Topics in Algebra, Chapter 3.4, Part 1

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 2](/journal/topics-in-algebra-chapter-3-section-4b) (3.4.11 through 3.4.16) 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.
### Topics covered: 3.3
* **Definition**: Let $R,R'$ be rings. A *ring homomorphism* is a mapping $\phi:R\to R'$ such that $\phi(a+b)=\phi(a)+\phi(b)$ and $\phi(ab)=\phi(a)\phi(b)$ for all $a,b\in R$.
* **Lemma 3.3.1**: If $\phi:R\to R'$ is a ring homomorphism, then $\phi(0)=0$ and $\phi(-a)=-\phi(a)$.
* **Definition**: With $\phi:R\to R'$ a ring homomorphism, the *kernel* of $\phi$ is the set $\{a\in R\mid\phi(a)=0\}$.
* **Lemma 3.3.2**: With $\phi:R\to R'$ a ring homomorphism, kernel of $\phi$ is a subgroup of $R$ under addition and, if $a\in\ker\phi$ and $r\in R$, 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 $\{m+n\sqrt{2}\mid m,n\in\mathbb{Z}\}$ is a ring under standard operations, and conjugation, $\phi:m+n\sqrt{2}\mapsto m-n\sqrt{2}$, is a ring homomorphism with trivial kernel.
* The natural map $\phi:\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ 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 $r\in R$, 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 $I\subset R$, we consider an equivalence relation $a\sim b$ if $a-b\in I$. The cosets are of the form $r+I=\{r+i\mid i\in I\}$ for elements $r\in R$. 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 $\phi:R\to R/I$ given by $\phi(r)=r+I$ is a ring homomorphism.
* **Theorem 3.4.1**: Let $\phi:R\to R'$ be a ring homomorphism with kernel $I=\ker\phi$. Then $R'\cong R/I$. 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 $J'\mapsto \phi^{-1}(J')$, and $R/\phi^{-1}(J')\cong R'/J'$.
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 $I\subset R$ be an ideal such that $1\in I$. Prove that $I=R$.
For any $r\in R$, we have that $r1=r\in I$ so that $R\subset I$ and hence $I=R$.
### Herstein 3.4.2: Prove that the only ideals of a field $F$ are trivial, $\{0\}$ and $F$ itself.
Let $I\subset F$ be an ideal. If there exists a non-zero element $r\in I$, then $r^{-1}r=1\in I$ 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 $aR=\{ar\mid r\in R\}$ 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 $ar+as=a(r+s)\in aR$ for any $r,s\in R$, so $aR$ forms a subgroup under addition by Lemma 2.4.2. Furthermore, $(ar)s=a(rs)\in aR$ and $s(ar)=a(sr)\in aR$, 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 $s(ar)\in aR$ for arbitrary $s$. We will look for an example in one of the simplest non-commutative rings, the $2\times 2$ matrices over $\mathbb{R}$. Put
$$a=\begin{pmatrix}1 & 0\\0 & 0\end{pmatrix}$$
from which we easily see that the generic element in $aR$ is
$$\begin{pmatrix}1 & 0\\0 & 0\end{pmatrix}\begin{pmatrix}x & y\\z & w\end{pmatrix}=\begin{pmatrix}x & y\\0 & 0\end{pmatrix},$$
a matrix whose bottom row is zero. However, consider left multiplication by
$$s=\begin{pmatrix}1 & 1\\1 & 1\end{pmatrix}.$$
We find, for instance,
$$saI=\begin{pmatrix}1 & 1\\1 & 1\end{pmatrix}\begin{pmatrix}1 & 0\\0 & 0\end{pmatrix}\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}=\begin{pmatrix}1 & 0\\1 & 0\end{pmatrix}\not\in aR.$$
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 $I+J=\{i+j\mid i\in I,j\in J\}$. Prove that $I+J$ is an ideal of $R$.
Let $i,i'\in I$ and $j,j'\in J$ 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 $r\in R$, we have $r(i+j)=(ri)+(rj)\in I+J$ and $(i+j)r=(ir)+(jr)\in I+J$. 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 $i\in I$ and $j\in J$. Prove that $IJ$ is an ideal of $R$.
That $IJ$ is an additive subgroup is clear. Let $r\in R$ and consider
$$r(i_1 j_1+\ldots+i_n j_n)=(ri_1)j_1+\ldots+(ri_n)j_n.$$
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 $IJ\subset I\cap J$.
Consider $i_1 j_1+\ldots+i_n j_n\in I+J$. Because $I$ is an ideal, $i_k j_k=i_k'\in I$ for each $k\in\{1,\ldots,n\}$ and therefore this element belongs to $I$. In precisely the same way, $i_k j_k=j_k'\in J$ by virtue of $J$ being an ideal. Then, again, the element belongs to $J$, and so $IJ\subset I\cap J$.
### Herstein 3.4.8: Let $R=\mathbb{Z}$ and let $I$ be the ideal of $R$ made up of all multiples of $17$. Prove that any ideal $J$ of $R$, with $I\subset J\subset R$, must be either $J=I$ or $J=R$. How does this generalize?
The set $I$ is an ideal because $17m+17n=17(m+n)\in I$ and $(17m)n=17(mn)\in I$ for all $m,n\in\mathbb{Z}$. 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 $\alpha,\beta\in\mathbb{Z}$ such that $a\alpha+17\beta=1$ and, by the closure properties of ideals, $J$ must contain it. Therefore, if $I$ is a proper subset of $J$ then $1\in J$ 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 $r(I)=\{x\in R\mid xi=0$ for each $i\in I\}$. Prove that $r(I)$ is an ideal of $R$.
Suppose $x,y\in r(I)$ and let $i\in I$ be arbitrary. Then $(x+y)i=xi+yi=0$ so that $x+y\in r(I)$. Also, for $r\in R$, $(rx)i=r(xi)=0$ and $(xr)i=x(ri)=0$ because $ri\in I$ 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 $[R:I]=\{x\in R\mid rx\in I$ for each $r\in R\}$. Prove that $[R:I]$ is an ideal of $R$ and that $I\subset [R:I]$.
The second statement follows directly from the fact that $I$ is an ideal, so, for any $i\in I$ and $r\in R$, $ri\in I$ which means that $i\in [R:I]$ and hence $I\subset [R:I]$.
Let $x,y\in [R:I]$ and $r\in R$. Then $r(x+y)=rx+ry\in I$ 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 $s\in R$, $r(sx)=(rs)x\in I$ and $r(xs)=is\in I$ where $i=rx\in I$ by assumption. Then $[R:I]$ is an ideal of $R$.