Irreducibility of Polynomials in R/I[x] and R[x]

[Math Amateur]

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

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

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

Peter

 

[jvicens]

,

Thank you for bringing this up. It is true that Proposition 2 guarantees an isomorphism between $R[x]/I$ and $(R/I)[x]$, but it does not guarantee a factorization of $p(x)$ in $(R/I)[x]$. However, we can use the fact that $(R/I)[x]$ is a unique factorization domain (UFD) to show that $p(x)$ must have a factorization in $(R/I)[x]$.

Since $p(x)$ cannot be factored in $(R/I)[x]$, it must be irreducible in $(R/I)[x]$ (by definition of a UFD). Therefore, $p(x)$ must be a prime element in $(R/I)[x]$. This means that $p(x)$ must generate a prime ideal in $(R/I)[x]$. By Proposition 2, this prime ideal corresponds to an ideal in $R[x]$ that is generated by $p(x)$ and elements of $I$. In other words, $p(x)$ must generate a prime ideal in $R[x]$.

Now, since $p(x)$ is monic and $R$ is an integral domain, we know that $p(x)$ is irreducible in $R[x]$. Therefore, $p(x)$ must be a prime element in $R[x]$. This means that $p(x)$ generates a prime ideal in $R[x]$. By Proposition 2, this prime ideal corresponds to an ideal in $(R/I)[x]$ that is generated by $p(x)$ and elements of $I$. In other words, $p(x)$ must generate a prime ideal in $(R/I)[x]$.

But since $p(x)$ generates a prime ideal in both $R[x]$ and $(R/I)[x]$, it must also generate a prime ideal in their intersection, which is $(R/I)[x]$. This means that $p(x)$ must have a factorization in $(R/I)[x]$, which contradicts our assumption that $p(x)$ cannot be factored in $(R/I)[x]$.

I hope this explanation helps clarify how Proposition 2, along with the fact that $(R/I)[x]$ is a UFD, guarantees a factorization of $p(x)$ in $(R/I)[x]$.