The Qualifying Examination

This piece was taken from the Web, where it was claimed that "this article was scanned from an old photocopy. No publication medium and date is know. Let the webmaster know if you have additional information about it." Please do email me if you know more details.

The Qualifying Examination

Richard Roth, University of Colorado

Drama in one act with 4 characters: The Grand Alpha, The Grand Beta, The Grand Omicron, and The Candidate.

As the curtain rises, Alpha, Beta, and Omicron are seated in a classroom of a large university and the Candidate comes in.

ALPHA: Who enters?

CANDIDATE: I am the Candidate.

ALPHA: State your purpose.

CANDIDATE: I come in pursuit of mathematical knowledge. I am prepared in the fundamental fields, Algebra, Analysis, Anti-derivation. You may question me.

ALPHA: The candidate will please define what is meant by a continuous denominator.

CANDIDATE: Consider the set of all doubly evocative singly homologous functions on the unit sphere. Introducing a continuous group structure in the usual way we may define the Skolem uniformity of automorphic cycles to be the theta relation on all sets of measure zero and the zeta function on left ideals whose valuation is Gaussian, uniformly on compacta. Then given any cardinal predicate, the continuous denominator is the corresponding normal quaternion for which the problem vanishes almost everywhere.

BETA: Could the candidate please give an example of a non-Skolem uniformity?

CANDIDATE: I believe the inversion of the reals under countable intersections is non-Skolem... at least almost everywhere.

BETA: That's correct. Now could you...

OMICRON: (Interrupting) I wish to contradict. It isn't a non-Skolem uniformity since the third axiom concerning the density of the seventh roots of unity is not in fact satisfied.

BETA: Ah, yes, but you see, in my paper on toxic algebras... 1957... Journal of Refined Mathematics and Statistical Dynamics of the University of Lompoc... I showed that the third axiom need not be satisfied if the basis is countably finite and the metric is Noetherian, hence...

ALPHA: (Interrupting) Ahem, excuse me. The candidate will please prove the hokus-locus theorem on uniform trivialities.

CANDIDATE: By the Heine-Borel Theorem we reduce the Hamilton-Cayley equation to the canonical Cauchy-Riemann form. The Bolzano-Weierstrass property then shows that the Radon-Nikodym derivative satisfies the Jordan-Holder relation. Hence by the Stone-Weierstrass approximation we can get the Schroeder-Bernstein map to be simply separable. The Lebesgue-Stieltjes integral then satisfies the Riemann-Roch result when extended by the Hahn-Banach method almost somewhere.

BETA: Please define a compact set.

CANDIDATE: A set is compact if every covering by open sets has a finite sub-opening. I mean every opening by finite sets has a compact subcovering. Er... rather, every compact by an open finite has a subcover. I mean a finact combine subopen if setcover set everything. That is, almost some of the time.

ALPHA: Leave that for a moment. Instead could you give us an example of a compact set.

CANDIDATE: Uh, you consider the real line and take any bounded subset, I mean closed subnet, er, I mean complete subsequence... bounded elements...

BETA: For example, is an interval compact?

CANDIDATE: Yes... er, I mean no... that is sometimes... almost everywhere?... if it is finite... or rational, I mean the irrationals -- given a Dedekind cut -- er, all the numbers less than square root of 2 have a limit, that is...

OMICRON: Never mind. Look... is square root of 2 rational or irrational?

CANDIDATE: It's rational... I mean it's not rational... n 2 = 2 m 2 and all that... n less than m or I mean prime to 2... they're all integers of course.

ALPHA: What do you mean by integers?

CANDIDATE: Well... there's Peano's postulates or axioms and there's this element 1 and s (1) is 2 and s ( s (1)) and so forth. I think almost everywhere and uh... yes.

BETA: We have a feeling that you are not quite sure of the material. For example, how much is 2 added to 2?

CANDIDATE: Well, we have a binary operation +, defined by induction and we let 2 denote...

BETA: Never mind the proof... Just tell us the ordinary name of the integer which results from adding the integer 2 to the integer 2.

CANDIDATE: Er... uh... that sounds familiar. I remember: 2 generates a prime ideal in a Dedekind domain, which is ramified when...

ALPHA, BETA, and OMICRON in chorus: How much is 2 and 2? You learned it in the first grade?

CANDIDATE: Yes, oh yes... I just can't think... I really know it... let me see... the first grade, you say. That's right... 2 plus 2 is... Now first one plus one is two, one plus two is three, 8 times 8 is 65... Stuff like that. 2 plus 2 is 2 plus 2 is 2 plus 2 is....

ALPHA: That is quite enough. The examination is over. The candidate will write his name on the board while the committee deliberates on its decision.

(The candidate, chalk in hand, stands facing the blackboard, writes a few letters on the board, erases them, looks blankly around the room as the curtain falls.)

Note: This is a shortened version of a piece written in January 1960, when the author was a graduate student at the University of California, Berkeley.)