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The definition of a square element can be made for arbitrary rings $R$:

$p\in R$ is a square if there is an $a\in R$ with $p = aa$.

The definition of a perfect square element for arbitary rings – inspired by perfect square numbers – would presuppose the concept of an integral element in arbitrary rings:

$p\in R$ is a perfect square if there is an integral element $a\in R$ with $p = aa$.

Note, that for $R = \mathbb{Z}$ being a square and being a perfect square are equivalent, while for $R = \mathbb{Q}$ it's not.

But the definition of integral elements:

$p\in R$ is integral over $A$, a subring of $R$, if there are $n \geq 1$ and $a_{j}\in A$ such that $p^{n}+a_{n-1}p^{n-1}+\cdots +a_{1}p+a_{0}=0$

depends on a subring $A \subset R$. So also the definition of perfect squares depends on a subring $A \subset R$.

My questions are:

  • Is there an "absolute" (not relative) definiton of being a perfect square (like there are absolute definitons of being a prime element or a square element)?

  • Does the concept of being a perfect square (with respect to a subring $A \subset R$) play an important role in ring theory/algebra?

  • Are there sensible reasons that the name "perfect square" had been chosen (instead of the more self-explanatory name "integral square")?

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    $\begingroup$ I don't really see the point of having such a notion. Whatever $A$ is, you're just talking about the notion of being a square in $A$, which you already have a word for, namely "being a square in $A$." So you might as well just focus on $A$; the discussion isn't really about $R$ anymore. $\endgroup$ Commented Nov 2, 2018 at 2:10
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    $\begingroup$ @hans I got the impression this was your own invention, but I guess in light of the last sentence I have to ask where you have seen it. $\endgroup$
    – rschwieb
    Commented Nov 2, 2018 at 2:39
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    $\begingroup$ But $p$ isn't required to be in $A$. So why should I call $p$ a square in $A$? $\endgroup$ Commented Nov 2, 2018 at 6:51
  • $\begingroup$ @HansStricker Perhaps using $a$ was too suggestive of being in $A$, even though you wrote $a\in R$. Perhaps it was perceived that $a\in A$ so that $p\in A$ as well. And by the way, do you have any information to answer my question about where this definition comes from? $\endgroup$
    – rschwieb
    Commented Nov 2, 2018 at 20:16
  • $\begingroup$ No, there's no information, because you are right: it's somehow my own invention. (That's why I wrote "inspired by".) But to be honest: I was sure it was not my invention, I was sure this definition had been made before (in analogy to other number-theoretic concepts transferred from $\mathbb{Z}$ or even $\mathbb{N}$ to arbitrary rings - like being integer, prime, and so on.). $\endgroup$ Commented Nov 2, 2018 at 20:21

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A perfect square (in normal use) is an element of $\Bbb Z$ which is the square of an element of $\Bbb Z$. That is, it is a square in $\Bbb Z$. Even if you look in some larger ring, say $\Bbb Q$, then the perfect squares are precisely the elements of $\Bbb Z$ which are squares in $\Bbb Z$. The fact that $\Bbb Z$ is a subring of $\Bbb Q$ is crucial to the definition - indeed, the point of the term "perfect square" is that for $\Bbb Z \subset R$ - e.g. $R = \Bbb C, \Bbb R, \Bbb Q$ - then the set of "perfect squares" in $R$ is the same for any choice of $R$ - it's the set of squares in $\Bbb Z$.

So why we have the term "perfect square" at all? Fundamentally, it's because if you call them "square numbers" then you get people saying "but $2$ is a square number it's the square of $\sqrt 2$". "Integral square" would work, except that this a concept you teach to 8 year olds and "perfect", unlike "integral", is a word they already know.

But if you're at the point where you can say "a square in the ring $R$", then the term "perfect square" isn't adding any mathematical content: it's just shorthand for "a square in the ring $\Bbb Z$".

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