Hi I just want to clear up some confusion regarding some notation.
If $R$ is a ring and $\mathfrak p$ is a prime ideal in $R$, does $(R/\mathfrak p)[x] = R[x]/\mathfrak p[x]$? (or perhaps they are isomorphic?)
I am getting this from Atiyah and MacDonald's Intro to Commutative Algebra.
Many thanks!
On
It depends on what you call $\mathfrak p[x]$.
By definition, $(R/\mathfrak p)[x]$ is the ring of polynomials with coefficients in $R/\mathfrak p$, which is a full-fledged ring. So it's $A[x]$ with $A = R/\mathfrak p$, there is nothing special about it.
If you call $\mathfrak p[x]$ the subset of $R[x]$ made of polynomials in $R[x]$ whose coefficients are all in $\mathfrak p$, then yes, this is a prime ideal in $R[x]$ and the quotient is canonically isomorphic to $(R/\mathfrak p)[x]$.
This rings are not the same, the LHS are polynomials with coefficients in a quotient ring and the RHS are equivalence classes of polynomials. However, the two rings are isomorphic. It is a nice exercise to work this out.
Hint: Compute the kernel of $$\phi: R[x] \rightarrow \left( R/\mathfrak p \right) [x], \ \sum_{i=0}^n a_i x^i \mapsto \sum_{i=0}^n \overline{a_i} x^i.$$ Use this to establish the desired isomorphism.