Theorem: Let I be a proper ideal in the integral domain R and let p(x) be a non-constant polynomial in R[x]. If p(x) is reducible in R[x] then the image of p(x) in (R/I)[x] is also reducible. But we will show the contrapositive of it
Contrapositive Statement of the Theorem: Let I be a proper ideal in the integral domain R and let p(x) be a non-constant polynomial in R[x]. If the image of p(x) in (R/I)[x] can not be factored into two polynomials of a smaller degree then p(x) is irreducible in R[x]
For that first, we need to prove this theorem
Theorem: Let I be an ideal of the ring R and let (I)=I[x] denote the ideal of R[x] generated by I (the set of polynomials with coefficients in I ). Then R[x]/(I)≅(R/I)[x]
Proof: There is a natural map φ:R[x]→(R/I)[x] given by reducing each of the coefficients of a polynomial modulo I. The definition of addition and multiplication in these two rings shows that φ is a ring homomorphism. The kernel is precisely the set of polynomials each of whose coefficients is an element of I, which is to say that kerφ=I[x]=(I), proving the theorem. ◼
Now we come back to the proof of the theorem
Proof of Contrapositive Statement of the Theorem: Suppose p(x) cannot be factored in (R/I)[x] but that p(x) is reducible in R[x]. This means there are monic, nonconstant polynomials a(x) and b(x) in R[x] such that p(x)=a(x)b(x). By the theorem above, reducing the coefficients modulo I gives a factorization in (R/I)[x] with nonconstant factors, a contradiction. Hence p(x) is irreducible in R[x] ◼
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