# Thread: Maximal Ideal Problem within a Field

1. ## Maximal Ideal Problem within a Field

Let M be a maximal ideal of a commutative ring R with identity. Prove that for each $a \in R, MR[X]+(X+a)R[X]$ is a meximal ideal of R.

Proof so far.

Suppose that $MR[X]+(X+a)R[X] \subseteq J \subseteq R$, and I want to show that J = R.

Or should I process another way, prove that J = $MR[X]+(X+a)R[X] \subseteq J \subseteq R$?

2. Originally Posted by tttcomrader
Let M be a maximal ideal of a commutative ring R with identity. Prove that for each $a \in R, MR[X]+(X+a)R[X]$ is a meximal ideal of R[X] .
$I=MR[X] + (X+a)R[X]$ is an ideal of $R[X]$ not $R$. suppose $J$ is an ideal of $R[X]$ and $I \subsetneq J \subseteq R[X].$ choose $f(X) \in J-I.$ then $f(X)=(X+a)g(X) + c,$ for some $c \in R$ and

$g(X) \in R[X].$ why? now $X+a \in I \subset J$ and $f(X) \in J.$ thus $c=f(X)-(X+a)g(X) \in J.$ but $c \notin M$ because otherwise $c \in I$ and thus $f(X) \in I,$ which is not possible. so since M

is maximal, we must have $M+Rc=R.$ hence $m + rc = 1,$ for some $r \in R, \ m \in M.$ but then: $1=rf(X)-r(X+a)g(X)+m \in J.$ why? therefore: $J=R[X]. \ \ \Box$

3. Thanks for your help!

So now, I want to show that $MR[X]+(X+a)R[X]=MR[X]+XR[X]$ if and only if $a \in M$

Proof so far.

Suppose that $a \in M$, $MR[X]+(X+a)R[X]=MR[X]+XR[X]+aR[X]$, now, does $aR[X]$ get absorbed by $MR[X]$ since a is in M?

The converse would be the same, isn't it?

Thanks

4. Originally Posted by tttcomrader
Thanks for your help!

So now, I want to show that $MR[X]+(X+a)R[X]=MR[X]+XR[X]$ if and only if $a \in M$

Proof so far.

Suppose that $a \in M$, $MR[X]+(X+a)R[X]=MR[X]+XR[X]+aR[X]$, now, does $aR[X]$ get absorbed by $MR[X]$ since a is in M?

The converse would be the same, isn't it?

Thanks
your proof is not correct, because in general: $(X+a)R[X] \neq XR[X]+aR[X].$ the reason is that an element of the LHS is in the form $(X+a)f(X)$ but an element of the RHS is in the form

$Xf(X) + ag(X).$ so in general we only have: $(X+a)R[X] \subseteq XR[X]+aR[x].$ here's how to solve your problem: first if $a \in M,$ then we'll have:

$MR[X]+(X+a)R[X] \subseteq MR[X] + XR[X]+aR[X] = MR[X]+XR[X],$ because $a \in M.$ on the other hand we have: $(-a)R[X] \subseteq MR[X].$ thus:

$MR[X]+XR[X] \subseteq MR[X]+(X+a)R[X]+(-a)R[X] = MR[X]+(X+a)R[X].$ this proves half of your problem. now suppose $MR[X]+(X+a)R[X]=MR[X]+XR[X].$ then since we

have $X+a \in MR[X]+(X+a)R[X],$ we must also have $X+a \in MR[X]+XR[X].$ hence: $X+a=\sum_{i=0}^k m_i X^i + X\sum_{i=0}^{\ell}q_iX^i,$ for some $q_i \in R, \ m_i \in M.$ so the constant coefficients in

both sides must be equal. thus: $a=m_0 \in M. \ \ \Box$