# Ring Theory/Subrings

Definition 1: Let (R,+,.) be a ring. A non empty subset S of R is called a subring of R if (S,+,.) is a ring.

For example the set ${\displaystyle m\mathbb {Z} }$ which stands for ${\displaystyle \{0,\pm m,\pm 2m\cdots \}}$ is a subring of the ring of integers, the set of Gaussian integers ${\displaystyle \mathbb {Z} [i]}$ is a subring of ${\displaystyle \mathbb {C} }$ and the set ${\displaystyle \mathbb {Z} _{4}}$ has the set ${\displaystyle \{{\overline {0}},{\overline {2}}\}}$ as a subring under addition and multiplication modulo 4.

Theorem 1.15: A non empty subset S of a ring R is a subring of R iff (i)${\displaystyle a-b\in S}$ and (ii)${\displaystyle ab\in S\forall a,b\in S}$.

Proof: The proof is an elementary consequence of a similar theorem about groups. Clearly necessity is clear. For sufficiency, note that 0=a-a is in S, -a=0-a is in S and a+b=a-(-b) is also in S. Other properties of a ring follow trivially.${\displaystyle \Box }$

Theorem 1.16: The intersection of two subrings of a ring R is a subring of R.

Proof: Let S1 and S2 be two subrings and ${\displaystyle a,b\in S_{1}\cap S_{2}}$. Then ${\displaystyle a-b,ab\in S_{1}\cap S_{2}}$ as they are also in S1 and S2. (Note that the intersection is nonempty as it defnitely contains 0). But then the result holds by the previous theorem.${\displaystyle \Box }$

Note that the corresponding result about unions may not be true. For example the union of ${\displaystyle 2\mathbb {Z} }$ and ${\displaystyle 3\mathbb {Z} }$ has 3 and 2 but not their difference 1 and so is not a subring of ${\displaystyle \mathbb {Z} }$.

Definition 2: The center of a ring R is defined as the set ${\displaystyle Z(R)=\{a\in R:xa=ax\forall x\in R\}}$.

Theorem 1.17: The center of a ring R is a subring of R.

Proof: Clearly if a and b are two elements in the center then for any x in R x(a-b)=xa-xb=ax-bx=(a-b)x and x(ab)=xab=axb=abx=(ab)x and so both a-b and ab are in the center. The result follows now from Theorem 1.15.${\displaystyle \Box }$

Theorem 1.18: The center of a division ring is a field.

Proof: If R is a division ring, then its center contains the identity 1 as x1=1x=x for all x. Also if a is in the center and ab=ba=1 then for any x, xb=x1b=xabb=axbb and so x=axb. Now bx=baxb=1xb=xb and so b is also in the center. Hence each element's inverse is also in the center. Finally note that that the elements commute with each other as they do so with all other elements of R. Other properties of a field follow by virtue of R being a division ring.${\displaystyle \Box }$.

## Some more properties

These problems should be first tried as exercises by the reader.

Theorem 1.19: If a is a fixed element of a ring R, show that ${\displaystyle I_{a}=\{x\in R:ax=0\}}$ is a subring of R.

Proof: Clearly if x,y are two elements in R then a(x-y)=ax-ay=0-0=0 and a(xy)=axy=0y=0 and so Ia is a subring.${\displaystyle \Box }$

Theorem 1.20:If A and B are two subrings of a ring R then their sum is defined as the set ${\displaystyle A+B=\{a+b:a\in A,b\in B\}}$. Show that the sum of two subrings need not be a subring.

Proof: Consider the ring M2 of 2x2 matrices with entries belonging to the integers. (Check that this is indeed a ring.) It is easy to verify that the sets ${\displaystyle S=\{{\bigl (}{\begin{smallmatrix}a&0\\b&0\end{smallmatrix}}{\bigr )}:a,b\in \mathbb {Z} \},T=\{{\bigl (}{\begin{smallmatrix}0&c\\0&0\end{smallmatrix}}{\bigr )}:c\in \mathbb {Z} \}}$ are subrings of M2. However their sum which contains the matrices ${\displaystyle {\bigl (}{\begin{smallmatrix}1&1\\1&0\end{smallmatrix}}{\bigr )}}$ and ${\displaystyle {\bigl (}{\begin{smallmatrix}2&2\\2&0\end{smallmatrix}}{\bigr )}}$ doesn't contain their product which is ${\displaystyle {\bigl (}{\begin{smallmatrix}4&2\\2&2\end{smallmatrix}}{\bigr )}}$. Hence sum of two subrings need not be a subring.${\displaystyle \Box }$.

Theorem 1.21: An element ${\displaystyle x\in R}$ is called idempotent if x2=x. Let e be an idempotent in a ring R. Show that ${\displaystyle eRe=\{eae:a\in R\}}$ is a subring of R with unity e.

Proof: Clearly if x,y are in eRe then x=eae and y=ebe for some a,b in R. Then x-y=eae-ebe=e(a-b)e and so x-y is in eRe. Also xy=eaeebe=eaebe=e(aeb)e and so xy is in eRe. Hence eRe is a subring. Finally note that xe=eaee=eae=x and ex=eeae=eae=x and so e is the unity of eRe.${\displaystyle \Box }$.

## Exercises

1. Show that the normalizer N(a) of a element a of a ring R defined by ${\displaystyle N(a)=\{x\in R:xa=ax\}}$ is a subring of R.

2. A non empty subset S of a field (F,+,.) is called a subfield of F is (S,+,.) is a field. Show that a subset S of a field F, containing at least two elements is a subfield of F iff (i)${\displaystyle a-b\in S\forall a,b\in S}$ and (ii)${\displaystyle ab^{-1}\in S\forall a\in S,b(\neq 0)\in S}$.