# Zero polynomial [duplicate]

Possible Duplicate:
Polynomial of degree $-\infty$?

Today in Abstract Algebra my instructor briefly mentioned that sometimes the zero polynomial is defined to have degree $-\infty$. What contexts have caused this to become convention?

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We want $deg(p\cdot q) = deg(p) + deg(q)$. However you need to reflect the fact $0\cdot p = 0$ for all polynomials $p$, so $deg(0)$ can at least not be a natural number. But $-\infty + k = -\infty$ as we'd like. –  Dario Sep 25 '12 at 14:57
Thanks all for your responses I see now the rationale behind this definition –  Zvpunry Sep 25 '12 at 21:03

## marked as duplicate by Hans Lundmark, M Turgeon, William, Ｊ. Ｍ., tomaszOct 5 '12 at 19:57

Let $f$ and $g$ be nonzero polynomials with coefficients in some integral domain (such as $\mathbb{Z}$ or $\mathbb{R}$ or $\mathbb{C}$ or whatever). Then we define $\deg f$ to be the greatest $n \ge 0$ such that the coefficient of $x^n$ in $f(x)$ is nonzero. Then $$\deg (f \cdot g) = \deg f + \deg g$$ and $$\deg (f + g) \le \max \{ \deg f, \deg g \}$$ Now if $f=0$ then $f \cdot g=0$. We'd like to define $\deg 0$ in such a way that it obeys the above rules. Well, for all polynomials $g$, we have $$\deg 0 = \deg (0 \cdot g) = \deg 0 + \deg g$$ This is only true for all $g$ if we take $\deg 0 = -\infty$ (or $\infty$, but the second equation above means that $-\infty$ makes more sense).

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Note that we can only establish $deg(f+g) \leq \max\{deg(f), deg(g)\}$ as we can e.g. choose $g=-f$. –  Dario Sep 25 '12 at 15:03
@Dario: Ack, of course! Thanks. –  Clive Newstead Sep 25 '12 at 15:06

Here’s a viewpoint that puts the question in much broader context. In a discretely valued field, like $\mathbb{Q}$ with the $p$-valuation $v_p$ that counts the divisibility by the prime $p$, defined so that $v_p(p)=1$ and $v_p(m)=0$ for integers prime to $p$, and additive in the sense that $v_p(\lambda\mu)=v_p(\lambda)+v_p(\mu)$, it’s universal to extend $v$ so that $v(0)=+\infty$. Then the relation $v(\lambda+\mu)\ge\min(v(\lambda),v(\mu))$ holds throughout the field.

Well, just as $\mathbb{Q}$ has many discrete valuations, so does the field $k(X)$ of rational functions over a field $k$. One nice valuation is defined by setting, for a polynomial $P(X)$, $v(P)=-\deg(P)$, and extending by additivity, so that that $v(P/Q)=\deg(Q)-\deg(P)$, both $P$ and $Q$ being polynomials. This $v$ measures the zeroness of a rational function at $\infty$. And in accordance with the standard practice for valued fields, $v(0)=+\infty$, fitting in perfectly with $\deg(0)=-\infty$.

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+1 Conceptually, this is the best way to view it, and is probably the answer that one would receive from most algebraists. –  Bill Dubuque Oct 6 '12 at 14:11
Persistance. You want formulas to make sense also when abusively applying them to cases involving the zero polynomial. For example, we have $\deg(f\cdot g)=\deg f +\deg g$ and $\deg (f+g)\le \max\{\deg f, \deg g\}$. Therefore we assign a symbolic value - and be it only for mnemonic purposes - of $-\infty$ as the degree of $0$, because that makes $-\infty =\deg(0\cdot g)=-\infty+\deg g$ and $\deg g = \deg (0+g)=\max\{-\infty,\deg g\}$ work.
We want $\deg(P\cdot Q)=\deg P+\deg Q$ for two polynomials. In particular $\deg\mathbf 0=\deg P+\deg\mathbf 0$, so we can't take an integer. $+\infty$ could be a choice as we want $\deg(P+Q)\leq \max(\deg P,\deg Q)$ but with the definition $\deg((\alpha_j)_{j\geq 0}:=\sup\{k\geq 0\mid \alpha\neq 0\}$, we take a supremum over an emptyset, so we take $-\infty$.