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(Context: rewriting a joint paper with a coauthor.)

We are defining the degree of a morphism $f:A^m\to A^{n}$ to be $\max_{1\leq i\leq n} \deg(f_i)$, for $f_1,f_2,\dotsc,f_{n}$ the polynomials defining $f$. Then we define the degree of $f:X\to Y$ (for affine varieties $X\subset A^m$, $Y\subset A^{n}$) to be the minimum of $\deg(\overline{f})$ over all $\overline{f}:\mathbb{A}^{m}\rightarrow\mathbb{A}^{n}$ with $\overline{f}|_{X}=f$. This is not the same as the standard definition of degree for finite morphisms.

(1) Is this evil/illiterate?

(2) Shall we call our "degree" something else (what exactly?)?

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  • $\begingroup$ Perhaps "complexity" or "degree of definition"? $\endgroup$
    – Will Sawin
    Commented Sep 19, 2022 at 14:32
  • $\begingroup$ Hm. "Degree of definition" (degdef?) makes it sound like the degree depends on the definition, but it doesn't; we are taking a minimum of $\deg(\overline{f})$. $\endgroup$
    – H A Helfgott
    Commented Sep 19, 2022 at 14:35
  • $\begingroup$ The name is by analogy with "field of definition", which is also a minimum, not depending on a particular definition. (en.wikipedia.org/wiki/Field_of_definition). So for readers familiar with that concept you should be OK, but I agree it might be problematic for others. $\endgroup$
    – Will Sawin
    Commented Sep 19, 2022 at 14:50
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    $\begingroup$ (1) Yes. (2) maxdegree (one word). $\endgroup$
    – user483792
    Commented Sep 20, 2022 at 11:21
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    $\begingroup$ Maybe not relevant, but you might want to clarify what $\operatorname{deg}(0)$ is. If all $f_i$ are zero, the map does not appear to be "worse" than any other constant map, but with typical definitions, you end up with a negative degree (as opposed to degree $0$ for other constant maps). $\endgroup$
    – red_trumpet
    Commented Sep 20, 2022 at 15:21

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