# Proof

### $(1)$

• Let $I\lhd F$ be an ideal of a field $(F, + ,\cdot).$
• By definition of ideals, $I$ is not empty, for all $a,b\in I$ also $a-b\in I,$ and for all $i\in I$ and all $r\in F$ we have $ir\in I$ and $ri\in I.$
• It is clear from the definition of ideals, that if $I=\{0\},$ then $I$ is an ideal.
• Therefore, assume, $I\neq \{0\}.$
• Then, we can also choose $a\in I$ such that $a\neq 0.$
• Since $F$ is a field, there is a multiplicative inverse $a^{-1}\in F.$
• Therefore, $aa^{-1}\in I$ or $1\in I.$
• It follows $I=F.$

### $(2)$

• Assume, ring $(R, + ,\cdot)$ has exactly two ideals.
• Note that every zero ring has only one ideal.
• Therefore, $R$ is not a zero ring.
• Therefore, the two ideals in question of $R$ must be $\{0\}\lhd R$ and $R\lhd R,$ since these two subsets of $R$ are trivially ideals, by definition.
• In particular, $1\neq 0$ holds in $R$ and there is an $a\in R$ with $a\neq 0.$
• Consider the principal ideal $(a)\lhd R.$
• Since $(a)\neq\{0\}$ then $1\in(a),$ therefore $1=a\cdot a^{-1}.$
• We have just shown that every $a\in R$ with $a\neq 0$ has a multiplicative inverse $a^{-1}\in R.$
• It follows that $R$ is a field.

Github: ### References

#### Bibliography

1. Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013