For a fixed element 'a' of a ring R, prove that the set {x exists in R such that ax=0)is a subring of R.
To prove that the set {x ∈ R such that ax = 0} is a subring of R for a fixed element 'a' of a ring R, we need to show that it satisfies the criteria for a subring.
First, let's verify that it's a non-empty subset. Since 0a = 0 for any element 'a', we have that 0 is always in the set.
Next, we need to prove closure under addition and multiplication. Let's consider two elements x and y in the subset.
Closure under addition:
If ax = 0 and ay = 0, then (ax + ay) = a(x + y). By the distributive property of rings, this becomes a(x + y) = ax + ay = 0 + 0 = 0.
Therefore, x + y is in the set.
Closure under multiplication:
If ax = 0 and ay = 0, then (ax)(ay) = (ax)(0) = a(x0) = a0 = 0.
Therefore, xy is in the set.
Lastly, we need to prove that the subset is closed under the additive inverse. If ax = 0, then (-a)x = -(ax) = -0 = 0.
Thus, the additive inverse of x is also in the set.
Since the subset satisfies closure under addition, multiplication, and additive inverse, we can conclude that {x ∈ R such that ax = 0} is a subring of R for any fixed element 'a' of the ring R.