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    Example of a UFD that is not a Euclidean domain

    P-VN6VF

    Area
    Algebra
    Topics
    Counterexamples, Euclidean Domains, Factorization
    Status
    draft

    Give an example of a UFD that is not a Euclidean domain.

    Solution

    Let \(k\) be a field. Then \[ k[x,y] \] is a UFD but not a Euclidean domain.

    First, \(k[x]\) is a PID and hence a UFD. By Gauss’s lemma, a polynomial ring over a UFD is again a UFD, so \[ k[x,y]=k[x][y] \] is a UFD.

    On the other hand, \(k[x,y]\) is not a PID. The ideal \[ (x,y) \] is not principal: if \[ (x,y)=(f), \] then \(f\) divides both \(x\) and \(y\). Since \(x\) and \(y\) have no common nonunit divisor, \(f\) must be a unit, which would give \[ (x,y)=k[x,y], \] contradicting \(1\notin(x,y)\).

    Every Euclidean domain is a PID. Since \(k[x,y]\) is not a PID, it cannot be Euclidean.

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