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?

  • 3
    $\begingroup$ The problem is not that $0$ is not assigned a degree, but that it is assigned too many degrees. $\endgroup$ – Emil Jeřábek May 2 '15 at 12:27
  • 3
    $\begingroup$ 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 ) $\endgroup$ – 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.

  • 5
    $\begingroup$ 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. $\endgroup$ – Marc van Leeuwen May 2 '15 at 13:46
  • 2
    $\begingroup$ 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. $\endgroup$ – David Richerby May 2 '15 at 14:44
  • $\begingroup$ @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. $\endgroup$ – Uncountable May 2 '15 at 22:36
  • $\begingroup$ 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". $\endgroup$ – Marc van Leeuwen May 3 '15 at 4:36
  • $\begingroup$ @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. $P(x)=a\prod (x-a_i)$. $\endgroup$ – 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 $$\text{deg }0= -\infty$$

  • 2
    $\begingroup$ Increasing in what variable ? $\endgroup$ – Belgi May 2 '15 at 10:17
  • 2
    $\begingroup$ 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. $\endgroup$ – user26486 May 2 '15 at 10:22
  • 1
    $\begingroup$ "increasing" in the sense $|P(x)| = o(|Q(x)|) \implies \text{deg }P \le \text{deg }Q$. $\endgroup$ – mookid May 2 '15 at 10:22
  • 9
    $\begingroup$ 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 $Q = -P \neq 0$. $\endgroup$ – Travis Willse May 2 '15 at 11:49
  • 6
    $\begingroup$ Another way to think of it is that $\operatorname{deg}$ is similar to a logarithm. $\endgroup$ – 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.

  • $\begingroup$ 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? $\endgroup$ – Sufyan Naeem May 8 '15 at 10:15
  • 1
    $\begingroup$ This is irrelevant. A polynomial is a symbolic expression that is independent of any value of the variable. $\endgroup$ – Yves Daoust May 8 '15 at 10:23
  • $\begingroup$ 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! $\endgroup$ – Sufyan Naeem May 8 '15 at 10:27
  • $\begingroup$ 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? $\endgroup$ – Sufyan Naeem May 8 '15 at 10:38
  • 1
    $\begingroup$ You are welcome. I will cleanup my comments, conversations aren't appropriate in comments. $\endgroup$ – Yves Daoust May 8 '15 at 10:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.