To show a certain ideal is maximal.

Question

I was asked the following question in my exam :

$R$ is a commutative ring with identity and $M$ is a maximal ideal of $R$. I needed to show that the ideal of $R[x]$ generated by $M$ and $x$, denoted as $(M,x)$ is maximal in $R[x]$.

My attempt : I kept trying to somehow have a homomorphism from $R$ to the quotient ring $R[x]/(M,x)$ whose kernel is $M$. If I would have succeeded, I could apply the 1st Isomorphism theorem which would imply that $R[x]/(M,x)$ is a field which in turn implies that $(M,x)$ is maximal. But, I kept struggling and couldn't come up with something constructive. Any help is appreciated. Thanks for your time.

Answer 1

Consider the homomorphism $$ R[x]\to R/M,\qquad f(x)\mapsto f(0)+M $$ and compute its kernel and image.

Answer 2

Consider the morphism $f:R[x]\to R[x]/(M,x)$ and the inclusion $g:R\to R[x]$. Then, $$ \ker(f\circ g) = g^{-1}(\ker(f))= \ker(f)\cap R=(M,x)\cap R=M $$ and as you explained, this gives you $R/M\cong R[x]/(M,x)$.

Answer 3

First we show $(M,x)$ is a proper ideal of $R[x]$.

Suppose instead that $(M,x) = (1)$.

Then $1 \in (M,x)$ implies $1 = m + xg(x)\;$for some $m \in M$, some $g \in R[x]$.

Since $M$ is maximal in $R,\, M$ is a proper ideal of $R$, hence $m \ne 1$.

Then

\begin{align*}&1 = m + xg(x)\\[4pt] \implies\;&1-m = xg(x) \end{align*}

contradiction, since $xg(x)$ can't be a nonzero constant.

It follows that $(M,x)$ is a proper ideal of $R[x]$, as claimed.

Let $f \in R[x]\setminus(M,x)$.

Then to show $(M,x)$ is maximal in $R[x]$, it suffices to show $(M,x,f) = (1)$.

Letting $f(x) = r + xg(x)$, for some $r \in R$, some $g \in R[x]$, we can't have $r \in M$, else $f \in (M,x)$.

Since $M$ is maximal in $R$, it follows that as an ideal of $R,\,(M,r) =(1),],$ hence also $(M,r) = (1)$ as an ideal of $R[x]$.

Then \begin{align*} &f(x) = r + xg(x)\\[4pt] \implies\;&r \in (x,f)\\[4pt] \implies\;&r \in (M,x,f)\\[4pt] \implies\;&(M,r) \subseteq (M,x,f)\\[4pt] \implies\;&(M,x,f) = (1) \end{align*}

It follows that $(M,x)$ is maximal in $R[x]$, as was to be shown.