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Is constant number a polynomial?

Is $5$ a polynomial?

My present understanding is no:

$0$ is a polynomial with no degree be defined,

But if $5$ is a polynomial, then $5$'s degree is $0$

Community
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User19912312
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    From what I understand is that you can look at both $5$ and $0$ as polynomials. The difference is that the degree of the constant polynomial $5$ is $0$, and here is what I am not sure about. I think that for matter of convinience one defines the degree of the $0$ polynomial to be negative infinity, but I would like to see what other people say. – Daniel Montealegre Aug 01 '13 at 07:23
  • yes, usually the degree of the "polynomial" $0$ is set as negative infinity, so that the degree of the product of two polynomials is the sum of the degrees of the polynomials. – mau Aug 01 '13 at 08:17

2 Answers2

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The algebra of polynomials $k[X]$ over a field $k$ can be built as the set of sentences $k^{(\mathbb{N})}$ (that is the sentences eventually zero) endowed with the following operations:

  • $(a_n)+(b_n)=(a_n+b_n)$,
  • $(a_n) \times (b_n)=(c_n)$ with $\displaystyle c_n= \sum\limits_{i+j=n} a_ib_j$,
  • $\lambda \cdot (a_n)=(\lambda \cdot a_n)$.

As a $k$-vector space, we can see that $\{X^0,X^1,X^2,\dots\}$ is a basis of $k[X]$ where $X=(0,1,0,\dots)$ (we suppose that $X^0=(1,0,\dots)$). Therefore, every polynomial can be written as $\displaystyle \sum\limits_{n \geq 0} a_nX^n$ where $(a_n)$ is eventually zero.

In particular, $k$ is canonically embedded into $k[X]$ by $a \mapsto (a,0,\dots)$.

The degree of a polynomial $P=(a_n)$ is defined as $\deg(P)= \max\{ n \geq 0 \mid a_n \neq 0\}$ if $P \neq 0$; otherwise, $\deg(P)=-\infty$.

Therefore, in $\mathbb{R}[X]$ for example, $5$ is a polynomial of degree zero and $0$ is a polynomial of degree $- \infty$.

Remark: It is convenient to suppose $\deg(0)=-\infty$ so that relations like $\deg(PQ)=\deg(P)+\deg(Q)$ still hold when $P$ or $Q$ is zero.

Seirios
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The degree of 0, as far as I know, is set equal to $-1$ (some practical reasons) and constants have degree $0$.

pepa.dvorak
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    I think $\deg0:=-\infty$ is even more practical. – anon Aug 01 '13 at 07:22
  • There reason to define $\deg_X(0)=-\infty$ is that (with the obvious definition for addition on the set ${-\infty}\cup\Bbb N$) one has the rule $\deg_X(PQ)=\deg_X(P)+\deg_X(Q)$ without any exception, and this is is useful in proofs. The rule does require that the coefficient ring is an integral domain (in particular the rule applies when the coefficients lie in a field). – Marc van Leeuwen Aug 01 '13 at 07:47