Let R be a commutative ring in with a^2=a for every element a. Show that a+a=0 for all a in R
To show that a+a=0 for all a in the ring R, we need to use the given condition that a^2 = a for every element a in R.
Let's begin by considering an arbitrary element a in R.
Using the given condition, we have a^2 = a. Now, we can multiply both sides of this equation by a to obtain a^3 = a^2.
Next, we can subtract a^2 from both sides of the equation to obtain a^3 - a^2 = a^2 - a.
Now, we can factor the left-hand side of the equation to get a^2(a - 1) = a(a - 1).
Since R is a commutative ring, we can rearrange the terms on both sides of the equation to get (a - 1)a^2 = (a - 1)a.
Now, we can cancel out the common factor of (a - 1) from both sides of the equation to get a^2 = a.
Using the given condition again, we know that a^2 = a, so we can substitute a for a^2 in the equation above to get a = a + a.
Finally, we can subtract a from both sides of the equation to get 0 = a + a - a.
But since addition is commutative in R, we can rearrange the terms to get 0 = a - a + a.
Finally, we can simplify the right-hand side of the equation to get 0 = a + 0, which further simplifies to 0 = a.
Therefore, we have shown that a + a = 0 for all a in the commutative ring R, given the condition that a^2 = a for every element a in R.