# What are perfect squares in general rings?

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")?

• 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. Commented Nov 2, 2018 at 2:10
• @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. Commented Nov 2, 2018 at 2:39
• But $p$ isn't required to be in $A$. So why should I call $p$ a square in $A$? Commented Nov 2, 2018 at 6:51
• @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? Commented Nov 2, 2018 at 20:16
• 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.). Commented Nov 2, 2018 at 20:21

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 $$\Bbb Z$$".