field-theoryring-theory
Let $P = \{a + b\sqrt[3]3 + c\sqrt[3]9, a, b, c \in \Bbb Z \}$
It is easy to prove that $(P, +, \cdot)$ is a ring considering ordinary addition and multiplication. How to prove that this set is or is not a field? I am stuck. I think that showing that there is an inverse element (or there is not) for all elements is the key, but I don't know how to do it. I tried "rationalizing"
$\frac{1}{a + b\sqrt[3]3 + c\sqrt[3]9}$ but got stuck. Not really good with this stuff. Any help appreciated.
As mentioned in the comments, the term 'ring' came from Hilbert (previous question). Bill's answer there is excellent.
Fields are a bit funnier. It started with Dedekind using the word "Zahlenkörper" (body of numbers). In a supplement that he wrote to Dirichlet's Vorlesungenueber Zahlentheorie, he used that word (looking for an image for that) instead of 'rationally known quantities.' It was Moore who coined 'field' - in 1893, he wrote on Galois fields for the Bulletin of the New York Mathematical Society. However, he was always careful to mention 'fields of order' something, as field had the additional meaning at the time of neighborhood (as we understand it today).
I believe he used 'body of numbers' because they stay together just as a group does. But the transition to 'field' is a harder one to understand. Perhaps he meant it as an extension of the neighborhood-field idea - these numbers stayed in a certain neighborhood of a set of numbers in the sense that they were only a couple of powers of s away (referring to his passage below). Or perhaps he meant in to be similar to a 'body', in the sense that they just stay together.
In any case, if there were a strong, direct metaphor for 'ring' or 'field,' it is no longer readily apparent. The origins of the terms are difficult to track and harder to understand completely, and different people used the same terms in different ways. It is even possible that little thought was given to the naming of these things - it is most often hard to judge the importance of something new. Reaching for the term 'body' or 'field' because it fits and doesn't already mean something else might be the real origin.
Here is part of his piece on Galois fields for the Bulletin of the NYMS
Later, Edward Huntington wrote about the development of the theory of fields (it's funny how today we think of fields and groups as being basic, though groups are far older)
Here is an excerpt from Huntington's piece:
Huntington:
with footnote - indicating the well-understood meaning of field today:
The axioms for a ring already imply that $0\cdot a=0$ for all $a\in S$, so you have almost all of the multiplication table:
$$\begin{array}{c|cc} \cdot&0&1&2\\ \hline 0&0&0&0\\ 1&0&1&2\\ 2&0&2 \end{array}$$
And $2$ has to have a multiplicative inverse, so we must have $2\cdot 2=1$.
For addition we automatically have this much:
$$\begin{array}{c|cc} +&0&1&2\\ \hline 0&0&1&2\\ 1&1&&\\ 2&2& \end{array}$$
Now $1$ must have an additive inverse, so either $1+1=0$, or $1+2=0$. Suppose that $1+1=0$; then what is $1+2$? If $1+2=0$, then $1=1+0=1+(1+2)=(1+1)+2=0+2=2$, which is absurd. If $1+2=1$, then $0=1+1=1+(1+2)=(1+1)+2=0+2=2$, which is also absurd. Thus, $1+2=2$. But we know that $2$ has an additive inverse $-2$, even if we don’t yet know what it is, so $1=1+\big(1+(-2)\big)=(1+2)+(-2)=2+(-2)=0$, which is also impossible. Thus, $1+1$ cannot be $0$. It also can’t be $1$ (why not?), so it must be $2$, and we have
$$\begin{array}{c|cc} +&0&1&2\\ \hline 0&0&1&2\\ 1&1&2&\\ 2&2& \end{array}$$
Clearly $2$ must be $-1$, giving us
$$\begin{array}{c|cc} +&0&1&2\\ \hline 0&0&1&2\\ 1&1&2&0\\ 2&2&0 \end{array}$$
and it’s not hard to check that $2+2$ can now only be $1$.
Added: By the way, that last bit can be shortened to practically nothing if you recall that up to isomorphism there is only one group of order $3$, $\Bbb Z/3\Bbb Z$: that gives you the addition table right away.
How to see that $S$ cannot be an ordered field depends on how you’ve defined ordered field. If you’ve defined it in terms of a positive cone, note that $1+1=2=-1$, so $1$ can’t be in the positive cone: it’s closed under addition and never contains a non-zero element and its additive inverse. Similarly, $(-1)+(-1)=2+2=1$, so $-1$ can’t be in the positive cone, either. But this is also impossible: one of them has to be in it.
If you’ve defined it in terms of an order relation, you can get essentially the same contradictions. If $0<1$, then $1=0+1<1+1=2$, $2=1+1<2+1=0$, so by transitivity $0<0$; OOPS! A similar problem arises if $1<0$.
Best Answer
$P = \mathbb Z[\alpha]$, where $\alpha=\sqrt[3]{3}$. Since $P$ is the image of $\mathbb Z[X]$ under $X\mapsto \alpha$, it is clear that $P$ is a ring. Now, $2\in P$ but $1/2 \not\in P$ because $1,\alpha,\alpha^2$ are linearly independent over $\mathbb Q$. So, $P$ is not a field.