 Ă I and so V pIq Ă V pI2 q. On the left below is the variety V pIq, on the right is V pI2 q.
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Given ideals I, J Ă R in a commutative ring, we define the sum, intersection, and product:
I ` J “ ta ` b ∶ a P I, b P Ju
I X J “ ta ∶ a P I, a P Ju
ř
IJ “ t ai bi ∶ ai P I, bj P Ju
These are again ideals in R, as you can check. If I “ paq and J “ pbq are principal, then
in fact IJ “ paqpbq “ pabq. We now explore the geometric interpretations of these operations.
§ For ideals I, J Ă Rrx, ys we have V pI ` Jq “ V pIq X V pJq.
2
MTH 461: Survey of Modern Algebra, Spring 2021
Lecture 26
Proof. Let px0 , y0 q P V pIq X V pJq. Then f px0 , y0 q “ 0 for all f P I and f P J. Let f P I ` J.
Then f “ f 1 ` f 2 where f 1 P I, f 2 P J. We compute
f px0 , y0 q “ f 1 px0 , y0 q ` f 2 px0 , y0 q “ 0 ` 0 “ 0
and therefore px0 , y0 q P V pI ` Jq, and we conclude V pIq X V pJq Ă V pI ` Jq.
For the reverse inclusion, since I Ă I ` J we have V pI ` Jq Ă V pIq. Similarly, V pI ` Jq Ă
V pJq. Thus V pI ` Jq Ă V pIq X V pJq. All together we obtain the statement.
The conclusion we draw from this is that the sum of ideals corresponds to the intersection
of varieties. For example, consider the ideals I1 “ IpS1 q, I2 “ IpS2 q, I3 “ IpS3 q from earlier.
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Then I1 ` I2 is the ideal which is the sum of the principal ideals px2 ` y 2 ´ 1q and ppx ` y ´
1qpx ´ y ´ 1qq. It is not a principal ideal itself. However the variety V pI1 ` I2 q is very simple:
V pI1 ` I2 q “ V pI1 q X V pI2 q “ tp0, 1q, p0, ´1q, p1, 0qu Ă R2
It consists of the 3 points of intersection between the circle V pI1 q and the two lines V pI2 q.
Similarly, we see that V pI2 ` I3 q is 2 points, and V pI3 ` I1 q is 4 points.
§ For ideals I, J Ă Rrx, ys we have V pI X Jq “ V pIq Y V pJq.
The proof is similar to the previous one. Thus the variety associated to the intersection
of two ideals is the union of the two varieties of each ideal. As a consequence, V pI1 X I2 q,
V pI2 X I3 q, V pI3 X I1 q are the three pictures above (not just the intersection points!).
In fact, the same property holds for the product of ideals!
§ For ideals I, J Ă Rrx, ys we have V pIJq “ V pIq Y V pJq.
Note in general IJ ‰ I X J. For example, consider I “ pxq And J “ pxyq in Rrx, ys. Then
IJ “ px2 yq while I X J “ pxyq “ J. In general, IJ Ă I X J, but not the other way around.
3
MTH 461: Survey of Modern Algebra, Spring 2021
Lecture 26
Thus the above two results give instances of the phenomenon that V pIq “ V pJq for two
ideals I ‰ J. In other words, the map given by
tideals in Rrx, ysu ÐÑ tvarieties in R2 u
I zÑ V pIq
is not 1-1. (From our earlier discussion, it is onto.) In fact it is very far from being 1-1!
Take any polynomial f P Rrx, ys which has no real solutions to f px, yq “ 0. Examples are
f px, yq “ 1 and f px, yq “ x2 ` 1. Then for any such polynomial,
V pf q “ H,
since f px0 , y0 q “ 0 does not hold for any px0 , y0 q P R2 . However the ideals p1q “ Rrx, ys and
p1 ` x2 q are not the same.
To fix this last sort of problem, we can work over a field such as C which has more solutions.
Then for the ideal p1 ` x2 q Ă Crx, ys the variety
V p1 ` x2 q “ tpx0 , y0 q P C2 ∶ 1 ` x20 “ 0u “ tpi, y0 q ∶ y0 P Cu Y tp´i, y0 q ∶ y0 P Cu
is two disjoint copies of C contained inside C2 . Thus it is distinguished from V p1q “ H.
Still, even after replacing R with C, the do not have a 1-1 correspondence between ideals
and varieties. The idea is then to try and shrink our focus on a particular class of ideals
which does make the assignment I ↦ V pIq into a 1-1 correspondence.
§ For any ideal I Ă R in a commutative ring R, the radical
of I is defined to be
?
I “ ta P R ∶ an P I for some positive n P Zu
?
?
It is instructive to verify that I Ă R is an ideal. A radical ideal is an ideal of the form I.
a
?
For example, consider I “ px2 q Ă Rrx, ys. Then I “ pxq.
On
the
other
hand,
pxq “ pxq.
?
?
A key property is the following: for ideals I, J we have IJ “ I X J.
§ The following assignment is a 1-1 and onto correspondence:
tradical ideals in Crx, ysu ÐÑ tvarieties in C2 u
I zÑ V pIq
This sets up a “dictionary” between certain ideals, which are algebraic objects, and varieties, which are geometric objects. The dictionary converts sums of ideals into intersections
of varieties, and intersections of ideals into unions of varieties.
This lecture is essentially an introduction to the subject of Algebraic Geometry, which is the
study of algebraic varieties in a much broader setting.
4
MTH 461: Survey of Modern Algebra, Spring 2021
Lecture 24
Principal ideals
In this lecture we continue studying ideals, focusing on the new concept of a principal ideal.
First we introduce some notation. Let R be a commutative ring and a P R. We write
aR “ tar ∶ r P Ru Ă R
Then aR is an ideal in R. To check this, note 0 “ a0 P paq; if ar, ar1 P paq then ar ´ ar1 “
apr ´ r1 q P aR; and if ar P aR and r1 P R then r1 parq “ aprr1 q P aR. The ideal aR Ă R is
called the principal ideal generated by a P R. Another common notation for aR is paq.
Example: The ideals pnq “ nZ Ă Z are principal ideals.
Example: Let R be any commutative ring. Then 0R “ p0q “ t0u is the zero ideal, and
1R “ p1q “ R is the ideal which is the whole ring. These are both principal ideals.
§ Let R be a commutative ring.
(i) If I Ă R is an ideal and a P I then paq Ă I.
(ii) If a “ ub for u a unit, then paq “ pbq.
These are straightforward from the definitions. For example, let us prove (ii). If a “ ub
where u is a unit, then if ar P aR we have ar “ pubqr “ bpurq P bR, so paq Ă pbq. And if
br P pbq then br “ pu´1 aqr “ apu´1 rq P paq, so pbq Ă paq. We conclude paq “ pbq.
?
Example: Consider the ring R “ Zr ´3s. Last lecture we defined a homomomorphism
φ ∶ R ÐÑ Z4
?
φpa ` b ´3q “ a ` b pmod 4q.
?
?
´3q
“
1
´
1
“
0
pmod
4q,
so
1
´
´3 P kerpφq.
Consider the ideal kerpφq. Note that φp1
´
?
We obtain an inclusion of ideals p1 ´ ´3q Ă kerpφq.
?
?
this
question,
let
a
`
b
Is it also true that kerpφq Ă p1 ´ ´3q? To rephrase
?
? ´3 P kerpφq,
i.e. a ` b ” 0 pmod 4q. Then, can we write a ` b ´3 as a multiple of 1 ´ ´3, i.e
?
?
?
a ` b ´3 “ p1 ´ ´3qpc ` d ´3q
for some c, d P Z? Note that the right hand side of this last equation becomes
?
pc ` 3dq ` pd ´ cq ´3
and so we must have a “ c ` 3d and b “ d ´ c. Solving for c, d we get c “ pa ´ 3bq{4,
d “ pa ` bq{4 which are
? integers because a ` b is a multiple of 4. Thus the answer is “yes”,
so that kerpφq Ă p1 ´ ´3q. Consequently we have
?
kerpφq “ p1 ´ ´3q
1
MTH 461: Survey of Modern Algebra, Spring 2021
Thus the kernel of this homomorphism is a principal ideal, generated by 1 ´
Lecture 24
?
´3.
The following gives an example of an ideal which is not principal.
Example (non-principal ideal): We modify our homomorphism from above: define
?
ψ ∶ R “ Zr ´3s ÐÑ Z2
?
ψpa ` b ´3q “ a ` b pmod 2q.
?
Note again 1 ´ ´3 P kerpψq. We will argue that kerpψq is not principal.
Suppose for a
?
contradiction that kerpψq is principal, i.e. there is some for some a ` b ´3 P R such that
?
kerpψq “ pa ` b ´3q
?
In other words, kerpψq is?the principal ideal generated by a ` b ´3. Necessarily a ` b ” 0
pmod 2q. Now since 1 ´ ´3 P kerpψq, we must have
?
?
?
1 ´ ´3 “ pa ` b ´3qpc ` d ´3q
?
for some c, d P Z. Note that since ´3 appears on the left side of this equation we can
assume b and d are not both zero, for if they are both zero then the right hand side is a real
number. Take the squared norms of both sides to obtain
?
?
?
4 “ |1 ´ ´3|2 “ |a ` b ´3|2 |c ` d ´3|2 “ pa2 ` 3b2 qpc2 ` 3d2 q
Noting that a, b, c, d P Z, a ` b ” 0 pmod 2q, and one of b, d is non-zero,
we must
?
? have a,?b, c P
t1, ´1u and d “ 0. The only choices which solve the equation 1´ ´3 “ pa`b ´3qpc`d ´3q
are a “ c “ 1, b “ ´1, d “ 0 and a “ c “ ´1, b “ 1, d “ 0. We conclude
?
?
a ` b ´3 “ ˘p1 ´ ´3q
Thus under our current assumption that kerpψq is principal, we obtain
?
kerpψq “ p1 ´ ´3q “ kerpφq
However, note that ψp2q “ 0 pmod 2q so that 2 P kerpψq, while φp2q “ 2 ı 0 pmod 4q, so
2 R kerpφq. We conclude that kerpψq ‰ kerpφq. We have a contradiction. Thus kerpψq cannot
be a not principal ideal.
2
MTH 461: Survey of Modern Algebra, Spring 2021
Lecture 27
Prime and maximal ideals
An ideal I in a commutative ring R is prime if I ‰ R and ab P I implies that either a P I or
b P I. The ideal I Ă R is maximal if I ‰ R, and I Ă J for some ideal J ‰ I implies J “ R.
§ A maximal ideal is a prime ideal.
Proof. Let I be a maximal ideal, and suppose it is not prime: let ab P I such that a R I and
b R I. Consider the ideals paq ` I and pbq ` I. These both contain I and are not equal to I.
Since I is maximal, it follows that paq ` I “ R and pbq ` I “ R. The product ideal of paq ` I
and pbq ` I is then on the one hand equal to p1q “ R and on the other hand contained in I,
implying R “ I, a contradiction.
The following characterizes prime and maximal ideals in terms of the quotient ring.
§ Let R be a commutative ring and I Ă R a proper ideal.
Then
(i) I is prime if and only if the ring R{I is an integral domain.
(ii) I is maximal if and only if the ring R{I is a field.
Proof. (i) Suppose I is prime. Suppose ab ` I “ pa ` Iqpb ` Iq “ I in R{I. Then ab P I.
Since I is prime, one of a or b is in I. Then one of a ` I or b ` I is equal to I. This proves
R{I is an integral domain. Conversely, suppose R{I is an integral domain. Now let ab P I.
Then ab ` I “ pa ` Iqpb ` Iq “ I, and since R{I is an integral domain, one of a ` I or b ` I
is equal to I. This implies one of a or b is in I. Thus I is prime.
(ii) Suppose I is maximal. We claim R{I is a field. This is equivalent to R{I having only
the ideals tIu, R{I. Suppose R{I has a non-zero proper ideal J 1 . Then J “ ta P R ∶ a ` I P
J 1 u Ă R is an ideal of R. To verify this: if a ` I, b ` I P J 1 then
pa ` Iq ` pb ` Iq “ pa ` bq ` I
is in J 1 because a ` b P J, using that J is an ideal. Similarly, if a P J and b P R then
pa ` Iqpb ` Iq “ ab ` I is in J 1 because ab P J. Next, since I is maximal and J contains I,
we must have either J “ I or J “ R. If J “ I then J 1 is zero, a contradiction. If J “ R then
J 1 “ R{I, contradicting our assumption that J 1 is proper.
Conversely, suppose R{I is a field. Suppose I is not maximal, and let I Ă J with I ‰ J
and J ‰ R. Then consider J 1 “ ta ` I ∶ a P Ju Ă R{I. This is an ideal of R{I. Since R{I
is a field, it must be either zero or R{I. If J 1 is zero then a ` I “ I for all a P J i.e. a P I
for all a P J; but this implies I Ă J, a contradiction. Thus J 1 “ R{I. Then for all a P R
we have a ` I “ b ` I for some b P J, i.e. a ` c “ b ` d for some c, d P I and b P J. Then
a “ b ` d ´ c P J. We have shown J “ R, a contradiction.
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MTH 461: Survey of Modern Algebra, Spring 2021
Lecture 27
Examples
1. Consider an ideal pnq Ă Z. Suppose it is a prime ideal. This means ab P pnq implies one
of a or b is in pnq. Note ab P pnq if and only if ab “ nk for some k P Z, i.e. n divides ab. So
pnq is prime if and only if “n divides ab” implies “n divides a or b”. For this to be true n
must be a prime number. Thus the prime ideals of Z are
p2q,
p3q,
p5q,
p7q,
p11q,
…
We recover that Zn is an integral domain if and only if n is prime.
We also know from earlier lectures that Zn is a field if and only if n is prime. This implies
that the prime ideals in Z are also the maximal ideals.
2. Consider the ring Crx, ys. The ideal I “ px2 ` y 2 ´ 1q is prime, but not maximal. To see
it is prime, suppose f px, yqgpx, yq “ px2 ` y 2 ´ 1qhpx, yq. Then since x2 ` y 2 ´ 1 is irreducible
(cannot be factored over Crx, ys) it must divide one of f px, yq or gpx, yq. To see that it is
not a maximal ideal, note that we have an inclusion
px2 ` y 2 ´ 1q Ă px ´ 1, yq
because x2 ` y 2 ´ 1 “ px ` 1qpx ´ 1q ` pyqpyq. This inclusion is proper since y R I and also
px ´ 1, yq ‰ Crx, ys. Thus I is not maximal. However, px ´ 1, yq is maximal.
3. In the ring Zrxs, the ideal p2q is prime but not maximal. To see this, consider
φ ∶ Zrxs ÐÑ Z2 rxs
defined by φpf pxqq “ f pxq pmod 2q, i.e. take the coefficients mod 2. Then φ is a homomorphism and the kernel is the principal ideal p2q Ă Zrxs. By the 1st Isomorphism Theorem,
Zrxs{kerpφq “ Zrxs{p2q – Z2 rxs
Now Z2 rxs is an integral domain, but not a field (e.g. x is not invertible). Thus p2q is a
prime ideal that is not maximal. On the other hand, we may consider the homomorphism
ψ ∶ Zrxs ÐÑ Z2
defined by ψpf pxqq “ f p0q pmod 2q. Then you can check that the kernel is the ideal
p2, xq “ p2q ` pxq. Since Z2 is a field, this ideal is maximal.
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