If (R, +, .) Is a ring and

nainamourya181

nainamourya181

Answered question

2022-08-25

If (R, +, .) Is a ring and a belongs to R then S = { x belongs to R : ax = 0} is a subring of R

Answer & Explanation

Vasquez

Vasquez

Expert2023-05-29Added 669 answers

To prove that the set S=xR:ax=0 is a subring of R, we need to show that S satisfies the three conditions for being a subring:
1. S is non-empty.
2. S is closed under addition.
3. S is closed under multiplication.
Let's go through each condition step by step using LaTeX markup:
1. S is non-empty:
Since a belongs to R, we know that a * 0 = 0, which implies that 0 is in S. Therefore, S is non-empty.
2. S is closed under addition:
Let x and y be elements of S. We need to show that x + y is also in S.
By definition, ax = 0 and ay = 0. Now, let's consider the sum (x + y)a:
(x+y)a=xa+ya
Since xa = 0 and ya = 0, we have:
(x+y)a=0+0=0
This shows that x + y is in S, and thus S is closed under addition.
3. S is closed under multiplication:
Let x and y be elements of S. We need to show that xy is also in S.
Again, by definition, ax = 0 and ay = 0. Now, let's consider the product (xy)a:
(xy)a=x(ya)
Since ya = 0, we have:
(xy)a=x(0)=0
This shows that xy is in S, and thus S is closed under multiplication.
Since S satisfies all three conditions, we can conclude that S=xR:ax=0 is a subring of R.

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