There are no exercises from 3.3, so this section covers both 3.3 (“Homomorphisms”) and 3.4 (“Ideals and Quotient Rings”).
Throughout, is a ring and is a prime integer.
Definition: Let be rings. A ring homomorphism is a mapping such that and for all .
Lemma 3.3.1: If is a ring homomorphism, then and .
Definition: With a ring homomorphism, the kernel of is the set .
Lemma 3.3.2: With a ring homomorphism, kernel of is a subgroup of under addition and, if and , then both and 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 is a ring under standard operations, and conjugation, , is a ring homomorphism with trivial kernel.
The natural map is a ring homomorphism whose kernel consists of the multiples of .
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.
Ideals are motivated in analogy to normal subgroups.
Definition: An ideal of is a subset of such that is a subgroup of under addition and, for any , both and are in .
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 , we consider an equivalence relation if . The cosets are of the form for elements . The set of cosets is denoted , again in complete analogy to normal subgroups. In fact, is a ring, and it is called a quotient ring.
If is unital, then is unital.
Lemma 3.4.1: If is an ideal of , then given by is a ring homomorphism.
Theorem 3.4.1: Let be a ring homomorphism with kernel . Then . There is also a bijection between the set of ideals of and the set of ideals of containing : if is an ideal of the mapping is the preimage , and .
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.
For any , we have that so that and hence .
Let be an ideal. If there exists a non-zero element , then so that by problem 3.4.1. Otherwise, there is only the zero element, so is the zero ideal.
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.
(a) For 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 for any , so forms a subgroup under addition by Lemma 2.4.2. Furthermore, and , the latter because is commutative. Thus, is an ideal.
(b) When is not commutative, the above reasoning will fail where we relied on commutativity, i.e. when we want for arbitrary . We will look for an example in one of the simplest non-commutative rings, the matrices over . Put
from which we easily see that the generic element in is
a matrix whose bottom row is zero. However, consider left multiplication by
We find, for instance,
Hence is not a two-sided ideal. It is, of course, a one-sided ideal.
Let and and consider . Because both and are closed under addition, this is again an element of . Also, for any , we have and . Therefore, is an ideal of .
That is an additive subgroup is clear. Let and consider
Because is closed under such multiplications, this is again an element of . Multiplication from the right also remains within because is closed under such multiplications. Therefore, is an ideal of .
Consider . Because is an ideal, for each and therefore this element belongs to . In precisely the same way, by virtue of being an ideal. Then, again, the element belongs to , and so .
The set is an ideal because and for all . Suppose that the ideal contains some integer which is not a multiple of . Then is coprime to , so there exist such that and, by the closure properties of ideals, must contain it. Therefore, if is a proper subset of then and . Otherwise, .
Of course, this argument holds with any prime integer in place of . In the integers, a prime ideal is maximal.
Suppose and let be arbitrary. Then so that . Also, for , and because so annihilates it just as well. Therefore is an ideal of .
The second statement follows directly from the fact that is an ideal, so, for any and , which means that and hence .
Let and . Then since each of and are elements of , which is an ideal. Therefore is a subgroup of under addition by Lemma 2.4.2. In addition, with , and where by assumption. Then is an ideal of .
(a) That is closed under and follows from the fact that is a ring. Let . The new addition is commutative: . It is associative:
The multiplication is associative:
The distributive law also holds,
In the next section, we will demonstrate the existence of a zero element and additive inverses.
(b) The zero element of is such that for all . Then we must take (i.e. the additive inverse in of the unit element of ). The additive inverse of is such that . Then so that . This is, of course, in , so we have shown that is a ring with these new operations.
(c) The multiplicative identity element is such that for all . This is satisfied by , the zero element of .
(d) We can explicitly construct a homomorphism . We must have that
for all . This forces and we may as well try the map . In fact, with ,
Therefore this map is truly a homomorphism. Its kernel is trivial, containing just the identity element , so it is also an isomorphism.
Let be an ideal of . If there exists an invertible matrix , then , so that . Therefore consider an ideal which consists of only non-invertible matrices. Suppose there is a non-zero matrix . The gist of the argument from here is that we can take and exhibit an invertible matrix in using the ideal’s closure under addition and external multiplication.
First note that we can make the following sorts of manipulations:
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 . 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 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
This is the desired invertible matrix which is in . Therefore if is to not be the entire ring, then it can contain nothing but the zero matrix.
(a) Compute a generic product of elements in :
This is again an element of , as is a generic sum. Thus forms a ring, and we can easily exhibit elements, with 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 elements. Conceivably there could be some elements that get duplicated, but we see that if
which forces , etc., so there is no duplication and there are truly elements.
Again we would like to consider an ideal of and show that the existence of any non-zero element in the ideal necessarily leads to the conclusion that . Let be a non-zero element, presumably with no inverse. We can play around with , computing things such as , , and so forth. Eventually, we hit upon
which almost reminds us of complex conjugation. Of course, and look similar. In fact, we prod a little more and find that the combination
Now, supposing , this is the desired invertible element belonging to (thankfully, we have assumed that our prime is not ). On the other hand, if , we can rotate a non-zero coefficient into its place by multiplying appropriately by , or , and then performing the same trick. To sum up, we have shown a way to take any non-zero element of the ideal and manipulate it into an invertible element which also belongs to , thus proving that a non-zero ideal is actually the entire ring.
(b) To show that is not a division ring, we must prove that there exists an element with no multiplicative inverse. Consider conjugation,
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 such that were zero modulo , then we could show that is not invertible in the ring. Suppose to the contrary that there was an element in the ring such that . Then we would have which is a contradiction.
Therefore we seek an element with (in other words, such that the sum of the squares of its coefficients is zero modulo ), and if we can produce it then we have shown that is not a division ring. As an example, with , such a non-invertible element is . 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 math.stackexchange.com post, we narrow our attention to solutions with one coefficient fixed as and one fixed as . By the lemma immediately below, we have the existence of and such that . Hence, for any odd prime , there will always exist a non-invertible element of the form , proving that is not a division ring.
The case is immediately satisfied by , . Therefore, let be an odd prime.
First consider the case of two distinct integers whose squares are to be congruent modulo . We have or . As is a prime, it must divide one of the factors. This isn’t possible for because of the restricted domain for and . Hence, for the two squares to be congruent, we must have . The set therefore pairs off into equivalence classes , , etc., with zero the odd man out, and there are exactly distinct values for modulo .
Now we would like to show the existence of such that . There are evidently possible values for the left hand side, and possible values for the right hand side, and both quantities reside in the interval which accomodates a total of numbers. Thus there is no way to fit the values of modulo alongside the values of modulo , a total of numbers, without some overlap (this is the pigeonhole principle). This overlap is the desired solution to the congruence , so we are done.
Let . Then so that is an additive subgroup of by Lemma 2.4.2. Also, if we multiply on the left by arbitrary , we find because . Therefore is a left ideal. See also H 3.4.4.
and are each additive subgroups of so their intersection is again an additive subgroup of . If and , then and , hence . Therefore, is a left ideal of .
If , then and , so . Thus is an additive subgroup of . Letting , we have that and , 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 the set of rational matrices,
and . We can easily see that consists of the matrices whose second column is zero, and consists of the matrices whose second row is zero. Their intersection 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:
We can see it’s also not a right ideal by taking the transpose of the above equation.
is the set of elements which annihilates from the left. If , then so that and is an additive subgroup of . With , we also have so that . Thus is a right ideal of .
is the set of elements which annihilate the entire left ideal . If and , then so that is an additive subgroup of . Let and consider , which shows to be a left ideal, and , which shows to be a right ideal, with because is a left ideal. Therefore is a two-sided ideal.
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 so that for every . This is not a statement about characteristic, because we do not assume to be an integral domain. We do not even assume that is unital, so recall that is simply shorthand for .
so that we have for any . In particular, plugging in gives
By the above comment, we can subtract from the equation for free, and this leaves us with
The simplest example of a ring with for all is which has characteristic , and we immediately see that this result is worthless in that case. Clearly we need an additional result to prove that is commutative even when we can’t “cancel” the three.
A natural thing to compute is
Subtracting the second () equation from the first () gives . If we multiply that equation on the left by we find
whereas if we multiply by on the right instead,
Subtracting one equation from the other again, we see that
Finally, subtracting that from the previous result, we arrive at the desired statement that for all , i.e. that 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 given by is a ring homomorphism, in a not-entirely-trivial way. This may lead somewhere useful.
For any , there exists such that , because is surjective. Then we also must have that
so that is the unit element of .
Note that will be the unit element of the sub-ring regardless, but if is not surjective then there may be elements of for which does not act as a unit.
Clearly there is a problem if , so first let be outside the kernel of and consider
This shows that . Now we also see from the above computation that acts as a unit element for (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 .
Again letting and letting be arbitrary, we have that so that
By assumption, , so, because is an integral domain, we conclude that for all .