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A commutative ring with an ideal $I$ such that $I=I^2$

This is not true in general. Consider the ring $R=\mathbb{Q}[x,x^{1/2},x^{1/4},x^{1/8},\dots]$ as a subring of the algebraic closure of $\mathbb{Q}[x]$, and let $I$ be the ideal $\langle x,x^{1/2},\...
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Why are $(0)$ and $(1)$ the only ideals in a field?

If $I$ is an ideal in a field $K$ that contains some nonzero element $x$ then (since it's an ideal) $1 = xx^{-1} \in I$ so $I = K$.
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