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  "body": "\n## Definition\n\nLet $A$ and $B$ be [[commutative rings|0GYGHKX834]] and let $f \\colon A \\to B$ be a [[ring homomorphism|0GGE0Z0JQ1]].\n\n1. For an [[ideal|0MYSA6MYNA]] $I$ of $A$, the *extension* of $I$ is the [[ideal generated|030TZ0DQWT]] by $f(I)$ in $B$:\n$$I^e = f(I) B = \\set{\\textstyle\\sum_{k=1}^{n} b_k f(a_k) : n \\ge 0,\\ a_k \\in I,\\ b_k \\in B}.$$\n2. For an ideal $J$ of $B$, the *contraction* of $J$ is the preimage $J^c = f^{-1}(J)$, an ideal of $A$.\n\n## Remarks\n\nThe relations $I \\subseteq I^{ec}$, $J^{ce} \\subseteq J$, $I^{e} = I^{ece}$ and $J^{c} = J^{cec}$ hold.\nIf $J$ is a [[prime ideal|06HC6HK0Z8]] of $B$, then $J^c$ is a prime ideal of $A$, because $A/J^c$ embeds into the integral domain $B/J$.\nExample: for the inclusion $\\Z \\to \\Z[i]$, $(2 + i)^c = 5\\Z$, since $(2 + i) \\cap \\Z$ contains $5 = (2+i)(2-i)$ and is a proper ideal containing $5\\Z$.\nNon-example of the analogous property for extension: $2\\Z$ is prime in $\\Z$, but $(2\\Z)^e = 2\\Z[i]$ is not prime, since $(1 + i)^2 = 2i$ lies in it while $1 + i$ does not.\nThe image $f(I)$ itself need not be an ideal of $B$; for $\\Z \\subseteq \\Q$, the image of $2\\Z$ is not an ideal of $\\Q$, and its extension is all of $\\Q$.\n",
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