Topics in Algebra, Chapter 3.4

2012-11-21 math algebra topics-in-algebra

There are no exercises from 3.3, so this section covers both 3.3 (“Homomorphisms”) and 3.4 (“Ideals and Quotient Rings”).

Throughout, R is a ring and p is a prime integer.

Topics covered: 3.3

  • Definition: Let R,R be rings. A ring homomorphism is a mapping ϕ:RR such that ϕ(a+b)=ϕ(a)+ϕ(b) and ϕ(ab)=ϕ(a)ϕ(b) for all a,bR.

  • Lemma 3.3.1: If ϕ:RR is a ring homomorphism, then ϕ(0)=0 and ϕ(a)=ϕ(a).

  • Definition: With ϕ:RR a ring homomorphism, the kernel of ϕ is the set {aRϕ(a)=0}.

  • Lemma 3.3.2: With ϕ:RR a ring homomorphism, kernel of ϕ is a subgroup of R under addition and, if akerϕ and rR, then both ar and ra are also in the kernel.

  • The zero map is a ring homomorphism whose kernel is the entire domain.

  • The identity map is a ring homomorphism with a trivial kernel.

  • The set {m+n2m,nZ} is a ring under standard operations, and conjugation, ϕ:m+n2mn2, is a ring homomorphism with trivial kernel.

  • The natural map ϕ:ZZ/nZ is a ring homomorphism whose kernel consists of the multiples of n.

  • Definition: A ring isomorphism is a bijective ring homomorphism.

  • Lemma 3.3.3: A ring homomorphism is an isomorphism if and only if its kernel is trivial.

Topics covered: 3.4

  • Ideals are motivated in analogy to normal subgroups.

  • Definition: An ideal of R is a subset I of R such that I is a subgroup of R under addition and, for any rR, both ur and ru are in I.

  • The kernel of a ring homomorphism is an ideal. In fact, the definition of an ideal is modeled around kernels of homomorphisms.

  • Given an ideal IR, we consider an equivalence relation ab if abI. The cosets are of the form r+I={r+iiI} for elements rR. The set of cosets is denoted R/I, again in complete analogy to normal subgroups. In fact, R/I is a ring, and it is called a quotient ring.

  • If R is unital, then R/I is unital.

  • Lemma 3.4.1: If I is an ideal of R, then ϕ:RR/I given by ϕ(r)=r+I is a ring homomorphism.

  • Theorem 3.4.1: Let ϕ:RR be a ring homomorphism with kernel I=kerϕ. Then RR/I. There is also a bijection between the set of ideals of R and the set of ideals of R containing I: if J is an ideal of R the mapping is the preimage Jϕ1(J), and R/ϕ1(J)R/J.

The problems below are paraphrased from/inspired by those given in Topics in Algebra by Herstein. The solutions are my own unless otherwise noted. I will generally try, in my solutions, to stick to the development in the text. This means that problems will not be solved using ideas and theorems presented further on in the book.

Herstein 3.4.1: Let IR be an ideal such that 1I. Prove that I=R.

For any rR, we have that r1=rI so that RI and hence I=R.

Herstein 3.4.2: Prove that the only ideals of a field F are trivial, {0} and F itself.

Let IF be an ideal. If there exists a non-zero element rI, then r1r=1I so that I=F by problem 3.4.1. Otherwise, there is only the zero element, so I={0} is the zero ideal.

Herstein 3.4.3: Prove that any ring homomorphism of a field is either the zero map or an isomorphism.

The kernel of such a homomorphism is an ideal of the field, and therefore it can only be trivial or the entire field, by problem 3.4.2. If the kernel is the entire field, then the homomorphism is the zero map. If the kernel is trivial, then the map is an isomorphism by Lemma 3.3.3.

Herstein 3.4.4: Let R be commutative.

(a) Show that aR={arrR} is an ideal of R.

(b) Show that aR is not necessarily an ideal if R is not commutative.

(a) For aR to be an ideal, it must be an additive subgroup of the ring and also closed under multiplication on the left or right by arbitrary ring elements. We have ar+as=a(r+s)aR for any r,sR, so aR forms a subgroup under addition by Lemma 2.4.2. Furthermore, (ar)s=a(rs)aR and s(ar)=a(sr)aR, the latter because R is commutative. Thus, aR is an ideal.

(b) When R is not commutative, the above reasoning will fail where we relied on commutativity, i.e. when we want s(ar)aR for arbitrary s. We will look for an example in one of the simplest non-commutative rings, the 2×2 matrices over R. Put a=(1000)

from which we easily see that the generic element in aR is (1000)(xyzw)=(xy00),

a matrix whose bottom row is zero. However, consider left multiplication by s=(1111).

We find, for instance, saI=(1111)(1000)(1001)=(1010)aR.

Hence aR is not a two-sided ideal. It is, of course, a one-sided ideal.

Herstein 3.4.5: Let I and J be ideals of R. Define I+J={i+jiI,jJ}. Prove that I+J is an ideal of R.

Let i,iI and j,jJ and consider (i+j)+(i+j)=(i+i)+(j+j). Because both I and J are closed under addition, this is again an element of I+J. Also, for any rR, we have r(i+j)=(ri)+(rj)I+J and (i+j)r=(ir)+(jr)I+J. Therefore, I+J is an ideal of R.

Herstein 3.4.6: Let I and J be ideals of R. Define IJ to be the set of all ring elements formed as finite sums of the form ij with iI and jJ. Prove that IJ is an ideal of R.

That IJ is an additive subgroup is clear. Let rR and consider r(i1j1++injn)=(ri1)j1++(rin)jn.

Because I is closed under such multiplications, this is again an element of IJ. Multiplication from the right also remains within IJ because J is closed under such multiplications. Therefore, IJ is an ideal of R.

Herstein 3.4.7: Let I and J be ideals of R. Show that IJIJ.

Consider i1j1++injnI+J. Because I is an ideal, ikjk=ikI for each k{1,,n} and therefore this element belongs to I. In precisely the same way, ikjk=jkJ by virtue of J being an ideal. Then, again, the element belongs to J, and so IJIJ.

Herstein 3.4.8: Let R=Z and let I be the ideal of R made up of all multiples of 17. Prove that any ideal J of R, with IJR, must be either J=I or J=R. How does this generalize?

The set I is an ideal because 17m+17n=17(m+n)I and (17m)n=17(mn)I for all m,nZ. Suppose that the ideal J contains some integer a which is not a multiple of 17. Then a is coprime to 17, so there exist α,βZ such that aα+17β=1 and, by the closure properties of ideals, J must contain it. Therefore, if I is a proper subset of J then 1J and J=R. Otherwise, J=I.

Of course, this argument holds with any prime integer in place of 17. In the integers, a prime ideal is maximal.

Herstein 3.4.9: Let I be an ideal of R and let r(I)={xRxi=0 for each iI}. Prove that r(I) is an ideal of R.

Suppose x,yr(I) and let iI be arbitrary. Then (x+y)i=xi+yi=0 so that x+yr(I). Also, for rR, (rx)i=r(xi)=0 and (xr)i=x(ri)=0 because riI so x annihilates it just as well. Therefore r(I) is an ideal of R.

Herstein 3.4.10: Let I be an ideal of R and let [R:I]={xRrxI for each rR}. Prove that [R:I] is an ideal of R and that I[R:I].

The second statement follows directly from the fact that I is an ideal, so, for any iI and rR, riI which means that i[R:I] and hence I[R:I].

Let x,y[R:I] and rR. Then r(x+y)=rx+ryI since each of rx and ry are elements of I, which is an ideal. Therefore [R:I] is a subgroup of R under addition by Lemma 2.4.2. In addition, with sR, r(sx)=(rs)xI and r(xs)=isI where i=rxI by assumption. Then [R:I] is an ideal of R.

Herstein 3.4.11: Let R be unital and define a new ring S with the same elements as R but with an addition ab=a+b+1 and a multiplication ab=ab+a+b for elements a,bR.

(a) Prove that S is a ring with the operations and .

(b) What is the zero element of S?

(c) What is the multiplicative identity element in S?

(d) Show that RS.

(a) That S is closed under and follows from the fact that R is a ring. Let a,b,cS. The new addition is commutative: ab=a+b+1=b+a+1=ba. It is associative: (ab)c=(a+b+1)c=a+b+c+1=a+(b+c+1)=a(b+c).

The multiplication is associative: (ab)c=(ab+a+b)c=(ab+a+b)c+ab+a+b+c=a(bc+b+c)+a+(bc+b+c)=a(bc).\begin{aligned} (a\cdot b)\cdot c&=(ab+a+b)\cdot c=(ab+a+b)c+ab+a+b+c\
&=a(bc+b+c)+a+(bc+b+c)=a\cdot(b\cdot c). \end{aligned}

The distributive law also holds, a(bc)=a(b+c+1)=a(b+c+1)+a+b+c+1=(ab+a+b)+(ac+a+c)+1=(ab)(ac).\begin{aligned} a\cdot(b\oplus c)&=a\cdot(b+c+1)=a(b+c+1)+a+b+c+1\
&=(ab+a+b)+(ac+a+c)+1=(a\cdot b)\oplus(a\cdot c). \end{aligned}

In the next section, we will demonstrate the existence of a zero element and additive inverses.

(b) The zero element of S is 0S such that a0S=a for all aS. Then we must take 0S=1 (i.e. the additive inverse in R of the unit element of R). The additive inverse α of aS is such that aα=0S=1. Then a+α+1=1 so that α=a2. This α is, of course, in S, so we have shown that S is a ring with these new operations.

(c) The multiplicative identity element 1S is such that a1S=a for all aS. This is satisfied by 1S=0, the zero element of R.

(d) We can explicitly construct a homomorphism ϕ:RS. We must have that ϕ(a)=ϕ(a+0)=ϕ(a)ϕ(0)=ϕ(a)+ϕ(0)+1

for all aR. This forces ϕ(0)=1 and we may as well try the map ϕ(a)=a1. In fact, with a,bR, ϕ(a+b)=a+b1=(a1)+(b1)+1=ϕ(a)ϕ(b).

Also, ϕ(ab)=ab1=(a1)(b1)+(a1)+(b1)=ϕ(a)ϕ(b).

Therefore this map is truly a homomorphism. Its kernel is trivial, containing just the identity element 1R, so it is also an isomorphism.

Herstein 3.4.12*: Let R=Mat2×2(Q), the ring of 2×2 rational matrices. Prove that R has no ideals besides the zero ideal and R itself.

Let I be an ideal of R. If there exists an invertible matrix aI, then a1aI, so that I=R. Therefore consider an ideal which consists of only non-invertible matrices. Suppose there is a non-zero matrix aI. The gist of the argument from here is that we can take a and exhibit an invertible matrix in I using the ideal’s closure under addition and external multiplication.

First note that we can make the following sorts of manipulations: (αβγδ)(1001)=(αβγδ),

so (by taking the sum of the original matrix and the one we’ve produced) we can always isolate a column of a matrix and find it to still be in I. In completely analogous ways, we will always be able to isolate an element of a matrix.

By assumption, we have a matrix with at least one non-zero entry. Suppose we follow the above prescription to produce a matrix in I with just one non-zero entry in isolation. For the sake of demonstration, assume that this entry is in the top-left corner (nearly identical methods work in the other three cases). Then we have (α000)+(0010)(α000)(0100)=α(1001)I.

This is the desired invertible matrix which is in I. Therefore if I is to not be the entire ring, then it can contain nothing but the zero matrix.

Herstein 3.4.13*: Consider the quaternions over Z/pZ, with p an odd prime, namely


where i2=j2=k2=ijk=1.

(a) Prove that R is a ring with p4 elements with no non-trivial ideals.

(b)** Prove that R is not a division ring.

(a) Compute a generic product of elements in R: (α0+α1i+α2j+α3k)(β0+β1i+β2j+β3k)=(α0β0α1β1α2β2α3β3)+(α0β1+α1β0+α2β3α3β2)i+(α0β2+α2β0+α3β1α1β3)j+(α0β3+α3β0+α1β2α2β1)k.\begin{aligned} (\alpha_0+\alpha_1 i+\alpha_2 j+\alpha_3 k)&(\beta_0+\beta_1 i+\beta_2 j+\beta_3 k)\
&+(\alpha_0\beta_2+\alpha_2\beta_0+\alpha_3\beta_1-\alpha_1\beta_3)j+(\alpha_0\beta_3+\alpha_3\beta_0+\alpha_1\beta_2-\alpha_2\beta_1)k. \end{aligned}

This is again an element of R, as is a generic sum. Thus R forms a ring, and we can easily exhibit p4 elements, with p choices for each of the four coefficients. There’s no way to break out of the ring using multiplication or addition, so there are at most p4 elements. Conceivably there could be some elements that get duplicated, but we see that if α0+α1i+α2j+α3k=β0+β1i+β2j+β3k,

then (α0β0)+(α1β1)i+(α2β2)j+(α3β3)k=0+0i+0j+0k

which forces α0=β0, etc., so there is no duplication and there are truly p4 elements.

Again we would like to consider an ideal I of R and show that the existence of any non-zero element in the ideal necessarily leads to the conclusion that I=R. Let a=α0+α1i+α2j+α3k be a non-zero element, presumably with no inverse. We can play around with a, computing things such as ia, aj, and so forth. Eventually, we hit upon iai=α0α1i+α2j+α3k

which almost reminds us of complex conjugation. Of course, jaj and kak look similar. In fact, we prod a little more and find that the combination aiaijajkak=4α0I.

Now, supposing α00, this is the desired invertible element belonging to I (thankfully, we have assumed that our prime p is not 2). On the other hand, if α0=0, we can rotate a non-zero coefficient into its place by multiplying appropriately by i, j or k, and then performing the same trick. To sum up, we have shown a way to take any non-zero element of the ideal I and manipulate it into an invertible element which also belongs to I, thus proving that a non-zero ideal is actually the entire ring.

(b) To show that R is not a division ring, we must prove that there exists an element with no multiplicative inverse. Consider conjugation, aaˉ=(α0+α1i+α2j+α3k)(α0α1iα2jα3k)=α12+α22+α32+α42.

This conjugation evidently offers a path to take a general quaternion and map it to an element of the base field. In particular, if we could find a non-zero element a such that aaˉ were zero modulo p, then we could show that a is not invertible in the ring. Suppose to the contrary that there was an element a1 in the ring such that a1a=1. Then we would have 0=a1aaˉ=aˉ which is a contradiction.

Therefore we seek an element aR with aaˉ=0 (in other words, a such that the sum of the squares of its coefficients is zero modulo p), and if we can produce it then we have shown that R is not a division ring. As an example, with p=3, such a non-invertible element is a=1+i+j. Can we prove the existence in general? There is a powerful theorem due to Lagrange which says that any positive integer is expressible as the sum of four integer squares. While that surely solves this problem, our statement is considerably weaker, so we search for a more direct way of proving it.

As suggested in this post, we narrow our attention to solutions with one coefficient fixed as 0 and one fixed as 1. By the lemma immediately below, we have the existence of α0 and α1 such that α02+α12+10modp. Hence, for any odd prime p, there will always exist a non-invertible element of the form α0+α1i+j, proving that R is not a division ring.

Lemma: For prime p, there exist integers x,y{0,,p1} such that


The case p=2 is immediately satisfied by x=1, y=0. Therefore, let p be an odd prime.

First consider the case of two distinct integers m,n{0,,p1} whose squares are to be congruent modulo p. We have m2n2modp or (m+n)(mn)0modp. As p is a prime, it must divide one of the factors. This isn’t possible for mn because of the restricted domain for m and n. Hence, for the two squares to be congruent, we must have m+n=p. The set {0,,p1} therefore pairs off into equivalence classes {1,p1}, {2,p2}, etc., with zero the odd man out, and there are exactly 1+(p1)/2=(p+1)/2 distinct values for m2 modulo p.

Now we would like to show the existence of x,y{0,,p1} such that x21y2modp. There are evidently (p+1)/2 possible values for the left hand side, and (p+1)/2 possible values for the right hand side, and both quantities reside in the interval {0,,p1} which accomodates a total of p numbers. Thus there is no way to fit the values of x2 modulo p alongside the values of 1y2 modulo p, a total of p+1 numbers, without some overlap (this is the pigeonhole principle). This overlap is the desired solution to the congruence x2+y2+10modp, so we are done.

Definition: A subset L of R is a left [right] ideal if L i a subgroup of R under addition and if rR, aL implies raL [arL].

Herstein 3.4.14: With aR, let Ra={xaxR}. Show that Ra is a left ideal of R.

Let x,yR. Then xa+ya=(x+y)aRa so that Ra is an additive subgroup of R by Lemma 2.4.2. Also, if we multiply on the left by arbitrary rR, we find r(xa)=(rx)aRa because rxR. Therefore Ra is a left ideal. See also H 3.4.4.

Herstein 3.4.15: Let L,MR be left ideals. Show that LM is a left ideal of R.

L and M are each additive subgroups of R so their intersection is again an additive subgroup of R. If xLM and rR, then rxL and rxM, hence rxLM. Therefore, LM is a left ideal of R.

Herstein 3.4.16: Let L be a left ideal of R and let M be a right ideal of R. What can be said of LM?

If x,yLM, then x+yL and x+yM, so x+yLM. Thus LM is an additive subgroup of R. Letting rR, we have that rxL and xrM, but there is no guarantee that either of those will be shared between the ideals, so we cannot say more in general.

To illustrate this, recall the example presented in problem 3.4.4b, with R the set of 2×2 rational matrices, a=(1000),

L=Ra and M=aR. We can easily see that L consists of the matrices whose second column is zero, and M consists of the matrices whose second row is zero. Their intersection LM is the set of matrices with the top-left corner free to vary but all other entries fixed as zero. This is not a left ideal: (0110)(x000)=(00x0)LM.

We can see it’s also not a right ideal by taking the transpose of the above equation.

Herstein 3.4.17: With aR, let r(a)={xRax=0}. Show that r(a) is a right ideal of R.

r(a) is the set of elements which a annihilates from the left. If x,yr(a), then a(x+y)=ax+ay=0 so that x+yr(a) and r(a) is an additive subgroup of R. With rR, we also have a(xr)=(ax)r=0 so that xrr(a). Thus r(a) is a right ideal of R.

Herstein 3.4.18: Let L be a left ideal of R and let λ(L)={xRxa=0 for each aL}. Show that λ(L) is an ideal (two-sided) of R.

λ(L) is the set of elements which annihilate the entire left ideal L. If x,yλ(L) and aL, then (x+y)a=xa+ya=0 so that λ(L) is an additive subgroup of R. Let rR and consider (rx)a=r(xa)=0, which shows λ(L) to be a left ideal, and (xr)a=x(ra)=xa=0, which shows λ(L) to be a right ideal, with a=raL because L is a left ideal. Therefore λ(L) is a two-sided ideal.

Herstein 3.4.19*: Suppose x3=x for every xR. Prove that R is commutative.

This problem is pretty tricky, and I must give credit to this page for filling in some gaps in my argument that I couldn’t fill myself even after considerable effort.

First observe that 2x=(2x)3=8x3=8x so that 6x=0 for every xR. This is not a statement about characteristic, because we do not assume R to be an integral domain. We do not even assume that R is unital, so recall that 6x is simply shorthand for x+x+x+x+x+x.

Consider x+x2=(x+x2)3=x3+3x4+3x5+x6=4x+4x2,

so that we have 3(x+x2)=0 for any xR. In particular, plugging in x+y gives 0=3(x+y+(x+y)2)=3(x+y+x2+xy+yx+y2)=3(x+x2)+3(y+y2)+3(xy+yx)=3(xy+yx).\begin{aligned} 0=3(x+y+(x+y)^2)&=3(x+y+x^2+xy+yx+y^2)\
&=3(x+x^2)+3(y+y^2)+3(xy+yx)=3(xy+yx). \end{aligned}

By the above comment, we can subtract 6yx=0 from the equation for free, and this leaves us with 3(xyyx)=0.

The simplest example of a ring with x3=x for all xR is R=Z/3Z which has characteristic 3, and we immediately see that this result is worthless in that case. Clearly we need an additional result to prove that R is commutative even when we can’t “cancel” the three.

A natural thing to compute is x±y=(x±y)3=x±y±x2y±yx2+xy2+y2x±xyx+yxy.

Subtracting the second () equation from the first (+) gives 2x2y+2yx2+xyx=0. If we multiply that equation on the left by x we find 2xy+2xyx2+2x2yx=0,

whereas if we multiply by x on the right instead, 2x2yx+2yx+2xyx2=0.

Subtracting one equation from the other again, we see that 2(xyyx)=0.

Finally, subtracting that from the previous result, we arrive at the desired statement that xy=yx for all x,yR, i.e. that R is commutative.

With a solution in hand, one still has to wonder why this problem is in section 3.4, ideals and quotient rings. It is worth noting that ϕ:RR given by ϕ(x)=3x is a ring homomorphism, in a not-entirely-trivial way. This may lead somewhere useful.

Herstein 3.4.20: Let R and S be rings, with R unital, and let ϕ:RS be a surjective ring homomorphism. Prove that ϕ(1) is the unit element of S.

For any sS, there exists rR such that s=ϕ(r), because ϕ is surjective. Then we also must have that s=ϕ(1r)=ϕ(1)ϕ(r)=ϕ(1)s

and that s=ϕ(r1)=ϕ(r)ϕ(1)=sϕ(1)

so that ϕ(1) is the unit element of S.

Note that ϕ(1) will be the unit element of the sub-ring ϕ(R)S regardless, but if ϕ is not surjective then there may be elements of S for which ϕ(1) does not act as a unit.

Herstein 3.4.21: Let R and S be rings with R unital and S an integral domain. Let ϕ:RS be homomorphism such that kerϕR. Prove that ϕ(1) is the unit element of S.

Clearly there is a problem if 1kerϕ, so first let rR be outside the kernel of ϕ and consider 0ϕ(r)=ϕ(1r)=ϕ(1)ϕ(r).

This shows that 1kerϕ. Now we also see from the above computation that ϕ(1) acts as a unit element for ϕ(R) (which is commutative by the definition of integral domain), and the question is whether it also acts as a unit element for the other elements of S.

Again letting rkerϕ and letting sS be arbitrary, we have that ϕ(r)s=ϕ(r)ϕ(1)s so that ϕ(r)(sϕ(1)s)=0.

By assumption, ϕ(r)0, so, because S is an integral domain, we conclude that s=ϕ(1)s for all sS.

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