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Irreducibility of Polynomials - Irresducibilty in R/I{x] and Irreducibility in R[x]

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I am reading Dummit and Foote on Irreducibility in Polynomial Rings. In particular I am studying Proposition 12 in Section 9.4 Irreducibility Criteria. Proposition 12 reads as follows:

Let $I$ be a proper ideal in the integral domain $R$ and let $p(x)$ be a non-constant monic polynomial in $R[x]$. If the image of $p(x)$ in $(R/I)[x]$ cannot be factored in $(R/I)[x]$ into two polynomials of smaller degree, then $p(x)$ is irreducible in $R[x]$.

The proof in D&F reads as follows:

Proof: Suppose $p(x)$ cannot be factored in $(R/I)[x]$. This means that there are monic non-constant polynomials $a(x)$ and $b(x)$ in $R[x]$ such that $p(x)=a(x) b(x)$. By Proposition 2, reducing the coefficients modulo $I$ gives a factorization in $(R/I)[x]$ with non-constant factors, a contradiction

BUT! Proposition 2 shows that there is an isomorphism between $R[x]/I$ and $(R/I)[x]$.

How does this guarantee a factorization of $p(x)$ in $(R/I)[x]$?


For your information, Proposition 2 in D&F reads as follows:

Proposition 2. 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

[TEX] R[x]/(I) \cong (R/I)[x] [/TEX]

In particular, if $I$ is a prime ideal of $R$ then $(I)$ is a prime ideal of $R[x]$.

Peter
 
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