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  "body": "\n## Statement\n\nLet $A \\subseteq B$ be commutative rings with $B$ [[integral over|B7X0KFY1S1]] $A$.\nLet $\\mathfrak{q} \\subseteq \\mathfrak{q}'$ be [[prime ideals|00MHHZ9SNE]] of $B$ with $\\mathfrak{q} \\cap A = \\mathfrak{q}' \\cap A$.\nThen $\\mathfrak{q} = \\mathfrak{q}'$.\n\n## Remarks\n\nLet $\\mathfrak{p} = \\mathfrak{q} \\cap A$ and localize at $\\mathfrak{p}$.\nThen $B_{\\mathfrak{p}}/\\mathfrak{q} B_{\\mathfrak{p}}$ is a domain integral over the field $A_{\\mathfrak{p}}/\\mathfrak{p} A_{\\mathfrak{p}}$, so it is a field by [[integral extensions of fields|5GCANV7337]].\nHence $\\mathfrak{q} B_{\\mathfrak{p}}$ is maximal, and the larger prime $\\mathfrak{q}' B_{\\mathfrak{p}}$ must equal it, which gives $\\mathfrak{q} = \\mathfrak{q}'$.\n\nA consequence: a prime $\\mathfrak{q}$ of $B$ is maximal if and only if $\\mathfrak{q} \\cap A$ is maximal in $A$.\n\nExample: in $\\Z \\subseteq \\Z[i]$, the primes $0 \\subsetneq (2 + i)$ contract to $0$ and $5\\Z$, which differ, as the theorem demands.\n\nIntegrality is needed: for $k[x] \\subseteq k[x, y]$, the primes $0 \\subsetneq (y)$ both contract to $0$.\n\n## Explanations\n\n### L1 human\n\nOver a fixed prime of the small ring there may be several primes of the big ring, but none of them lies inside another.\nIn geometry: the fibres of an integral map have dimension zero.\n\n### L2 human\n\nIncomparability says that chains of primes cannot get longer when one passes to an integral extension.\nWith [[going-up|GHKYR5PBQC]] it gives $\\dim B = \\dim A$.\nIt is also the reason that the maximal ideals of $B$ are exactly the primes over maximal ideals of $A$.\n",
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  "name": "Incomparability theorem",
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