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I tried to prove that for a commutative ring $R$ with identity, every maximal ideal $I$ is prime. I think my proof was sort of going in the right direction, but this MSE answer was what I now realize that I was going for. I am wondering where exactly my proof goes wrong; I was already suspicious when I had realized that I never used the fact $ab \in I$ anywhere.

Let $I \leq R$ be a maximal ideal, and let $ab \in I$ where $a, b \in R$. Suppose that $a$ and $b$ are contained in $R$ but not in $I$. Without loss of generality, let $J = I \cup \{a\}$. Since $I$ is maximal and does not contain $a$, we must have $J = R$, so $1 \in J$. This means that $1 = rj$ for some $j \in J$ and $r \in R$. But since $1 \neq rx$ for any $x \in I$, we must have $1 = r(ax) = a(rx)$, which implies that $a$ and $rx$ are multiplicative inverses. But $a^{-1} = rx$ implies that $a^{-1}$ exists in $I$, meaning that $1 \in I$. This is a contradiction, so $J = I$ and $a \in I$. Therefore, $I$ is prime.

The initial set-up is identical to the MSE answer, so I think it is the part where I state that $1 = r(ax)$.

user1181399
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  • It’s not clear that $J$ is an ideal. – Daniel Feb 13 '24 at 20:32
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    $I\cup {a}$ is not generally an ideal. If $x\in I$, $x\neq 0$, then $a+x\notin I\cup{a}$, so it will not be even closed under addition (it is closed under addition if and only if $I={0}$ and $a=-a$). – Arturo Magidin Feb 13 '24 at 20:35
  • Oh, now I see why the linked MSE answer said "the ideal generated by $I \cap {a}$. Thank you! – user1181399 Feb 13 '24 at 22:13

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The proof fails at the step that $1=r(ax)$ for some $x \in I$. The first reason why this does not need to hold is that $(I,a)$ is the set of all elements $i+ax$, where $i \in I$, $x \in R$. We cannot conclude that $i=0$, which I believe is what you are doing implicitly here. Secondly, even if this was true we would only require $x \in R$, not $x \in I$.

Inbo Gottlieb-Fenves
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