My conjecture is motivated by the desire, given a field $F$, to find an extension field by "adjoining" elements. Let's say $\forall x\in F[x*x+1\neq0]$ and we want to "adjoin" an element $α$ satisfying $α*α+1=0$.

A typical abstract algebra book (like Artin, see p. 444) would probably accomplish this by defining the "extension field" $F(α):=F[x]/(x^2+1)$. The problem with this approach, however, is that we will not have $F(α)\supseteq F$.

Therefore, I propose the following conjecture and would like to know whether it is true or false.

Let $(F,+_F,*_F)$ be a field.

Then there exists a set $K\supseteq F$ and functions $+,*:K^2\to K$ satisfying the following properties:

(i) $+|_{F^2}=+_F$

(ii) $*|_{F^2}=*_F$

(iii) $(K,+,*)$ is a field

(iv) $\exists α\in K[α*α+1=0]$ (where 1 and 0 denote the multiplicative/additive identities of $F$)

I have included the set-theory tag because my question regards the existence of sets as described by the ZFC axioms.

  • $\begingroup$ Just asking to be sure. So for you we also do not have $\mathbb{N} \subset \mathbb{Z}$ as you do not want to identify things? Why is it a problem to identify? I am really curious. $\endgroup$ – ThorWittich Jul 8 at 18:41
  • $\begingroup$ @ThorWittich I guess it depends how we define $\mathbb{N}$ and $\mathbb{Z}$. It might seem too "pedantic" but, unless we define them in a way that guarantees every set in $\mathbb{N}$ is also in $\mathbb{Z}$, I would rather use more precise notation. If I remember correctly, Edwin Moise distinguishes $\mathbb{Q}$ from $\mathbb{Q^*}$ by defining $\mathbb{R}:=$ (dedekind cuts of elements of $\mathbb{Q^*}$) and $\mathbb{Q} \subseteq \mathbb{R}$ $\endgroup$ – Pascal's Wager Jul 8 at 18:58
  • $\begingroup$ You could also do the latter in your situation. $\endgroup$ – ThorWittich Jul 8 at 19:03
  • $\begingroup$ When writing a GUI code in Visual Basic, you're not particularly interested in the assembler opcodes which ultimately run it. Just that they exist, and that the code compiles to them. $\endgroup$ – Asaf Karagila Jul 8 at 19:56
  • $\begingroup$ @AsafKaragila Personally, I am more interested in assembly than Visual Basic (both in your analogy and for real!) For most purposes, the Visual Basic programmer does not need to know what the machine code for an instruction is. But there are some limitations. For instance, self-modifying code is much easier to write in assembly. And in Visual Basic, the programmer has no idea what will happen if a hacker increments his instruction by 1. Just because two programs do the same thing (most of the time) doesn't mean they are the same program! $\endgroup$ – Pascal's Wager Jul 8 at 20:29

You don't need to insist that a field extension $K/F$ actually satisfies $F \subseteq K$. It is enough that $F$ is embedded in $K$, via a ring homomorphism $F \to K$. This homomorphism must be injective because $F$ is a field. The theory of field extensions can be done beautifully and instructively using embeddings.

If you insist on strict inclusion, consider $K= (F \times F^*)\cup F$. Then $F \subseteq K$ and you can set $\alpha = (0,1)$. This construction is artificial, because it effectively identifies $F$ with the removed $F \times 0$.

  • $\begingroup$ @Ihf Wonderful! So $K$ is almost like the vector space $F^2$. Forgive me for being perhaps overly pedantic, but I do have a question about this approach. How do we know $F \times F^*$ and $F$ are disjoint? If they aren't disjoint, does the proof still work? $\endgroup$ – Pascal's Wager Jul 8 at 20:15

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