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Why cannot $\frac{0}{0}$ be added to a field extension to complex field $\Bbb C$? I know the complex field $\Bbb C$ is closed. But can we define additional elements such as $1*\frac{0}{0}$, $2*\frac{0}{0}$, $i*\frac{0}{0}$ and extend $\Bbb C$ and study its properties?

zipirovich
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Gopal Anantharaman
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  • @BranimirĆaćić I'm guessing that OP wants to define $0/0$ as a separate element. But I still feel like that would be fraught with issues. – Jam Aug 13 '18 at 17:28
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    Argh, thanks, I need to think more carefully for a second. An element $\frac{0}{0}$ would at least need to satisfy $\frac{0}{0} \cdot 0 = 0$ to justify the notation, but $z \cdot 0 =0$ in the field of formal Laurent series $\mathbb{C}((z))$, which is a perfectly good field extension of $\mathbb{C}$. I guess the million dollar question for the OP is: do you want $w \cdot \frac{0}{0} = \frac{w \cdot 0}{0} = \frac{0}{0}$ to hold for any $w \in \mathbb{C}$, for some $w$, or for no $w$? – Branimir Ćaćić Aug 13 '18 at 17:36
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    @BranimirĆaćić it seems galling at first but $\frac 00$ would have the trivial property that $\frac 00* 0 = 0$ and nothing else. I think zambawithkolbasa's answer is correct. We can do this but it doesn't "mean" anything. If we let $w\frac 00 = \frac{w0}0$ then $\frac 00 = 0$. But we could just as easily say $w\frac 00=w$ and $\frac 00 = 1$. heck for that matter we could say $w\frac 00 = 2w$ and $\frac 00 = 2$. What we can't do I think is define any $\frac 10$ or any $0*w = k \ne 0$. – fleablood Aug 13 '18 at 17:44
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    what would be the multiplicative inverse of $0/0$? – AlvinL Aug 13 '18 at 17:49
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    You might be interested in wheel theory. – Wojowu Aug 13 '18 at 18:10
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    Duplicate of https://math.stackexchange.com/q/1370214/124095. – mweiss Aug 13 '18 at 18:12
  • The answer at https://math.stackexchange.com/a/1370342/124095 is relevant here. – mweiss Aug 13 '18 at 18:13
  • Thanks a lot for elucidating me. I was merely trying to find out if these "indeterminate holes" in continuous, possibly multivariate functions could be plugged by including $\frac{0}{0}$ in the Complex field. If we could then we could find the "uber Field" which has truly continuous functions – Gopal Anantharaman Aug 13 '18 at 18:29
  • The Complex field extensions, if definable, can be for Cantor's Transfinite numbers as well. This would possibly pave the way for computing problems such as P=NP to be addressed as I strongly feel that these sets could have transfinite cardinality and proving P = NP could be merely proving that that the cardinality are either equal or not... – Gopal Anantharaman Aug 13 '18 at 18:42

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Yes, you can add that as an element and get $C\left(\frac{0}{0}\right)$, the field of rational functions over $C$. That is what you obtain by adding an "indeterminate element" to a field.

The question is if you want to require any relations between $\frac{0}{0}$ or its powers and other elements of $C$.

Some people prefer to call it $x$ if no other relations are required.

  • Thanks! Yes I think it is meaningful to have relations of $\frac{0}{0}$ with other elements impossible with the traditional Field operators i.e "" and "+" under traditional sense. The question is, whether there is an operator that can work with all elements and still can extend to indeterminate. I know it is silly but I don't know if the "operator" defined can be a conglomeration or composition of the numerical operators(+, , , etc.) performed in certain order on the two elements that finally results in a closed Group( it's no longer a Field in the traditional sense) with Inverse. – Gopal Anantharaman Aug 13 '18 at 19:36
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If you consider $\frac00$ to be an independent symbol, you can adjoin it to a field. As noted by zambawithkolbasa, this would give the same field as just adjoining an indeterminate, $x$.

On the other hand, if you want $\frac00$ to behave like it looks like it should notationally, you will run into problems: For instance, in a field, by $\frac{a}{b}$, we mean $a\cdot b^{-1}$, but then $\frac00=0\cdot 0^{-1}$, and this is meaningless since $0$ in a field does not have a multiplicative inverse.

Also, the inverse of $\frac00$ notationally should be the reciprocal, $\frac00$. But then $\left(\frac00\right) ^2=1$. However then in your extension field, the polynomial $x^2=1$ has more than two roots ($1, -1, \frac00$). This is even true in characteristic $2$ where $1$ is a double root.

paw88789
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