Extension and contraction of ideals

For a ring homomorphism $f:R\to S$, contraction sends a two-sided ideal $J\subseteq S$ to $f^{-1}(J)$. Extension sends a two-sided ideal $I\subseteq R$ to the two-sided ideal generated by $f(I)$, consisting of finite sums ${\sum }_i{s}_{i}f(a_i)t_i$ with $a_i\in I$ and $s_i,t_i\in S$. For commutative rings, this simplifies to $Sf(I)$.

They are different operations: extension moves generators forward, while contraction takes a preimage. One always has $I\subseteq f^{-1}(I^e)$ and $(J^c{)}^e\subseteq J$, but equality need not hold. For example, extending $(2)$ along $\mathbb{Z}\to \mathbb{Q}$ gives the unit ideal, whose contraction is all of $\mathbb{Z}$. The prime ideal correspondence for localization gives a setting where the appropriate extension and contraction operations are inverse.

For a ring homomorphism $f:R\to A$, the extension $I^e$ of a two-sided ideal $I\subseteq R$ is the smallest two-sided ideal of $A$ containing $f(I)$. Thus its elements are finite sums ${\sum }_j{a}_{j}f(i_j)b_j$ with $i_j\in I$ and $a_j,b_j\in A$.

When $A$ is commutative, the formula becomes $I^e=IA=Af(I)$. For example, extending $(2)$ from $\mathbb{Z}$ to $\mathbb{Z}[i]$ gives the multiples of two in the Gaussian integers; extending it to $\mathbb{Q}$ gives the whole field because two is invertible there. Extension can therefore enlarge an ideal substantially even when the original ring map is injective.

For a ring homomorphism $f:R\to A$ and an ideal $J\subseteq A$, its contraction is $J^c=f^{-1}(J)$.