1:51 AM
not in particular. it was the general case
"Now that we have discussed the concept of second order polynomial rings we will attempt to extend the concept of a polynomial ring to the third order to see whether or not they are still rings. We suppose that $x$ is any solution to the polynomial equation $x^3 + \alpha x^2 + \beta x + \tau = 0$. By the first fundamental theorem of algebra, there are three solutions. We can get the"
"relationship between this two solutions by letting $t$, $y$, and $z$ denote the three solution and plugging them in to obtain $(x - t)(x - y)(x - z) = x^3 + \alpha x^2 + \beta x + \tau$. Rearranging terms we then obtain $x^3 + (-t - y - z)x^2 + (ty + tz + yz)x - tyz = x^3 + \alpha x^2 + \beta x + \tau$. By comparing the coefficients of the polynomials we get that $-t - y - z = \alpha$, $ty + tz + yz = \beta$, and $-tyz = \tau$."
Now we can examine $Z[t]$. We've already established that for any number set formed by appending elements to the integers, we obtain a set for which the negation and addition rules hold true. So we only need to determine whether or not $Z[t]$ is closed under multiplication."
We know by the definition of appending elements that we can write two elements $m$ and $n$ as $m = a + bt$ and $n = c + dt$, where $a$, $b$, $c$, and $d$ are integers. Taking their product we get that $mn = ac + adt + bct + cdt^2$. Using the identity $t = -y - z - \alpha$ we get that $mn = ac + adt + bct + cd(-ty - tz) - \alpha cdt$. Then, using the second identity $-ty - tz = yz - \beta$ we get that $mn = ac + adt + bct + cdyz - \beta cd - \alpha cdt$.
However, we are now stuck. There is no conceivable way to write $yz$ as an integer multiple of $t$. However, while we played this out to see exactly how far we could get, in hindsight it makes sense that this might fail.
Going back to the first proof we know that either a solution is an integer or irrational so when we use the polynomial $x^3 - 3 = 0$ as an example, we can quickly conclude that $(3)^{\frac{1}{3}}$ is an irrational number. In this situation $yz$ would be $\pm (3)^{\frac{2}{3}}$ which cannot be an integer multiple (or even a rational multiple) of $(3)^{\frac{1}{3}}$ as that would imply that the solution to the polynomial was in fact an integer.
However, that does not exclude that $(3)^{\frac{2}{3}} = a + b(3)^{\frac{1}{3}}$, where $a$ and $b$ are integers.
However, the attempt above shows that it is highly unlikely to be the case and so we conclude that not all sets formed by appending solutions of polynomials of order greater than $2$ to the integers or the rational numbers form either rings or fields. Of course, the complexity on trying to reduce the multiplication into the form warranted by $Z[x]$ will only increase the more the more the order is increased.
It is highly likely that there is an order at which the only rings formed by appending solutions is the integers themselves and with the same being true for all higher orders after that.
@AkivaWeinberger read my above quote from my paper.
The odd town puzzle.
You have a town with $m$ clubs formed by $n$ citizens of the town.
The clubs are so formed that
Each club has an odd number of members.
Any two clubs have an even number of common members. (Could be zero too).
Show that $m \le n$.
