Extension of Contraction of Extension of Ideal is Extension

Theorem

Let $f : A \to B$ be a ring homomorphism.

Let $\mathfrak a$ be an ideal of $A$.

Let ${\mathfrak a}^e$ be the extension of $\mathfrak a$ by $f$.

Let ${\mathfrak a}^{ec}$ be the contraction of ${\mathfrak a}^e$ by $f$.

Let ${\mathfrak a}^{ece}$ be the extension of ${\mathfrak a}^{ec}$ by $f$.

Then:

${\mathfrak a}^e = {\mathfrak a}^{ece}$

Since $\mathfrak a \subseteq {\mathfrak a}^{ec}$ by Ideal is Contained in Contraction of Extension:

$\ds {\mathfrak a}^e$

$\subseteq$

$\ds \paren { {\mathfrak a}^{ec} }^e$

$\ds $

$=$

$\ds {\mathfrak a}^{ece}$

$\ds $

$=$

$\ds \paren { {\mathfrak a}^e }^{ce}$

$\ds $

$\subseteq$

$\ds {\mathfrak a}^e$

$\blacksquare$