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In a polynomial ring, the ideal generated by the indeterminate is prime precisely when the coefficient ring is an integral domain

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.7

Solution: We begin with a lemma.

Lemma: Let $R$ be a ring. Then $\varphi : R[x] \rightarrow R$ defined by $\varphi(p) = p(0)$ is a surjective ring homomorphism and $\mathsf{ker}\ \varphi = (x)$.

Proof: $\varphi$ is clearly surjective. Let $p(x) = \sum a_i x^i$ and $q(x) = \sum b_ix^i$. Now \begin{align*}\varphi(p+q) = &\ \varphi(\sum (a_i + b_i)x^i) \\=&\  a_0 + b_0 \\=&\ \varphi(p) + \varphi(q)\end{align*} and \begin{align*}\varphi(pq) =\ \varphi(\sum_k (\sum_{i+j=k} a_ib_j)x^k)\\ = a_0b_0 = \varphi(p)\varphi(q),\end{align*} so that $\varphi$ is a ring homomorphism. Now Suppose $p \in \mathsf{ker}\ \varphi$. Then we can write $p(x) = a_0 + xp^\prime(x)$, and $p(0) = a_0 = 0$. Thus $p(x) = xp^\prime(x) \in (x)$. Conversely, if $p(x) = xp^\prime(x)$, then $p(0) = 0 \cdot p^\prime(x) = 0$, and $\varphi(p) = 0$. Thus $\mathsf{ker}\ \varphi = (x)$. $\square$

By the First Isomorphism Theorem for rings, we have $R[x]/(x) \cong R$. This exercise then follows from Propositions 12 and 13 in the text.


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