# 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 which stands for is a subring of the ring of integers, the set of Gaussian integers is a subring of and the set has the set 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) and (ii).

**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.

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

**Proof**: Let S_{1} and S_{2} be two subrings and . Then as they are also in S_{1} and S_{2}. (Note that the intersection is nonempty as it defnitely contains 0). But then the result holds by the previous theorem.

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

**Definition 2**: The **center** of a ring R is defined as the set .

**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.

**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..

## Some more propertiesEdit

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 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 I_{a} is a subring.

**Theorem 1.20**:If A and B are two subrings of a ring R then their **sum** is defined as the set . Show that the sum of two subrings need not be a subring.

**Proof**: Consider the ring M_{2} of 2x2 matrices with entries belonging to the integers. (Check that this is indeed a ring.) It is easy to verify that the sets are subrings of M_{2}. However their sum which contains the matrices and doesn't contain their product which is . Hence sum of two subrings need not be a subring. .

**Theorem 1.21**: An element is called **idempotent** if x^{2}=x. Let e be an idempotent in a ring R. Show that 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. .

## ExercisesEdit

1. Show that the **normalizer** N(a) of a element a of a ring R defined by 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) and (ii) .