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EDIT: I know that there is a commonly known way of proving this, namely using the isomorphism theorem and the quotient $R[x]/(x)$ but I didn't get any feedback whatsoever on my proof, why is it valid/invalid? Can somebody please comment on that?

Let $R$ be a commutative unital ring.

The question is to prove that for an $x\in R$ we have "(x) is a maximal ideal" $\iff$ "$R$ is a field".

$(x)$ is the ideal generated by $x$ in $R[x]$, so $(x)=\{p_1x+...+p_nx^n\ :\ n\in\mathbb{N},\ p_i\in R\}$

My attempt:

If $R$ is not a field $\exists \tilde p_0\in R$ s.t. $\tilde p_0$ is not a unit.

Now if we look at (x), we have that since there is no constant element in $(x)\ \ $ $(x)\subsetneq\{k\tilde p_0+p_1x+...+p_nx^n\ :\ k,p_1,...,p_n\in R\}:=I\subsetneq R$ which is an ideal that is different from $R[x]$ because $k\tilde p_0$ can never be equal to $1$. So $(x)$ is not a maximal ideal.

for the other way:

If $(x)$ is not a maximal ideal, $\exists I\ne R$ s.t. $(x)\subsetneq I\triangleleft R[x]$ and because $I$ is a maximal ideal it cannot contain $1$ (otherwise we get $I$=$R[x]$) so the constant element of every element of $I$ has to be $\ne 1$ so there exists some element that stays $\ne 1$ no matter by what you multiply it $\implies R$ is not a field

I am not really sure about those reasonings so any comment is welcome.

Community
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John Cataldo
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    Of course you mean $(x)$ is a maximal ideal in $R[x]$. – Wojowu Mar 03 '18 at 21:34
  • @Wojowu yes, sorry – John Cataldo Mar 03 '18 at 21:35
  • @StanislasHildebrandt If it is not very obvious why polynomials with constant terms are not in the ideal generated by $x$, you should consider what causes you to doubt that and revisit the definitions of ideal, ideal generated by a set, and/or polynomial that you're are using if necessary. Then "rigorously" prove the statement. You really want to be at a point where that statement is intuitively obvious. – Derek Elkins left SE Mar 03 '18 at 21:52
  • @DerekElkins yes, I re-examined the definition of an ideal generated by a set and I have no doubt about that anymore – John Cataldo Apr 04 '18 at 23:26

1 Answers1

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There is a map $R[X]\to R$ that sends a polynomial $p$ to $p(0)$. The kernel of this map is $(X)$, and this map is onto, so there is an isomophism $R[X]/(X) \simeq R$. Thus $R$ is a field iff $R[X]/(X)$ is a field iff $(X)$ is maximal.

Pedro
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