Prove that a finite subring R of a field F is itself a field. Hint: if x is an element of R and x is not equal to 0 show the function f:R->R with f(r) = xr is injective. From finiteness of R, deduce that its image includes 1© BrainMass Inc. brainmass.com March 21, 2019, 8:56 pm ad1c9bdddf
Let x be in R, and x is not equal to 0.
Then, the function f: R->R defined by f(r) = xr is injective.
First, show that f is well-defined. If r is in R and x is in R, then xr is in R, since R is a ring.
Now, show that f is injective. Suppose, there are two elements r1 and r2 in R such that f(r1) = f(r2).
But f(r1) = xr1, f(r2)=xr2, and so
xr1 = xr2
Since x is a non-zero element of a ...
We give a rigorous proof that a finite subring of a field is a field itself.