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May 20th

May 20th

# Math 4606 Special Problem 1 - a partial solution

Assignments are due at the start of class on the given Due date.

Please Note! Special problems are liketerm papers .” They must be well-written, in ink, on standard 8.5 x 11
paper, and must be succint - with exactly enough detail.

Paper torn from spiral notebooks is not acceptable.
Err in the direction of slightly excessive detail at first,
but prolixity is not acceptable.

Solution for the secondpart .

We have to define addition and multiplication so that the field axioms are satisfied. There are some things we can

• 0 and 1 have to act as identities for + and ×
thus each table will have an appropriate “repeated row”
• Since 0 × a = 0 for all a, the × table will have
a row and column of zeroes
• Inverses are needed, so each (appropriate) row and column
• has to contain just one copy of each (appropriate) element

This means we get the following tables, with ◦ denoting an element to be filled in:

In the × table, we have to put a 1 or 3 in the first place to fill in (row 4, column 4)., because there’s already a 2
in the row and the column. If we try to put a 1 there, then we have to put a 3 in the last column, and that won’t
work, because there’s already a 3 in column 5. So we have to put a 3 then a 1. Then in the last row we have to put
1 and 2. So now we know what the multiplication table has to be:

Now, how do we fill in the addition table so that it works, and so that both tables work together? We can take
advantage of the fact that proved, in the first part of the problem, that we had a field with two elements . That field
is usually called Z2, and its operations are exactly what we get when we add and multiply in the usual way, but we
only record whether the answers are even (i.e., zero) or odd (i.e., one). We are now going to use the idea of how to
make the complex numbers , by considering polynomials ax+b, ax^3+bx^2+cx+d and so on, where x is a variable
that we will later give a meaning to. (In the complex numbers, we do this, using i instead of x, and we require
that i^2 = −1.) We define equality of two of these polynomials by demanding that the coefficients of corresponding
powers are equal .

Now we define addition and multiplication of these polynomials by:

We followed this rule when writing these definitions: treat all quantities, including x, as commuting and associating
quantities, with respect to addition and multiplication, then group them as a sum of coefficients (from Z2) times
powers of x. This immediately gives commutativity, by inspection of the formulas, keeping in mind the question
“what happens when we interchange a and c, b and d?” In particular, we may have to write down the steps
showing that ad+bc = cb+da. Later we will replace x2 by Ax+B and rearrange again, to get a “quantity” of the
form gx+h. If necessary, we replace x3 by xx2 = x(Ax+B) = Ax^2 +Bx = A(Ax+B)+Bx = A2 +Bx+AB =
Bx+(A2+AB). Higher powers would be treated similarly, but we won’t encounter them. All our coefficients here are
elements of the field z2 from the first part of the problem, and the ope rations of addition and multiplication involving
pairs of those elements are the operations from the field z2. For example, (x+1)+(0x+1) = x+(1+1) = x+0 (= x)
and (x+1)×x = x2+x. When we later replace x2 by Ax+B, we will have x2+x = (Ax+B)+x = (A+1)x+B.

Because of the way we defined addition, we can take advantage of associativity in Z2 to check associativity here:

and, grouping the other way,

Either grouping thus gives the same result, so addition of these polynomials is associative.
We can handle multiplication the same way, but we get higher powers of x:

and, grouping the other way, following the rules,

Either grouping thus gives the same result, so multiplication of the polynomials of the form ax + b is associative.

Next we match polynomials with our symbols 0, 1, 2, 3. We will match 0x + 0 = 0 with 0 and 0x + 1 = 1 with
1. It seems natural to match 1x + 0 = x with 2 and match 1x + 1 = x + 1 with 3. Does this work – when we
multiply these polynomials together, do we get the right matches with the multiplication table that we that we know
is the only one possible? Because of commutativity we only have to check that the matchings for 2 × 2, 2 × 3 and
3 × 3 result in matches for 3, 1 and 2, respectively.

We get:2 × 2 corresponds to x2, 2 × 3 corresponds to x(x + 1) = x2 + x and 3 × 3 corresponds to

This is not a match yet! What we need is to see whether we can arrange for what we got, x2, x2 + x and x2 + 1,
respectively, to match 3, 1 and 2, respectively. Considering the matches we started with, we thus ask:is it true
that

x2 = x + 1 and x2 + x = 1 and x2 + 1 = x,

because the left-side polynomials are what we got by multiplying the polynomials as signed to the products 2×2 = 3
and 2 × 3 = 1 and 3 × 3 = 2, and the right-side polynomials are the polynomials assigned to 3, 1 and 2,
respectively.

So now we will replace x2, whenever it appears, by x + 1 (from the first equation). This makes x2 = x + 1 true.
In the second equation, the left side becomes x2 + x = (x + 1) + x = (1 + 1)x + 1 = 0x+1 = 1, which agrees with
the right side. Thus setting x2 = x + 1 makes x2 + x = 1. We next substitute x + 1 for x2 on the left side in
the third (hoped for) equation: x2 + 1 = (x + 1) + 1 = x + (1 + 1) = x + 0 = x, and so the equation x2 + 1 = x
is also true, so the multiplication table for our polynomials matches. We can now fill in the missing spots in the
addition table by adding polynomials in the usual way: we get 1+1 = 0, 1+2 ↔ (0x+1) + (1x+0) = x+1 ↔ 3,
1+3 ↔ (0x+1)+(1x+1) = x+(1+1) = x+0 = x ↔ 2, 2+1 = 1+2 = 3 by commutativity, 2+2 ↔ x+x = 0,
2 + 3 ↔ x + (x+1) = (x + x) + 1 = 1, and 3 + 3 ↔ (x + 1) + (x+1) = (x + x) + (1 + 1) = 0. The skipped cases
are OK, by commutativity.

On the set {0, 1, 2, 3} we can now define addition and multiplication by the following tables:

We have shown already that addition and multiplication are commutative and associative, and this is reflecred in
the tables. The addition and multiplication tables show the existence of additive and multiplicative identities, and
the existence of inverses (each row and column in the addition table contains all 4 elements, each row and column
of the multiplication table that involves non-zero elements contains all three non-zero elements). We know also that
1 ≠ 0, from the tables. It remains to prove the distributive law . We do this abstractly, returning to the polynomials,
remembering that we have to replace x2 by x + 1:

which simplifies to

simplifies to
(ax + b)((cx + d) + (ex + f)) = (ac + ae)(x + 1) + (ad + af + bc + be)x + b(d + f)
= (ac + ae + ad + af + bc + be)x + (ac + ae + bd + bf).

And we have

By tedious inspection of the two (using “extended” associativity and repeated commutativity), we see that distributivity
holds. Hence this system is a field.

There are other possibilities for an addition table, for example, arithmetic modulo 4. But that does not satisfy the
distributive law, since 3(2 + 2) = 3 · 0 = 0 and 3 · 2 + 3 · 2 = 1 + 1 = 2 if we use addition modulo 4 instead of the
addition from our table, which gives 3(2 + 2) = 3 · 0 = 0 and 3 · 2 + 3 · 2 = 1 + 1 = 0.

There is a field with N elements if and only if N is a power of a prime. Thus there is no field with six elements.

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