# Why is the zero polynomial not assigned a degree?

Yesterday, I read in my textbook,

We assign degree to every polynomial and even a non-zero constant is assigned a degree $0$ but $0$ itself is not assigned a degree.

Why is that? Why we don't assign degree $0$ to the zero polynomial?

• The problem is not that $0$ is not assigned a degree, but that it is assigned too many degrees. – Emil Jeřábek May 2 '15 at 12:27
• Whether or not it is practical to associate a degree to the zero polynomial depends on the context. Two popular choices are -infinity and -1 (according to mathworld.wolfram.com/ZeroPolynomial.html ) – soegaard May 2 '15 at 17:00

Assigning a degree to the zero polynomial will cause trouble with important and useful theorems that relate the degree of a polynomial to its roots:

If $F$ is a field (examples of fields are $\mathbb{R}$, $\mathbb{C}$, $\mathbb{Q}$, $\mathbb{Z}/p\mathbb{Z}$), a polynomial $P$ with coefficients in $F$ (the set/ring of these polynomials is usually denoted by $F[x]$) of degree $n$, has at most $n$ distinct points $\alpha\in F$ such that $P(\alpha)=0$.

This theorem follows from the fact that we can repeatedly factor out terms of the form $x-\alpha$ (where $\alpha$ is a root) from $P(x)$, lowering the degree of the remaining polynomial by $1$ in each step. See also: http://en.wikipedia.org/wiki/Factor_theorem.

When we restrict to polynomials with coefficients in $\mathbb{C}$, the statement is related to the Fundamental Theorem of Algebra.

Now since the zero polynomial in $F[x]$ has a root at every point in $F$, at least for infinite fields ($\mathbb{R},\mathbb{C},\mathbb{Q}$), we cannot assign a finite value to the degree of the zero polynomial without getting into trouble.

• The fundamental theorem of algebra says any non constant complex polynomial has a root. It implies that any monic polynomial can be factored as a product of monic polynomials of degree$~1$ (i.e., of factors of the form $X-a$), which less elementary statement is often more directly useful. (The first statement must exclude constants, the second can be extended to cater for constant factors but get a bit more complicated doing so.) Note that neither formulation ever involves $\deg(0)$. Your statement is unclear (multiplicity is not defined) and at best means the second statement. – Marc van Leeuwen May 2 '15 at 13:46
• This doesn't seem to be much of a reason. One could just adjust the Fundamental Theorem of Algebra to talk about non-constant polynomials, in the same way that the Fundamental Theorem of Arithmetic excludes 1. – David Richerby May 2 '15 at 14:44
• @Marc van Leeuwen Thanks, I have tried to modify my answer to be a bit more clear about why I think this creates a problem with important theorems about roots of polynomials. – Uncountable May 2 '15 at 22:36
• The edit has unfortunately made the answer completely wrong (though I would say it makes clear that the link with FTA evaporates when trying to make precise statements about it). (1) You factorisation statement is not true: the RHS is monic, so it can only work for monic polynomials $P$. But then the zero polynomial is excluded from the start. (2) Putting "at most" into the statement of the FTA destroys it original meaning; the resulting statement is just as true replacing "complex" by "real" or "rational". – Marc van Leeuwen May 3 '15 at 4:36
• @MarcvanLeeuwen Of course the one-directional FTA with "at most $n$ roots" instead of "exactly $n$ roots" is also useful and true in any integral domain. (I think the theorem in this form has Lagrange's name on it.) And the factorization is easily fixed by adding a leading coefficient, i.e. $P(x)=a\prod (x-a_i)$. – Mario Carneiro May 3 '15 at 5:54

This is to make nice rules such as $$\text{deg }(PQ) = \text{deg }P + \text{deg }Q\\ \text{deg }(P+Q) \le \max(\text{deg }P , \text{deg }Q)$$

So the only value that makes it possible is $$\text{deg }0= -\infty$$

• Increasing in what variable ? – Belgi May 2 '15 at 10:17
• It doesn't make sense to say the polynomial $P(x)=0$ has $-\infty$ roots (fundamental theorem of algebra), so $+\infty$ would fit better in this sense. – user26486 May 2 '15 at 10:22
• "increasing" in the sense $|P(x)| = o(|Q(x)|) \implies \text{deg }P \le \text{deg }Q$. – mookid May 2 '15 at 10:22
• Alternatively, it is a consequences of asking that the sum identity $\deg (P + Q) \leq \max\{\deg P, \deg Q\}$ to hold for all $P, Q$. This always holds if we declare $\deg 0$ to be $-\infty$ but not if we declare $\deg 0 = +\infty$; to see this, consider the special case $Q = -P \neq 0$. – Travis Willse May 2 '15 at 11:49
• Another way to think of it is that $\operatorname{deg}$ is similar to a logarithm. – Ali Caglayan May 2 '15 at 17:40

The degree of a polynomial is the exponent of the term with the highest power (and non-zero coefficient).

$$5x-4x^4+2\to x^4\to 4$$ $$0x^{45}+1x^3+2x^2\to x^3\to3$$ $$21\to x^0\to0$$ $$0\to ??$$

The null polynomial contains no power of the variable.

• For $21x^n=21$ you assumed that $n=0, x^0=1$ and hence, $deg(21)=0$. What if the domain of the variable contains 1? – Sufyan Naeem May 8 '15 at 10:15
• This is irrelevant. A polynomial is a symbolic expression that is independent of any value of the variable. – Yves Daoust May 8 '15 at 10:23
• Are you sure that a polynomial expression is independent of any value of variable? I have read in my textbook that a variable in expression contains a domain and domains contains all the values for which expression is defined! – Sufyan Naeem May 8 '15 at 10:27
• That is, if I have a polynomial $33$ then why should I assume that in $x^n$, $n=0$ however the case remains same if an specific element of the domain of the variable is $1$. viz., $$33x^0=33$$ and, $$33(1)^n=33$$ also. I know, there may be a mistake but can you explain it to me? – Sufyan Naeem May 8 '15 at 10:38
• You are welcome. I will cleanup my comments, conversations aren't appropriate in comments. – Yves Daoust May 8 '15 at 10:51