In this post, we learn about rings and fields. These are part of the usual algebraic structure. We mention that fields are important in the study of vector spaces, the concept of dimension, and other properties.
An introduction to rings and fields
Definition: A ring is a set $\mathscr{R}$ endowed with two internal composition lows $+$ and $\times$ such that
- $(\mathscr{R},+)$ is a commutative group, where we denote $0_{\mathscr{R}}$ its neutral element,
- the law $\times$ is associative, i.e. for any $a,b,c\in \mathscr{R}$, $a\times (b\times c)=(a\times b)\times c,$
- the law $\times$ has a neutral element denoted by $1_{\mathscr{R}}$,
- for any $a,b,c\in\mathscr{R},$ we have \begin{align*} a\times (b+c)=a\times b+a\times c\quad (b+c)\times a=b\times a+c\times a.\end{align*}
Let $(\mathscr{R},+,\times)$ be a ring. If in addition, the law $\times$ is commutative, then we say that the ring is commutative.
Remark: Let us mention that the element $a$ of a ring $\mathscr{R}$ does not necessarily have inverses with respect to the law $\times$. When this is the case, we say that $a$ is invertible or it is a unit of $\mathscr{R},$ and the inverse will be denoted by $a^{-1}$.
The product of rings: Let $R$ and $W$ be two rings. On $R\times W$ we define the following structure \begin{align*}(a,b)+(c,d)=(a+c,b+d),\quad (a,b)\times(c,d)=(a\times c,b\times d).\end{align*}Then $(R\times W,+,\times)$ is a ring.
Subrings: Let $(\mathscr{R},+,\times)$ be a ring. A subset $S$ of $\mathscr{R}$ is called a subring of $\mathscr{R}$ if $(S,+,\times)$ is a ring. We have the following characterization: $S$ is a subring of $\mathscr{R}$ if and only if the following assertions hold:
- $1_{\mathscr{R}}\in S,$
- for any $x,y\in S,$ $x+(-y)\in S,$
- for any $x,y\in S$, $x\times y\in S$.
Binomial expansion formula: If $\mathscr{R}$ is a commutative ring, then for any $a,b\in \mathscr{R},$ and $n\in\mathbb{N},$ we have \begin{align*}(a+b)^n=\sum_{k=0}^n (\begin{smallmatrix}n\\ k\end{smallmatrix}) a^{k}\times b^{n-k}.\end{align*} Since the ring of matrices is not commutative, the binomial formula is not applicable to the matrix unless the matrices commute.
Definition: A field is a commutative ring in which every nonzero element is invertible.
A selection of exercises on rings
In what follows in this section, we propose some exercises on rings and fields with detailed solutions.
Exercise: Prove that the set of dyadic numbers $H=\{n2^{-p}:(n,p)\in\mathbb{Z}\times \mathbb{N}\}$ endowed with usual laws is a ring. Can $H$ be a field?
Solution: It suffices to show that $H$ is a subring of $\mathbb{R}$. Observe that $1=1\times 2^{-0}\in H$. Let $(x,y)\in H^2$. There exist $(n,m)\in\mathbb{Z}^2$ and $(p,q)\in\mathbb{N}^2$ such that \begin{align*} x=n2^{-p}\quad\text{and}\quad y=m2^{-q}. \end{align*} We select $r=\max\{p,q\}$. Then \begin{align*} x-y&=\frac{n2^{r-p}}{2^r}-\frac{m2^{r-q}}{2^r}\cr &= \frac{n2^{r-p}-m2^{r-q}}{2^r}. \end{align*} As $n2^{r-p}-m2^{r-q} \in \mathbb{Z}$ and $r\in \mathbb{N}$ then we have $x-y\in H$. On the other hand, we have $xy=nm 2^{-(p+q)}\in H$ because $nm\in\mathbb{Z}$ and $p+q\in\mathbb{N}$. Hence $H$ is a subring of $\mathbb{R},$ then it is a ring for the usual laws.
Let us show that $H$ is not a field. In fact, we have $3=6\times 2^{-1}\in H$. But $\frac{1}{3}\notin H$. In fact if there exists $(n,p)\in\mathbb{Z}\times \mathbb{N}$ such that $\frac{1}{3}=n2^{-p},$ then $2^p=3n$ which means that $3|2^p$, absurd.
Exercise: Prove that the set \begin{align*} K=\{x+y\sqrt{2}:(x,y)\in\mathbb{Q}\} \end{align*} is a field with respect to the usual laws.
Solution: It suffices to show that $K$ is a subfield of $\mathbb{R}$. Observe that $1=1+0\sqrt{2}\in K$. Let $(x,y)\in K^2$. There exist $(p,q,r,s)\in\mathbb{Q}^4$ such that \begin{align*} x=p+q\sqrt{2}\quad\text{and}\quad y=r+s\sqrt{2}. \end{align*} We select $r=\max\{p,q\}$. Then \begin{align*} x-y&=(p-r)+(q-s)\sqrt{2}\in K \end{align*} On the other hand, \begin{align*} xy&=(p+q\sqrt{2})r+s\sqrt{2}\cr & =(pr+2qs)+(qr+ps)\sqrt{2}\in K. \end{align*} Now if $x\neq 0,$ we have $p-q\sqrt{2}\neq 0$ because $\sqrt{2}\notin \mathbb{Q}$. We can write \begin{align*} x^{-1}&=\frac{p-q\sqrt{2}}{x(p-q\sqrt{2})}\cr &= \frac{p-q\sqrt{2}}{p+q\sqrt{2}(p-q\sqrt{2})}\cr & = \frac{p-q\sqrt{2}}{p^2-2q^2}\cr & =\frac{p}{p^2-2q^2}-\frac{q}{p^2-2q^2}\sqrt{2}\in K. \end{align*} This shows that $K$ is a subfield of $\mathbb{R}$, then it is a field.
Here come other exercises of rings and fields, in particular, ideals.
Exercise: Let $(\mathcal{R},+,\times)$ be a commutative ring. An ideal of $I$ of $\mathcal{R}$ is called prime ideal if $I$ is not the whole ring $\mathcal{R}$ and for any $x,y\in \mathcal{R}$ such that $xy\in I$ we have $x\in I$ or $y\in I$.
- Let $d\in \mathbb{Z}$. Prove that $d$ is prime if and only if for all $(x,y)\in\mathbb{Z}^2$, $d|(xy)$ implies that $d|x$ or $d|y$.
- Determine the prime ideals of $\mathbb{Z}$.
- Determine the prime ideal of $\mathbb{K}[X],$ where $\mathbb{K}$ is a subfield of $\mathbb{C}$.
- Let $J$ and $K$ be two ideals of $\mathcal{R}$, and $I$ is a prime ideal such that $J\cap K=I$. Prove that $J=I$ or $K=I$.
- Assume that any ideal of $\mathcal{R}$ is prime. Prove that $\mathcal{R}$ is an integer. Prove that $\mathcal{R}$ is a field.
Solution: 1) Assume that $d$ is prime, then for all $(x,y)\in\mathbb{Z}^2,$ $d|(xy)$ implies that $d|x$ or $d|y$. Conversely, if for all $(x,y)\in\mathbb{Z}^2,$ $d|(xy)$ implies that $d|x$ or $d|y$, and if $ell$ is a strict divisor of $d$. We take $x=\ell$ and $y=\frac{d}{\ell}$. Then $d|(xy)$ but $d$ does not divide either $x$ or $y$, absurd. So $d$ does not have a strict divisor. Hence $d$ is prime.
2) It is well known that the ideals of $\mathbb{Z}$ have the forme $d\mathbb{Z}$ with $d\in\mathbb{N}$. Moreover, $x\in d\mathbb{Z}$ if and only if $d|x$. So $d\mathbb{Z}$ is prime if and only if for all $(x,y)\in\mathbb{Z}^2$, $d|(xy)$ implies that $d|x$ or $d|y$. This is equivalent to $p$ is a prime number.
3) We know that the ideals of $\mathbb{K}[X]$ are of the forme $P\mathbb{K}[X]$ with $P\in \mathbb{K}[X]$. By analogy with the previous question, it seems that $P\mathbb{K}[X]$ is prime if and only if $P$ is irreducible. In fact, if $P$ is irreducible, then for $(Q,R)\in \mathbb{K}[X]^2$ such that $QR\in P\mathbb{K}[X]$, we have $P|QR$. So P is an element in the irreducible factor decomposition of $QR$ obtained from the product of the decomposition of $Q$ and that of $R$. Hence $P$ is present in the irreducible factor decomposition of $Q$ or of that of $P$. This implies that $P|Q$ or $P|R$. This means that $Q\in P\mathbb{K}[X]$ or $R\in P\mathbb{K}[X]$, then $P\mathbb{K}[X]$ is prime. Conversely, let $P\in \mathbb{K}[X]$ such that $P\mathbb{K}[X]$ is prime. Assume that $P$ is not irreducible. Thus $P$ has a strict divisor $Q\in \mathbb{K}[X]$. Now Let $R\in \mathbb{K}[X]$ such that $P=QR$, we then have $QR\in P\mathbb{K}[X]$. But $P$ does not divide $Q$ nor $R$, then $Q\notin P\mathbb{K}[X]$ and $R\notin P\mathbb{K}[X]$, absurd. Finally, the prime ideals of $\mathbb{K}[X]$ are $P\mathbb{K}[X]$ with $P$ irreducible.
4) As $J\cap K=I$ then we have $I\subset J$ and $I\subset K$. Assume that $J\neq I$ and let $x\in J\setminus I$. For $y\in K$ we have $xy\in J\cap K$, because $J$ and $K$ are absorbent, so $xy\in I$. As $x\notin I,$ and $I$ is prime, we have $y\in I$. This implies that $Ksubset I,$ and then $I=K$.
5) Let $(a,b)\in\mathcal{R}$ such that $ab=0$. Then $ab\in\{0\},$ which is an ideal, and then it is prime by hypothesis. This implies that $a\in\{0\}$ or $b\in\{0\}$. This means that $a=0$ or $b=0$. So $\mathcal{R}$ is an integer.
Let $a\in \mathcal{R}\setminus\{0\}$ and $I=a^2\mathcal{R}$. As $I$ is prime and $aa\in I,$ then $a\in I$. Then there exists $y\in \mathcal{R}$ such that $a=a^2 y$, which implies $a(1-ay)=0$. As $\mathcal{R}$ is integer, we have $1-ay=0$. Then $ay=1$. Now as $\mathcal{R}$ is commutative, $a$ is invertible. We have proved that every null element of $\mathcal{R}$ is invertible. Then $\mathcal{R}$ is a field.
