# Why is rational exponentiation an algebraic operation?

I can't manage to understand what is the main criterion for an operation to be algebraic.
I orginally thought it was an operation that could be expressed via the standard arithmetic operations (addition,subtraction,multiplication,division). However rational exponention fail to satisfy this property.
I looked into the wikipedia article but no insights on this specific are presented.

• "Algebraic operation" is not a rigorously defined concept. And most of its rigorous versions do not allow $x^{2/3}$. – darij grinberg Jan 22 at 14:53
• @darijgrinberg Since according to wipedia page it is a pretty well defined concept, I would accept an answer that claim it to not be that well defined. Just for people to have a second source to look for when wikipedia definitions look a little bit nosense. – Gabriele Scarlatti Jan 22 at 14:58
• @GabrieleScarlatti This seems worth mentioning on the Talk page for the Wikipedia article. Maybe with a link to here if a good answer is posted. – timtfj Jan 22 at 15:11
• @GabrieleScarlatti Would you consider "$\sqrt{x}$" to be an algebraic operation ? – Peter Jan 22 at 15:20
• My first guess of the definition of algebraic operation was one that sends algebraic numbers to algebraic numbers. This concept is not mentioned on the wikipedia page but I think this would make rational exponention an algebraic operation but irrational exponentiation would not be algebraic. – quarague Jan 22 at 15:31