Subrings, Ideals, Quotient Rings & Euclidean Rings Ring Of Endomorphisms Of Abelian Solved Problems

Subrings, Ideals, Quotient Rings & Euclidean Rings Subrings Solved Problems

Example. 1. Prove that S1 = {0, 3}, S2 ={0,2, 4}  are subrings of Z6 = {0,1, 2,3,4 5} with respect to addition and multiplication of residue classes.

Solution. Since (Z6,+,•) is a ring, from the property R4  of the ring we have -0 = 6,-2 = 4,-3 = 3, -4 = 2. ={ 0, 3.} is a non-empty subset of Z6

From the above tables a,b ∈ S1 =>a-b∈ S1 and a.’b ∈ S1

By the theorem (1), S1 is a subring of Z6.

S2 = {0, 2, 4} is a non-empty subset of Z6

From the above tables ; a,b ∈ S2 =>a-b ∈ S2 and a.b ∈ S2

S2 is a subring of Z6

We can see that S1 ∩S2 = { 6 } is the trivial subring. But \(S_1 \cup S_2=\{\overline{0}, \overline{2},\overline{3}, \overline{4}\} \text { is not a }\)

subring of Z6, because 2,3\( S_1 \cup S_2 \Rightarrow \overline{2}+\overline{3}=\overline{5} \& S_1 \cup S_2 \)

Subrings And Ideals In Ring Of Endomorphisms Of Abelian Groups

Example. 2. Show that the set of matrices \(\left(\begin{array}{ll}
a & b \\
d & c
\end{array}\right)\) is a subring of the ring of 2×2 matrices whose elements are integers

Solution. Let R = \(\left\{\left(\begin{array}{ll}
a & b \\
d & c
\end{array}\right) \mid a, b, c, d \in Z\right\}\) be the ring of 2×2 matrices and

S = \(\left\{\left(\begin{array}{ll}
a & b \\
0 & c
\end{array}\right) \mid 0, a, b, c \in Z\right\}\) Then 5 # <|) and S cR .

Let A, B eS so that A= \(\left(\begin{array}{ll}
a_1 & b_1 \\
0 & c_1
\end{array}\right), B=\left(\begin{array}{cc}
a_2 & b_2 \\
0 & c_2
\end{array}\right)\) where 0, ah by, c{, a2, b2i c2 e Z.

∴ \(A-B=\left(\begin{array}{cc}
a_1-a_2 & b_1-b_2 \\
0 & c_1-c_2
\end{array}\right) \text { and } A B=\left(\begin{array}{cc}
a_1 a_2 & a_1 b_2+b_1 c_2 \\
0 & c_1 c_2
\end{array}\right) \text {. }\)

Sincea \(1-a_2, b_1-b_2, c_1-c_2, a_1 a_2, a_1 b_2+b_1 c_2, c_1 c_2 \in Z\)

Note: If R is a commutative ring then S is an ideal of R.

Understanding Euclidean Rings And Endomorphism Rings

Example. 3. Let R be a ring and a ∈ R be a fixed element. Then prove that S =(x  ∈ R | ax = 0} [is a subring of R.

Solution. If 0 ∈R is the zero element of r and a ∈ R, we have a0 = 0 => 0 ∈ S

S ≠ Φ and S⊂ R

Let x,y∈ S . Then x, y e R and ax = 0, ay= 0.

Now a(x-y) = ax-ay = 0- 0 = x-y ∈ S.

Also a (xy) = (ax)y = 0y = 0 => xy ∈ S. Hence S is a subring of R.

Notation. Let R be a ring and a∈ R be a fixed element. The intersection of the family of subrings containing ‘a’ is a subring of R. This subring is denoted by Ra and is called the subring of R generated by ‘a’.

Example. 4. If R is a ring and C (R) = (x € R \ xa — ax ∀ a∈ R] then prove that C (R) is a subring of R.

Solution.  For 0 ∈ R, the zero element of the ring, we have 0a = a0 ∀ a ∈R.

By the definition of C (R), 0 ∈C (R). C(R)≠ Φ and C (R)⊂R

Let x,y ∈ C(R)

Then x,y ∈ R and xa = ax, ya = a y V a ∈R

∀a∈R, a(x-y)-ax-ay =xa-ya =(x-y) a

Also, ∀ a ∈R,a(xy) = (ax)y = (xa)y = x (ay) = x (ya) = (xy) a …. (1)