If $I$ is a maximal ideal of $R$, why is $R/I$ a field?

If $I$ is a maximal ideal of $R$, why is $R/I$ a field?

I'm trying to use the fact that $I$ is maximal to show that $R/I$ only have ideals $\{0\}$ and $R/I$. Can anyone help me with this method. Many thanks.

I believe several people have already told you of some relation between ideals in the ring $R$ and the quotient $R/I$ of a maximal ideal $I$ in $R$. It is called the Lattice Theorem:

If $R$ is a commutative ring and $I$ an ideal of $R$, let $\phi : R \longrightarrow\!\!\!\!\!\!\!\!\to R/I$. Then there is a one-to-one order-preserving correspondence between ideals $\mathfrak{b}$ of $R$ that contain the $I$ and ideals $\mathfrak{\overline{b}}$ of $R/I$. The correspondence is given by $\mathfrak{b}$ being the inverse image of $\mathfrak{\overline{b}}$.

Now you already know that a commutative ring $R$ is a field iff it has no non-trivial ideals. So suppose that $R/I$ is not a field and $I$ a maximal ideal of $R$. Then since $R/I$ is not a field there is a proper ideal $\mathfrak{\overline{a}}$ of $R/I$. But then the inverse image of $\mathfrak{\overline{a}}$ in $R$, say $\mathfrak{a}$ must contain $I$ by the Lattice Theorem above. But then by maximality of $I$ either $\mathfrak{a} = I$ or $\mathfrak{a} = R$. This contradicts $\mathfrak{\bar{a}}$ being a proper ideal of $R/I$ so that $R/I$ is a field.

You should now attempt the following problems to strengthen your understanding of taking quotients of maximal ideals:

All rings are commutative with a unit.

1) Suppose that $R$ is a domain and $p$ a prime element of $R$. Prove that $p$ is also a prime element in $R[x]$. Hint: look at $R/(p)$.

2)Atiyah - Macdonald Problem 1.12 Prove that a local ring has no non-trivial idempotents.

Regards.