Topics in Algebra, Chapter 3.4, Part 2

There are no exercises from 3.3, so this section covers both 3.3 ("Homomorphisms") and 3.4 ("Ideals and Quotient Rings"). This page is split into parts so as to not get excessively long.
**See also**: [part 1](/journal/topics-in-algebra-chapter-3-section-4a) (3.4.1 through 3.4.10) and [part 3](/journal/topics-in-algebra-chapter-3-section-4c) (3.4.17 through 3.4.21).
Throughout, $R$ is a ring and $p$ is a prime integer.
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 3.4.11: Let $R$ be unital and define a new ring $S$ with the same elements as $R$ but with an addition $a\oplus b=a+b+1$ and a multiplication $a\cdot b=ab+a+b$ for elements $a,b\in R$.
### (a) Prove that $S$ is a ring with the operations $\oplus$ and $\cdot$.
### (b) What is the zero element of $S$?
### (c) What is the multiplicative identity element in $S$?
### (d) Show that $R\cong S$.
**(a)** That $S$ is closed under $\oplus$ and $\cdot$ follows from the fact that $R$ is a ring. Let $a,b,c\in S$. The new addition is commutative: $a\oplus b=a+b+1=b+a+1=b\oplus a$. It is associative:
$$(a\oplus b)\oplus c=(a+b+1)\oplus c=a+b+c+1=a+(b+c+1)=a\oplus (b+c).$$
The multiplication is associative:
\begin{multline*}
(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{multline*}
The distributive law also holds,
\begin{multline*}
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{multline*}
In the next section, we will demonstrate the existence of a zero element and additive inverses.
**(b)** The zero element of $S$ is $0_S$ such that $a\oplus 0_S=a$ for all $a\in S$. Then we must take $0_S=-1$ (i.e. the additive inverse in $R$ of the unit element of $R$). The additive inverse $\alpha$ of $a\in S$ is such that $a\oplus\alpha=0_S=-1$. Then $a+\alpha+1=-1$ so that $\alpha=-a-2$. This $\alpha$ is, of course, in $S$, so we have shown that $S$ is a ring with these new operations.
**(c)** The multiplicative identity element $1_S$ is such that $a\cdot 1_S=a$ for all $a\in S$. This is satisfied by $1_S=0$, the zero element of $R$.
**(d)** We can explicitly construct a homomorphism $\phi:R\to S$. We must have that
$$\phi(a)=\phi(a+0)=\phi(a)\oplus\phi(0)=\phi(a)+\phi(0)+1$$
for all $a\in R$. This forces $\phi(0)=-1$ and we may as well try the map $\phi(a)=a-1$. In fact, with $a,b\in R$,
$$\phi(a+b)=a+b-1=(a-1)+(b-1)+1=\phi(a)\oplus\phi(b).$$
Also,
$$\phi(ab)=ab-1=(a-1)(b-1)+(a-1)+(b-1)=\phi(a)\cdot\phi(b).$$
Therefore this map is truly a homomorphism. Its kernel is trivial, containing just the identity element $1_R$, so it is also an isomorphism.
### Herstein 3.4.12*: Let $R={\rm Mat}_{2\times 2}(\mathbb{Q})$, the ring of $2\times 2$ 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 $a\in I$, then $a^{-1}a\in I$, so that $I=R$. Therefore consider an ideal which consists of only non-invertible matrices. Suppose there is a non-zero matrix $a\in I$. 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:
$$\begin{pmatrix}\alpha & \beta\\\gamma & \delta\end{pmatrix}\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}=\begin{pmatrix}\alpha & -\beta\\\gamma & -\delta\end{pmatrix},$$
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
$$\begin{pmatrix}\alpha & 0\\0 & 0\end{pmatrix}+\begin{pmatrix}0 & 0\\1 & 0\end{pmatrix}\begin{pmatrix}\alpha & 0\\0 & 0\end{pmatrix}\begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}=\alpha\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\in I.$$
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 $\mathbb{Z}/p\mathbb{Z}$, with $p$ an odd prime, namely
$$R=\{\alpha_0+\alpha_1 i+\alpha_2 j+\alpha_3 k\mid \alpha_0,\alpha_1,\alpha_2,\alpha_3\in\mathbb{Z}/p\mathbb{Z}\},$$
where $i^2=j^2=k^2=ijk=-1$.
### (a) Prove that $R$ is a ring with $p^4$ elements with no non-trivial ideals.
### (b)** Prove that $R$ is not a division ring.
**(a)** Compute a generic product of elements in $R$:
\begin{multline*}
(\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_0-\alpha_1\beta_1-\alpha_2\beta_2-\alpha_3\beta_3)+(\alpha_0\beta_1+\alpha_1\beta_0+\alpha_2\beta_3-\alpha_3\beta_2)i\\
+(\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{multline*}
This is again an element of $R$, as is a generic sum. Thus $R$ forms a ring, and we can easily exhibit $p^4$ 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 $p^4$ elements. Conceivably there could be some elements that get duplicated, but we see that if
$$\alpha_0+\alpha_1 i+\alpha_2 j+\alpha_3 k=\beta_0+\beta_1 i+\beta_2 j+\beta_3 k,$$
then
$$(\alpha_0-\beta_0)+(\alpha_1-\beta_1)i+(\alpha_2-\beta_2)j+(\alpha_3-\beta_3)k=0+0i+0j+0k$$
which forces $\alpha_0=\beta_0$, etc., so there is no duplication and there are truly $p^4$ 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 $a=\alpha_0+\alpha_1 i+\alpha_2 j+\alpha_3 k$ 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
$$iai=-\alpha_0-\alpha_1 i+\alpha_2 j+\alpha_3 k$$
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
$$a-iai-jaj-kak=4\alpha_0\in I.$$
Now, supposing $\alpha_0\ne 0$, 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 $\alpha_0=0$, 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,
$$a\bar{a}=(\alpha_0+\alpha_1 i+\alpha_2 j+\alpha_3 k)(\alpha_0-\alpha_1 i-\alpha_2 j-\alpha_3 k)=\alpha_1^2+\alpha_2^2+\alpha_3^2+\alpha_4^2.$$
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 $a\bar{a}$ 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 $a^{-1}$ in the ring such that $a^{-1}a=1$. Then we would have $0=a^{-1}a\bar{a}=\bar{a}$ which is a contradiction.
Therefore we seek an element $a\in R$ with $a\bar{a}=0$ (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](http://en.wikipedia.org/wiki/Lagrange's_four-square_theorem) 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](http://math.stackexchange.com/a/241806/37832), 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 $\alpha_0$ and $\alpha_1$ such that $\alpha_0^2+\alpha_1^2+1\equiv 0\mod p$. Hence, for any odd prime $p$, there will always exist a non-invertible element of the form $\alpha_0+\alpha_1 i+j$, proving that $R$ is not a division ring.
#### Lemma: For prime $p$, there exist integers $x,y\in\{0,\ldots,p-1\}$ such that
$$x^2+y^2+1\equiv 0\mod p.$$
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 $m,n\in\{0,\ldots,p-1\}$ whose squares are to be congruent modulo $p$. We have $m^2\equiv n^2\mod p$ or $(m+n)(m-n)\equiv 0\mod p$. As $p$ is a prime, it must divide one of the factors. This isn't possible for $m-n$ 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 $\{0,\ldots,p-1\}$ therefore pairs off into equivalence classes $\{1,p-1\}$, $\{2,p-2\}$, etc., with zero the odd man out, and there are exactly $1+(p-1)/2=(p+1)/2$ distinct values for $m^2$ modulo $p$.
Now we would like to show the existence of $x,y\in\{0,\ldots,p-1\}$ such that $x^2\equiv -1-y^2\mod p$. 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 $\{0,\ldots,p-1\}$ which accomodates a total of $p$ numbers. Thus there is no way to fit the values of $x^2$ modulo $p$ alongside the values of $-1-y^2$ 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 $x^2+y^2+1\equiv 0\mod p$, 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 $r\in R$, $a\in L$ implies $ra\in L$ [$ar\in L$].
### Herstein 3.4.14: With $a\in R$, let $Ra=\{xa\mid x\in R\}$. Show that $Ra$ is a left ideal of $R$.
Let $x,y\in R$. Then $xa+ya=(x+y)a\in Ra$ so that $Ra$ is an additive subgroup of $R$ by Lemma 2.4.2. Also, if we multiply on the left by arbitrary $r\in R$, we find $r(xa)=(rx)a\in Ra$ because $rx\in R$. Therefore $Ra$ is a left ideal. See also H 3.4.4.
### Herstein 3.4.15: Let $L,M\subset R$ be left ideals. Show that $L\cap M$ 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 $x\in L\cap M$ and $r\in R$, then $rx\in L$ and $rx\in M$, hence $rx\in L\cap M$. Therefore, $L\cap M$ 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 $L\cap M$?
If $x,y\in L\cap M$, then $x+y\in L$ and $x+y\in M$, so $x+y\in L\cap M$. Thus $L\cap M$ is an additive subgroup of $R$. Letting $r\in R$, we have that $rx\in L$ and $xr\in M$, 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 $2\times 2$ rational matrices,
$$a=\begin{pmatrix}1 & 0\\0 & 0\end{pmatrix},$$
$L=Ra$ 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 $L\cap M$ 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:
$$\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}\begin{pmatrix}x & 0\\0 & 0\end{pmatrix}=\begin{pmatrix}0 & 0\\x & 0\end{pmatrix}\not\in L\cap M.$$
We can see it's also not a right ideal by taking the transpose of the above equation.