Check if the solution is correct. Correct the mistakes. Prove that a finite integral domain is a division ring Proof: Let R be finite integral domain. Take a nonzero element a ∈ R. Consider a map f: R → R defined as f(x) = ax for all x ∈ R. Since R is a commutative ring and a ≠ 0, the map f is a ring homomorphism. Let us show that f is an injection. Suppose that f(x) = f(y) for some x,y ∈ R. Then ax = ay. Since there are no zero divisors in the integrity region, we can subtract a and get x = y. So f is an injection. Since R is a finite set, the injection f is also a surjection. This means that for every nonzero element b ∈ R, there exists an element x ∈ R such that f(x) = ax = b. In particular, there is an element a⁻¹ ∈ R such that a * a⁻¹ = 1. Thus, every nonzero element in a finite integrity domain is invertible, and R is a division ring
Check if the solution is correct. Correct the mistakes.
Prove that a finite
Proof: Let R be finite integral domain. Take a nonzero element a ∈ R. Consider a map f: R → R defined as f(x) = ax for all x ∈ R. Since R is a commutative ring and a ≠ 0, the map f is a ring homomorphism.
Let us show that f is an injection. Suppose that f(x) = f(y) for some x,y ∈ R. Then ax = ay. Since there are no zero divisors in the integrity region, we can subtract a and get x = y. So f is an injection.
Since R is a finite set, the injection f is also a surjection. This means that for every nonzero element b ∈ R, there exists an element x ∈ R such that f(x) = ax = b. In particular, there is an element a⁻¹ ∈ R such that a * a⁻¹ = 1.
Thus, every nonzero element in a finite integrity domain is invertible, and R is a division ring
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