I've recently been reading Serge Lang's Math Talks for Undergraduates, specifically a section about the abc conjecture. Lang starts by stating and proving the Mason-Stothers Theorem:

Let $f,g \in \mathbf{C}[t]$ be nonconstant and relatively prime. Then $ \text{deg}(f+g) \leq n_0[fg(f+g)]-1$, where $n_0$ gives the number of distinct roots of a polynomial.

Lang then translates the Mason-Stothers theorem into a theorem about the integers. In doing so, he states "Experience shows that the analogue of the degree is the logarithm of the absolute value of the integer." Why is this so? For an integer $a$ and a polynomial $f$, how is $\log(|a|)$ analogous to $\text{deg}(f)$?

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    $\begingroup$ Any positive integer $n$ can be expanded in base $b$: $n = a_kb^k + a_{k-1}b^{k-1} + \ldots + a_0$ with $0 \le a_i < b$, which has "degree" $k = \lfloor log_b n \rfloor$ $\endgroup$
    – math54321
    Jun 5, 2023 at 21:53

5 Answers 5


Lang says that experience shows this. I only want to mention the following analogy. Writing an integer in decimal notation is like writing out a polynomial. For example, $$ 1045 = 1\cdot10^3+0\cdot10^2+4\cdot10^1+5\cdot1. $$ Here the degree is $3$, which is the logarithm of the leading term, $1000$.

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    $\begingroup$ I see, I hadn't thought about it like that - thank you! $\endgroup$ Jun 4, 2023 at 18:21
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    $\begingroup$ Put succincly, an integer is a polynomial (as we write it). $\endgroup$
    – Dúthomhas
    Jun 5, 2023 at 3:42

Here's a concrete way in which they're analogous:

  • $\mathrm{deg}\,(f \times g) = \mathrm{deg}\,f + \mathrm{deg}\,g$
  • $\log |xy| = \log |x| + \log |y|$

Also: the degree of the constant zero polynomial is often either left undefined, or said to be $-\infty$. This is required to make the above identity hold, but it's also directly analogous to the fact that $\log 0$ is either undefined or $-\infty$.

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    $\begingroup$ Analogously, you can argue that $\log 0 =-\infty$ is required to make the above identity hold. $\endgroup$ Jun 5, 2023 at 15:09
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    $\begingroup$ @GiuseppeNegro Indeed, though it can also be justified as the limit of $\log x$ as $x \to 0$ (which can't be done for polynomial degree). $\endgroup$
    – kaya3
    Jun 5, 2023 at 15:10

There is a certain analogy between the ring of polynomials over a field, $k[x]$ and the ring of integers, $\mathbb{Z}$. Note that both are Euclidean domains.

Now, for a finite field, $k=\mathbb{F}_q$, with $q$ elements and a polynomial $f\in k[x]$ of degree $d$, the number of elements in the quotient is $q^d$: $$ | k[x] / f\cdot k[x] | = q^d. $$ In other words, $d = \log_q |k[x]/f\cdot k[x]|$.

How about the cardinality of a quotient of $\mathbb{Z}$? If $a\in\mathbb{Z}$ is an integer, then we have $$ | \mathbb{Z}/a\cdot \mathbb{Z}| = |a|. $$ Hence, $\log |a|$ is supposed to be the integer analogue of the degree of a polynomial.

Indeed, there are many theorems where this analogy turns out to be quite useful. You might like the wonderful book "Number Theory in Function Fields" by Rosen which contains many examples of this phenomenon.

  • $\begingroup$ That makes a lot more sense. And the book by Rosen looks like a very helpful resource... I have a lot of learning to do! Thank you for the response! $\endgroup$ Jun 4, 2023 at 18:22

I want to expand on the comment of @math54321 and the answer of @Dietrich Burde.

If we write a non-zero natural number $n$ as $n = \sum_{j = 0}^{m} a_j 10^j$, where $a_j \in \{ 0,1,2,3,4,5,6,7,8,9 \}$ and $a_m \neq 0$, i.e. as its decimal expansion, then we get that

$$ m \leq \log_{10} \left( \sum_{j = 0}^{m} a_j 10^j \right) < m +1,$$

which can be rewritten as

$$ \lfloor \log_{10}(n) \rfloor = m. $$

This is of course analogous to the degree of a polynomial $\sum_{j = 0}^m a_j X^j$, which is precisely $m$ (under the assumption that $a_m \neq 0$).

For completeness, let's prove this: First notice that the logarithm is non-decreasing. So since $a_m 10^m \leq \sum_{j = 0}^{m} a_j 10^j$, we get

$$ \log_{10} \left( a_m 10^m \right) \leq \log_{10} \left( \sum_{j = 0}^{m} a_j 10^j \right). $$

The lower bound is equal to $\log_{10} (a_m) + \log_{10} (10^m) $, which is always larger than $\log_{10}(10^m)$, since $a_m \geq 1$. In total

$$ m = \log_{10} (10^m) \leq \log_{10} \left( \sum_{j = 0}^{m} a_j 10^j \right). $$

On the other hand, we can estimate $\sum_{j = 0}^{m} a_j 10^j$ as follows: The worst case is that all $a_j$ are equal to $9$. So $ \sum_{j = 0}^{m} a_j 10^j \leq \sum_{j = 0}^{m} 9 \cdot 10^j $ is the best we can do. But this is a geometric sum and can be simplified to

$$ \sum_{j = 0}^{m} 9 \cdot 10^j = 9 \frac{1 - 10^{m+1}}{1 - 10} = 10^{m+1} - 1. $$

For the logarithm we then get

$$ \log_{10} \left( {\sum_{j = 0}^{m} a_j 10^j} \right) \leq \log_{10} (10^{m+1} - 1) < \log_{10} (10^{m+1} ) = m +1.$$

This concludes the proof!

If $n = 0$, then $\log_{10}(0)$ can be seen as $\lim_{x \to 0} \log_{10} (x) = - \infty$, which also coincides with the degree of the zero polynomial being $- \infty$.

So here we have a precise analogy between the logarithm and the degree of a polynomial. This of course also works for all other bases other than $10$.


The degree of a monomial is (putting the coefficient aside) the log.

For a log, we have $\log(ab) = \log(a)+\log(b)$, and for degrees, we have $\deg(fg) = \deg(f)+\deg(g)$.

For addition, there's a slight discrepancy, in that $\log(a+b) = \log(a)+ r - \frac {r^2}2 + \frac{r^3}3 ...$, where $r = \frac a b$, while $\deg(f+g) = \max(\deg(f), \deg(g)$. This may seem like very different behavior, but if $a \gt \gt b$, $\log(a+b) \approx a$. \

You can think of $\deg(f)$ as being $\lim_{x \rightarrow \infty} \log_xf(x)$. That is, the degree tells you to what power of $x$ the polynomial asymptotically approaches.

  • $\begingroup$ You appear to have some addition symbols missing on line 2. $\endgroup$
    – Yakk
    Jun 7, 2023 at 18:24

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