# Much Ado about Noetherian Rings

During a lecture series last year, I wanted to spice up the hour-long sessions we had. I spent a lot of time drawing, and tried writing notes in different languages. But during one such lecture I wrote my notes like a play, in rhyming couplets. After some vague interest from others, and for the sake of posterity, I thought I'd put them up here.

It's not a masterpiece. The meter is all over the place, there are a lot of half-rhymes and I'm not certain the maths holds up. But considering that I wrote it in fifty minutes with no preparation, I'm really proud of it. I was tempted to clean up some of the off-bits, but I've chosen to leave it totally unedited. This is very much the "without makeup" version.

So please enjoy what I've since named
**Much Ado about Noetherian Rings**. A play in one Act.

(Alternative titles: Measure for Zero Measure, The Comedy of Mathematical Misconceptions, The Taming of the Discontinuous Functions)

## Dramatis personæ

ANTONIO, a merchant

GAIUS, his brother

HERMIUS, a
nobleman

SERAFINA, a gentlewoman

## Prologue

From the next hour we hope you yield,

Indiscriminate knowledge of
the finite field.

And cyclic codes we hope you know,

For
their properties we're soon to show.

## Scene 1: A port of Florence

*(Enter ANTONIO and GAIUS)*

Antonio: Recall, brother, the characteristic of a field.

Gaius: The smallest positive integer such that as many ones that are summed make nought?

Antonio: The exact same. Remember thee that it is prime or infinite?

Gaius: Alas, I missed such a fact.

Antonio: Now form a field of sums of one, up until they add to
none.

The suppose that n is p times q, both primes I should be
telling you.

Gaius: Then a sum of n is a sum of primes, but these be nought in any
time!

Hence we find a contradiction, and conclude with some
conviction.

That n cannot be p times q, and only if it's
prime will do!

Antonio: Now theorise that in a field, then in it thy fate is
sealed.

There exists a number pn, with p a prime, the other in
ℕ.

I shan't prove this in great detail, but an outline surely
cannot fail.

Note ℤp is a linear subspace, and that it works with
an F-basis.

So any a in F can be written unique, amongst the
basis ℤp of which we speak.

Gaius: I'm afraid brother that I don't get it!

Antonio: Perhaps t'would help to see it writ.

Here I show you
an example, with f(x) an irreducible polynomial,

which has as n
its degree, and coefficients in ℤp.

Now let me fall and show to
you, the following is certainly true.

The answer, it is sealed,
this ℤp[x] / f(x)ℤp[x] is a field.

See, brother, that the powers of x can form a basis of cosets.

1, [x], [x2], [xn-1] certainly form a linear space on,

ℤp, our
set of coefficients, and to show they span you must be proficient.

Each
element it is in the sum (
αi[xi], αi ∈ ℤ), so the cosets span everyone.

Since each αi has
options p, there are pn choices we can see.

Now find a field with
elements 9, and waste not any of our time.

*(Enter HERMIUS, a nobleman)*

Hermius: What trivial games you merchants play, sitting by the docks
all day.

With 9 elements, take p as 3, and n as 2 so we can
see

that ℤ3[x] / f(x)ℤ3[x] is irreducible, in all f(x) of degree
2.

Gaius: And we know a polynomial with degree ≤ 3 only displays
irreducibility,

if it has no factor in F, and hence no root but
alone, bereft.

Hermius: Indeed, so after all this fun, take f(x) to be x2 1.

Antonio: But can you form me one with eight?

Gaius: Nay in time, unless we have 'til late.

Hermius: Foolish boy, this is easy, take p as 2 and n as 3.

Now
form the field shown here (ℤ3[x] / f(x)ℤ3[x]) and let f(x) appear.

This
f(x) is of degree 3 and in ℤ2[x] shows irreducibility.

Try
f(x)=x3 x 1, and see that f(x)=0 gives none.

In integers of
course!

*(Exit HERMIUS and enter SERAFINA)*

Serafina: Merchant men, if you'll hear me, let me provide to you a
little theory.

That if n is a number we call natural, then an
isomorphism will certainly fall.

Upon ℤ/nℤ and ℤn where ℤn has
modulus addition.

And that if F in K are both fields, with f(x)
irreducible in F[X]'s shield,

take α in K so that it
evaluated gives none, then note again an isomorphism.

Here of
F[x]/f(x)F[x] and the F[α] polynomial set.

Antonio: May we call F[α] a simple extension?

Serafina: Yes, of F. I forgot to mention.

I bid you proof, that
much I owe, and with the 1st isomorphism theorem I shall show.

Form
g from F[x] to F[α], and by ker(g) we can solve her.

Say F were ℝ
and K be ℂ, take α=√1 instantly,

Then ℝ[x] / (x2 1)ℝ[x] to ℂ they
will be isomorphics.

Gaius: Such odd remarks you make today, of field extensions you
continue to say.

But let us return to the matter of codes, and
see that cyclic ones can be showed.

Take a word in a code and if
it may be, that that at the end at the start we may see,

If this
be true for any code word, then to a cyclic code we have referred.

Note
too that there are more criteria, a cyclic code it must be linear!

Antonio: Tell me more of cyclic shift, a concept I've heard is of some thrift.

Gaius: It's much what it sounds like brother, a smooth connection
as that between lovers.

Take the last digit of a word and add it
to the start, a cyclic shift has occurred in such a lover's heart.

Serafina: Could a polynomial be moved by such a lover?

Gaius: Aye, my lady, and we'd discover

the coefficients are
what make the code, so 1 x2 x4 is 10101 showed.

Then a=10101 is a
ring, Rn (=F[x]/(xn-1)F[x]) is the notation we bring.

Serafina: What really can we see from this?

Gaius: That x*a(x) is a cyclic shift.

Serafina: You speak with such determination, could you show some explanation?

Gaius: Alas, time falls but trust my lady, I believe you can see it may be.

*(A priest gives Serafina a potion, making her appear dead. Gaius
flees, wrought with grief, and Serafina awakes cold and alone.)*

Image: Edwin Landseer – Scene from A Midsummer Night's Dream