Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Pick out the true statements:

a. 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.

b. Let $R$ be as above and let M be an ideal such that $R/M$ is an integral domain. Then $M$ is a prime ideal.

c. Let $R = C[0, 1]$ be the ring of real-valued continuous functions on $[0, 1]$ with respect to pointwise addition and pointwise multiplication. Let $M = \{ f ∈ R \mid f(0) = f(1) = 0 \}$. Then $M$ is a maximal ideal.

Certainly (b) is true but no idea about (a) and (c).

share|cite|improve this question
up vote 2 down vote accepted


(a) For this one, use the fact that $R/M$ is a field $\iff$ $M$ is a maximal ideal. What happens if an ideal $I$ in $R$ contains a unit?

(c) For this part, remember that any maximal ideal is also a prime ideal.

Your ideal $M = \{f ∈ R \mid f(0) = f(1) = 0 \}$ is just the set of continuous function $f: [0, 1] \to \mathbb{R}$ such that $f(0) = f(1) = 0$.

Now, try to think of two continuous functions $f, g \in R$ such that their product satisfies $f(0)g(0) = f(1)g(1) = 0$, that is, $f\cdot g \in M$. Does this mean that $f \in M$ or $g \in M$?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.