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I understand that we do not want to say that the degree of the zero polynomial is zero, since deg($pq$) = deg($p$) + deg($q$), but this does not convince me that negative infinity is a better choice than infinity for the degree of the zero polynomial.

Is it because the zero polynomial still needs to have a lower degree than non-zero, non-constant polynomials? Or are there other reasons?

Nika
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BarbaricGlass
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  • How about "undefined" instead of negative infinity? Does that work for you? – Lee Mosher Nov 24 '19 at 22:21
  • A lot of proofs work by inducting on the degree, which would be ruined (or at least have more complicated bookkeeping) in that case. – anomaly Nov 24 '19 at 22:23
  • You'll find some authors saying that $0$ has no degree and some other autors saying that $0$ has degree $-\infty$. The convetion $\deg 0=-\infty$ preserves both inequalities $\deg(pq)=\deg p+\deg q$ and $\deg(p+q)\le \max{\deg p,\deg q}$. If you use the convention of not defining $\deg 0$ (I remember this having some upside to it, but I can't recall which was), then those inequalities are meant to hold for $p,q\ne 0$. I don't see $\deg 0=\infty$ having any upside. –  Nov 24 '19 at 22:26

2 Answers2

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You also have $deg(p+q)\leq \max\{\deg(p),\deg(q)\}$ and $\deg(0+p)=\deg(p)$.

Jonas Linssen
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Tsemo Aristide
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If you define the norm $N(p)$ of a polynomial $p$ by $$ N(p) = \begin{cases} 0 & \text{if $p = 0$}\\ 2^{\deg(p)} & \text{if $p \ne 0$} \end{cases} $$ you will get a function which satisfies $N(p q) = N(p) N(q)$ for all polynomials $p, q$. This will allow you, for instance, to state polynomial division (over a field) as

Given polynomials $a, b$, with $b \ne 0$, there exist polynomials $q, r$ such that $$ \begin{cases} a = b q + r\\ N(r) < N(b) \end{cases} $$

with no need to distinguish the case when $r = 0$. Clearly this is suggestive of the zero polynomial having degree $- \infty$. If you can handle this properly, this is ok, otherwise you can use this norm.

Andreas Caranti
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