I wonder why the degree of the zero polynomial is $-\infty$ ?

I heard that, it is $-\infty$ to make the formula $\deg(fg)=\deg(f)+\deg(g)$ hold when one of these polynomials is zero. However, if that was the only reason we could have said that it is $\infty$ instead.

  • 7
    $\begingroup$ Well, if you add higher order terms than what you already have then the degree should go up, right? So $+\infty$ is not a good candidate for the degree. $\endgroup$ Sep 16, 2013 at 12:43
  • $\begingroup$ BTW, the degree of the zero polynomial is undefined, −∞ is simply out of convenience. $\endgroup$ Sep 16, 2013 at 15:12
  • $\begingroup$ @RamchandraApte You are right some authors says it is undefined but look for example link $\endgroup$
    – Math137
    Sep 16, 2013 at 15:20
  • 3
    $\begingroup$ Having degrees of all polynomials defined is very convenient, as it avoids having to stop every time you write down $\deg$ and ask if the argument could be the zero polynomial. The things one does with degrees are usually quite limited: comparing, adding and very rarely multiplying. For these purposes $-\infty$ is not much of a pain. Subtraction of degrees arises with division of polynomials, but then $0$ is explicitly ruled out. $\endgroup$ Sep 16, 2013 at 15:45

4 Answers 4


One also wants $\deg(P+Q)\leq\max(\deg P,\deg Q)$ to hold, even if $P=-Q$.

Added much later: and maybe more importantly, in Euclidean division of some polynomial $A$ by $B\neq0$ we want the remainder $R$ to satisfy $\deg(R)<\deg(B)$, even if the division is exact (i.e., if $R=0$).


Convention 0. Marc has already explained why $\mathrm{deg}(0) = -\infty$ is a good convention.

Convention 1. On the other hand, if we want $\mathrm{deg}(PQ) \geq \mathrm{deg}(Q)$, well this implies $\mathrm{deg}(0) \geq \mathrm{deg}(Q)$ for all $Q$, so therefore $\mathrm{deg}(0) = +\infty$ is a good convention. Just take a look at the following divisibility sequence, which suggests that $\mathrm{deg}(0)$ should be "as large as possible."

$$1 \mid x \mid x^2 \mid x^3 \mid \ldots \mid 0$$

Bottom line: don't assume the reader understands your own personal preferred conventions regarding $\mathrm{deg}$ unless and until you've told them.

Convention 2. By the way, if you're thinking of $\mathbb{R}[x,y]$ as a graded algebra, you probably want the homogeneous polynomials of each degree to form an $\mathbb{R}$-module. In this case, its best to think of $0$ as having every possible degree, so that it belongs to each $\mathbb{R}$-module. So we might write

$$\mathrm{deg}(0) = \{0,1,2,3\ldots\},$$

or say something like "degree isn't a function, its a relation."

  • 2
    $\begingroup$ Why the downvote? $\endgroup$ Jun 7, 2016 at 8:23
  • 1
    $\begingroup$ Great Answer. The moral of the story is "We define it conveniently so that we get the proof of our desired theorems" $\endgroup$
    – Babai
    Aug 17, 2018 at 5:59
  • 1
    $\begingroup$ @Babai, thanks :) Yes, that's definitely a philosophy I support. I see young people getting marked down sometime behind they didn't define things quite the way the teacher or lecturer is defining them and it's a bit disheartening. Better to let definitions be flexible. As long as people are clear, it's okay. $\endgroup$ Aug 17, 2018 at 9:54
  • $\begingroup$ true. My idea of defining things as we require comes from the prospective of agebraic geometry. For example many times we define equivalence relation so that as a result the quotient space has nice structure. $\endgroup$
    – Babai
    Aug 17, 2018 at 9:57
  • 1
    $\begingroup$ @Babai, sounds interesting - feel free to elaborate. $\endgroup$ Aug 19, 2018 at 8:01

This is old but there is one very straightforward explanation no one gave in my understanding. A polynomial is some $ P = \sum_{k=0}^{+\infty} a_k X^k$ with the sequence $(a_k)$ being ultimately constant equal to 0. Then we define $$\deg P = \sup\left\{k\in\mathbb N,\ a_k \neq 0\right\} $$

When $P=0$ the sequence $(a_k)$ is constant equal to $0$ so $\deg P$ is the maximum of the empty set. In the extended real number line, this is $-\infty$.


If you define the degree of a polynomial as the degree of the highest non-zero power, then if the polynomial is zero, the degree is undefined. You could define it by convention, to make sense of some general rules, and then as the other answers explain, you get different results.

Now, think of why $x^0 =1$. The proof is by using the laws of exponents, $\frac{x^n}{x^n} = 1 = x^{n-n}$. However, $0^0$ is undefined, since $x^y$ is not continuous at 0 (as a function of two variables). However, we often define $0^0=1$. It is a convention, to make some multiplications easier.

Similarly, the degree of the zero polynomial is not defined, but we use different conventions to be able to generalise formulas (like the product of polynomials).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .