Here is the definition my professor gave to us:
A commutative ring $R$ is local if it has a unique maximal ideal $\mathfrak{m}.$In this case, we say $(R, \mathfrak{m})$ is a local ring. For example, if $R$ is a field, then $(R,(0))$ is a local ring, since the only proper ideal of a field is $(0).$
My question is:
Why the only proper ideal of a field is $(0),$ could anyone explain this to me please?
If $0\ne a\in\mathfrak a\subset F$ then for all $c\in F$, we have $c=\frac caa\in \mathfrak a$ and so $\mathfrak a=F$.
If $F$ is a field and $I$ is an ideal, suppose $j \neq 0 \in I$. Then since $I$ is an ideal, for any $x \in F$, we have $$ x = (xj^{-1})j \in I. $$ So either $I = (0)$ or $I = F$.
Suppose $I$ is an ideal of a field $F$. We want to show that either $I = F$ or $I = {0}$. We have two cases to consider:
Case1: Suppose $I$ has no non-zero element, then $I=0$
Case2: Suppose $I$ has a non-zero element, say $x$. Then, of course, $x$ lies in the field $F$. Hence, $x$ has a multiplicative inverse. Since, the ideal is closed under multiplication $x \times x^{-1} = 1$ must in the ideal. This means that, $I = F$, again due to closedness of the ideal with respect to the multiplication.
