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Monday, May 13, 2013

Prime Ideal Squared (9.1.12)

Dummit and Foote Abstract Algebra, section 9.1, exercise 12:

MathJax TeX Test Page Let the overbar denote passage from Q[x,y,z] into Q/(xyz2). Prove that ¯P=(¯x,¯z) is a prime ideal. Show that ¯xy¯P2 but that no power of ¯y lies in ¯P2, showing that ¯P is a prime ideal whose square is not a primary ideal.

Proof: It is easy to check that for a surjective homomorphism φ, we have (φ(a1),φ(a2),...)=φ(a1,a2,...), so that (¯x,¯z)=(x,z)/(xyz2). Evidently (xyz2)(x,z) so that ¯Q[x,y,z]/¯(x,z)Q[x,y,z]/(x,z)Q[y] is an integral domain, so that ¯P is a prime ideal. Furthermore, ¯xyz2=¯0 so that ¯xy=¯z2¯P2 as claimed. Now, if ¯yn¯P2¯P for some nZ+ then we have the natural image of ¯yn in the previously mentioned quotient isomorphic to Q[y] is zero, an impossibility by the commutativity of natural projections. 

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