correct parent for square root of constant polynomial

[DaveWitteMorris]

📚 Description

Fixes #35860.

If f is a polynomial that is a perfect square, then sqrt(f) should also be a polynomial. However, this has not been the case for polynomials that happen to be constant: the code returned a constant in the base ring, not a constant polynomial. The PR fixes this (and adds a doctest for this bug).

This change fixes the serious bug that was pointed out (and diagnosed) by @amithazi in issue #35860. A single-variable Laurent polynomial is stored as a pair (u, n), where u is an ordinary polynomial, and n is an integer that represents an offset (i.e., multiplication by a [usually negative] power of the variable). To add two Laurent polynomials, one of them needs to be shifted so they have the same offset:

f2 = right.__u << right.__n - left.__n

This code assumes that the parent of right.__u is a polynomial ring: if the parent of right.__u is ZZ, then the shift operator << multiplies by a power of 2, instead of shifting the polynomial, so the result is nonsense. The PR also adds a doctest for this bug.

📝 Checklist

• The title is concise, informative, and self-explanatory.
• The description explains in detail what this PR is about.
• I have linked a relevant issue or discussion.
• I have created tests covering the changes.

⌛ Dependencies

[mezzarobba]

Lgtm, thank you!

[DaveWitteMorris]

Thanks!