# 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 $\displaystyle a \in R, MR[X]+(X+a)R[X]$ is a meximal ideal of R.

Proof so far.

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

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

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

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

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

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

Proof so far.

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

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

Thanks

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

Proof so far.

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

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

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

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

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

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

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

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