If $1\in M$, could $M$ be a maximal ideal of a commutative ring with identity $R$?

Question

If $1\in M$, could $M$ be a maximal ideal of a commutative ring with identity $R$?

I know that this is a very silly question. I think that the answer is that $M$ can't contain the identity for the product, because if it does, then $M=R$ and then it wouldn't be a maximal ideal.

Thanks in advance.

Answer 1

Your thoughts are on the mark.

For any ideal $I$ in any ring $R$, commutative or not, we have

$1_R \in I \Rightarrow I = R, \tag{1}$

since in the case $1_R \in I$ we have, for any $r \in R$,

$r = 1_R r = r1_R \in I. \tag{2}$

Now maximal ideals $M$ are generally defined to be proper, that is, $M \subsetneq R$; this precludes $1_R \in M$.

Answer 2

Yes, you are correct. Here is an algebraic manipulation of this.

Let $r \in R$,

$$1 \in M \Rightarrow 1 \cdot r = r \in M$$ So $R \subseteq M \Rightarrow R = M$