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Let $R$ be a commutative ring with identity. Show that if there exists a monic polynomial $p(x)\in R[x]$ of degree $\geq 1$ such that the ideal $(p(x))\subset R[x]$ is maximal, then $R$ is a field.

My attempt: I understand that since $(p(x))$ is a maximal ideal, then $R[x]/(p(x))$ is a field. I also understand that $p(x)$ must be irreducible, else if $p(x)=g(x)h(x)$ then the ideal $(p(x))$ could not be maximal as it is contained in $(g(x))$.

Working with this, I want to show that for every element $x\in R$, there exists some $y$ such that $xy=1$. I'm thinking I can use a map from $R$ to $R[x]/(p(x))$ and somehow show that the product $xy$ gets mapped to the identity. I'm not at all sure about this though, is this a good direction?

Any help appreciated!

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Hint Suppose $\,0\ne r\in R.\,$ Since $p$ is monic it cannot divide a smaller degree nonzero polynomial, so $\,p\nmid r,\,$ so $(p)$ max $\,\Rightarrow (p,r) = 1\,$ $\Rightarrow$ $\, pf + r q =1\, $ $\Rightarrow$ $\,p\mid 1\!-\!rq\,$ $\Rightarrow$ $\,p\mid 1\!-\!r\bar q,\,$ $\,\bar q = q\bmod p.\,$ $\,1\!-\!r\bar q$ has smaller degree than $p\,$ so $\,r\bar q = 1,\,$ so $\,r\bar q(0) = 1.\,$ Thus $\,r\neq 0\,\Rightarrow\, r\,$ is invertible.