| For polynomial equations, the construction works in quite some generality, and is known as quotient ring: https://en.wikipedia.org/wiki/Quotient_ring Given any polynomial P (e.g. x^2 + 1) over a filed F (e.g. reals) we can form: `R = F[X]/P` This is an algebraic "set" that supports addition, substraction, multiplication and has 0,1 but not division in general. Elements are elements of F and a new symbol X that satisfies "P(X) = 0". Examples: R[X]/(x^2 + 1) = C
R[X]/x = R
C[X]/(x^2 + 1) = C + C.x
R[X]/1 = 0
# Properties- If the polynomial P is invertible, i.e. has degree 0 and is not zero, then the resulting ring is zero R[X]/P = 0. This is what happens in the example x = x-1 (which corresponds to P = x - 1 - x = -1). - If the polynomial P has degree 1 (i.e. P=aX+b), then the equation P=0 is equivalent to x=-b/a, representing an element already present in R, hence the ring R[X]/P is equal to R. - If the polynomial P is irreducible (i.e. not a product of two proper polynomials) then the quotient R[X]/P is a field. This happens in the case R[x]/(x^2 + 1) which results in the complex numbers. - If the polynomial P is a product of two polynomials P1,P2 which don't have common divisors, then R[X]/P = R[X]/P1 + R[X]/P2, this happens in the case that C[X]/(x^2+1), since P = x^2 + 1 factors as (x+i)*(x-i) in C. The equivalent result for integers is known as Chinese Remainder Theorem. |
Should be degree 0: only constant polynomials are invertible. E.g. x+1 is not invertible, and modding it out doesn't result in the zero ring.
The example is a bit confusing, because $x=x+1$ is equivalent to $0=1$, which has degree 0.