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Is constant number a polynomial?

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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 $0$

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2 Answers

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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, $\deg(P)=-\infty$.

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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.

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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 $0$.

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