Exercise 2.7

Question

Let $\mathfrak{p}$ be a prime ideal in $A$. Show that $\mathfrak{p}[x]$ is a prime ideal in $A[x]$. If $\mathfrak{m}$ is a maximal ideal in $A$, is $\mathfrak{m}[x]$ a maximal ideal in $A[x]$?

Amit Awasthi · Accepted Answer

\begin{proof}
  Let $\pi\colon A[x] 	o (A/\mathfrak{p})[x]$ be the ring homomorphism defined by reducing coefficients $\bmod \mathfrak{p}$. This is clearly a surjective ring homomorphism, with kernel $\mathfrak{p}[x]$, hence $A[x]/\mathfrak{p}[x] \cong (A/\mathfrak{p})[x]$ as rings. Since $A/\mathfrak{p}$ is a domain, $(A/\mathfrak{p})[x]$ is a domain, hence $\mathfrak{p}[x] \subset A[x]$ is a prime ideal.
  \par Finally, $\mathfrak{m}[x]$ is not a maximal ideal in $A[x]$, for by the above, $A[x]/\mathfrak{m}[x] \cong (A/\mathfrak{m})[x]$, and $A/\mathfrak{m}$ is a field $k$, but $A[x]/\mathfrak{m}[x] \cong k[x]$ is not a field so $\mathfrak{m}[x] \subset A[x]$ is not a maximal ideal.
\end{proof}

Exercise 2.7

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Exercise 2.7

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