Let $R$ be a commutative ring with identity. Let $M$ be an ideal such that every element of $R-M$ is a unit. Then $R/M$ is a field.
Solution 1
If $f:A\to B$ is an homomorphism of rings, then $f(a)$ is a unit in $B$ whenever $a\in A$ is a unit —indeed, it is clear that $f(a^{-1})$ is an inverse for $f(a)$.
Now, in your situation, we have the canonical projection $\phi:R\to R/M$, which is a ring homomorphism, and every non-zero element of $R/M$ is the image under $\phi$ of an element in $R$ which is not in $M$. The above observation then immediately shows that $R/M$ is a field.
As you probably know, this implies that $M$ is maximal.
Solution 2
Hint: an ideal containing a unit is the whole ring; if an ideal properly contains $M$, it contains a unit. Therefore $M$ is…
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niti
Updated on May 31, 2020Comments
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niti over 3 years
Let $R$ be a commutative ring with identity. Let $M$ be an ideal such that every element of $R$ not in $M$ is a unit. Then $R/M$ is a field.
I am solving this question of NBHM 2011. To solve this is $M$ must be maximal ideal? If so, how we can solve. Please help.
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coolpapa over 9 yearsYou are correct that $M$ must be a maximal ideal in this case. How could you show that the given condition implies that $M$ is maximal? (Hint: start with the definition of maximal ideals.)
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