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I have not found this question on the website, maybe because it easy, so if it has been asked before I will delete my question.

Let $A,B$ be two rings, $\mathfrak{a}$,an ideal of $A$ and $\mathfrak{b}$ is an ideal of $B$. Lastly, let $f: A \rightarrow B$ be a ring homomorphism. How can I show the following properties?

1.$\mathfrak{b}^{ce} \subset \mathfrak{b}$

2.$\mathfrak{a} \subset \mathfrak{a}^{ec}$

  1. seems trivial since $\mathfrak{a}^e=B\mathfrak{a}$ $ \implies f^{-1}(B \mathfrak{a}) =\mathfrak{a}^{ec}= kerf\mathfrak{a} \supset \mathfrak{a} $.

However, how can I show the first inclusion? I think it does not look "natural".

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Hint: use the fact that $f(f^{-1}(\mathfrak b))\subseteq \mathfrak b$.

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    Okay I take $x \in f(f^{-1}(\mathfrak{b}))$. Then $x= f(y)$ for some $y$ such that $y \in f^{-1}(\mathfrak{b})$ and $y \in f^{-1}(\mathfrak{b}) \implies f(y) \in \mathfrak{b}.$ Thank you! – 2017-01-01