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I have seen a lot the notation "$\mathfrak{p}R[x]$", where $\mathfrak{p}$ is a prime ideal and $R[x]$ is a polynomial ring with coefficients in $R$.

My first question is: What is $\mathfrak{p}R[x]$ exactly? I think that it is not the same as $\mathfrak{p}[x]$, whose elements are polynomials with coefficients in $\mathfrak{p}$.

Just to give some context, I am trying to prove that $1 + \dim(R) ≤ \dim(R[X]) ≤ 1 + 2 \dim(R)$. And the first thing that appears in every single proof I've seen is the prime ideals chain $$\mathfrak{p}_0R[x]\subsetneq\mathfrak{p}_1R[x]\subsetneq\dots\subsetneq\mathfrak{p}_nR[x]\subsetneq\mathfrak{p}_nR[x]+(x).$$ Somehow, there is a relationship between the ideals of $R$ and the ideals of $R[x]$, but I am not able to spot it.

Bernard
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Gerardo Vargas Flores
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    It denotes the extension of the prime ideal $\mathfrak p \in \operatorname{Spec}R$ to the polynomial ring $R[X]$, just as it can be extended to any $R$-algebra B (=(extension denoted $\mathfrak p B$). You can check, in the case of the polynomial ring, that it consists of the polynomials which have all their coefficients in $\mathfrak p$. – Bernard Nov 30 '20 at 18:38
  • But how is the extension defined? I mean, we have a particular ring homomorphism $\phi$ from R to R[x]? – Gerardo Vargas Flores Nov 30 '20 at 18:46
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    It's just the canonical injection. A constant is a polynomial of degree $0$. – Bernard Nov 30 '20 at 18:47

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That is a notation for the ideal generated by the set $\mathfrak{p}$. Let me specify that:

It gets used when you have a ring extension $R \rightarrow S$ and want to describe the ideal generated by the image of the a prime ideal $\mathfrak{p} \subset R$.

In this case, the extension is the inclusion $R \subset R[x]$.

Con
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