Any ideal of a field $F$ is $0$ or $F$ itself 
Prove that the only ideals of a field are $\left\{ 0 \right\}$ and the field itself.

Let $F$ be a field and $I$ be an Ideal of $F$.
Let $0 \ne x \in I$. Since $I$ is an Ideal of $F$, it is true that $I \subseteq F$ and so $x \in F$. $F$ is a field so there exists an element, say, $x^{-1}$ s.t
$x \cdot x^{-1} = e = 1 \in F$.
Any hints to keep me going?
Thanks in advance.
 A: Hint: Ideals are closed under external multiplication. For all $r \in F$, $x \in I$, $rx \in I$. Just apply this to the elements you obtained.
Complete Solution:
Let $I$ be any non-zero ideal of a field $F$. Then, $I$ contains at least one non-zero element $x$. Since $F$ is a field, $x^{-1} \in F$, which implies that $x^{-1} x = 1 \in I$ (since $I$ is an ideal of $F$). Therefore, $I = F$.
Appendix:


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*In any ring $R$, $\{0\}$ and $R$ are always ideals: $\{0\}$ and $(R, +)$ are always subgroups of $(R, +)$. For closure under external multiplication, observe that $\forall r \in R, r \cdot 0 = 0 \cdot r = 0 \in \{0\}$, and $\forall r \in R$ (the ring) and $s \in R$ (the ideal), $rs \in R$ (the ring and the ideal!).

*An ideal $I$ of a unital ring $R$ is a proper ideal of $R$ if and only if $1 \not\in I$. If $1 \not\in I$, $I$ is clearly a proper ideal (given that it is an ideal). Conversely, if $1 \in I$, then $\forall r \in R$, $r \cdot 1 = 1 \cdot r = r \in I$, which implies that $I = R$.

