For the learned mathematician it may be obvious and not worth mentioning: that the fundamental theorems of arithmetic and algebra look very similar and have to do with each other, in abbreviated form:

$$ n = p_1\cdot p_2 \cdots p_k$$

$$ P(z) = z_0\cdot(z_1 -z)\cdot (z_2 -z) \cdots (z_k -z)$$

which makes obvious that the irreducible polynoms of first degree play the same role in $\mathbb{C}[X]$ as do the prime numbers in $\mathbb{Z}$ (which both are unitary rings). It also gives — in this special case — the wording "fundamental theorem" a specific meaning: It is stated that and how some irreducible elements build the fundaments of a structure.

Is this analogy helpful, or is it superficial and maybe misleading? If the former, can it be formalised? If the latter, what are the differences that make it merely superficial?

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    $\begingroup$ The analogy doesn't generalize to other fundamental theorems: the fundamental theorem of calculus, for instance, is that differentiation and integration are inverse operations of each other, which has nothing to do with "how some irreducible elements build the fundaments of a structure" as far as I can see. $\endgroup$
    – zwol
    Commented Sep 28, 2018 at 11:50
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    $\begingroup$ I restricted it to "this special case". $\endgroup$ Commented Sep 28, 2018 at 11:54
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    $\begingroup$ On the other hand, some other fundamental theorems, like the one on finite(ly generated) abelian groups, also describe how certain objects factor. Many of those statements describe how general objects decompose into "primes" in that way, but this is not a general rule. $\endgroup$
    – Wojowu
    Commented Sep 28, 2018 at 12:16
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    $\begingroup$ Why would the analogy be made more explicit and stressed? I can't think of any reason why you'd do that. $\endgroup$ Commented Sep 28, 2018 at 14:35
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    $\begingroup$ Because it's a "deep" analogy? $\endgroup$ Commented Sep 28, 2018 at 14:38

4 Answers 4


In both cases, the theorem says "This ring is a unique factorisation domain and these are its irreducible elements". So in this sense, they are similar.

However, there are significant differences. In the case of $\Bbb Z$, the content is the unique factorization: since in any UFD, irreducible elements are prime, saying "and the irreducible elements are the prime numbers" doesn't add anything.

On the other hand, in the case of $\Bbb C[x]$, the content is what the irreducible elements are: given any field $K$, $K[x]$ is a UFD, and yet we know that the fundamental theorem of algebra doesn't hold over most fields (including $\Bbb R$, $\Bbb Q$, all finite fields, etc). So here the interesting part is that the polynomials of degree $1$ are the only irreducibles.

  • $\begingroup$ Sorry, I misread. I agree it's not interesting after another reading; deleting. $\endgroup$ Commented Sep 28, 2018 at 16:26
  • $\begingroup$ I disagree. For ${\mathbf Z}$, it just says "this ring is a unique factorisation domain". It does not give any description of primes. On the other hand, we have an explicit form of primes for ${\mathbf C}[x]$. The analogue of the fundamental theorem of arithmetic is simply "for any field $K$, the ring $K[X]$ is an UFD". I'm not a number theorist, but I suspect that the analogy can be made quite a bit deeper if you think of the integers as a variety over the one-element field, or some such nonsense. $\endgroup$
    – tomasz
    Commented Mar 25, 2019 at 19:04

Some other answers already make very good points. I just want to add that I think the truly amazing analogy is

"polynomials are the integers among the functions"; "polynomials behave like integers, and integers behave like polynomials"

which I could formally just state as: both $\Bbb Z$ and a polynomial ring $k[x]$ (over any field $k$) are Euclidean domains.

This realisation (that one can do division with remainder, hence has unique factorisation, what this means about the fraction field and its extensions, localisation, sheaves, ...) is indeed a profound insight, and arguably the analogy (and its generalisations) are a cornerstone of modern algebraic geometry and number theory.

That analogy, I think, should indeed be stressed more often. I sometimes mention it to my undergrad students, saying:

  • You learnt factoring numbers, then you learnt factoring polynomials, have you ever wondered what is the relation? Specifically, for numbers you end up with primes which you cannot factor anymore; are there polynomials like that? Which ones?
  • Or: In middle school you divided integers with remainder, in high school you divided polynomials with remainder. Well: the integer part of a fraction tells you its size, the polynomial part of a rational function tells you its behaviour for $x\to \infty$ ...
  • Or: Rational functions = quotients of polynomials, like rational numbers = quotients of integers. But they are not complete, limits of them are analytic or meromorphic functions (like $\sin, \tan$) -- just like rational numbers are not complete, limits of them are possibly transcendental numbers (like $\pi$) ...).

But this has nothing to do with $\Bbb C$. The Fundamental Theorem of Algebra, completely sidestepping that analogy, rather focusses on the fact (surely remarkable, but unrelated to all that) that in $\Bbb C[x]$ the "primes" are very easy.

  • $\begingroup$ When you said "limits of [rational functions] are analytic functions (like $\sin$, $\cos$)", were you referring to the polynomial expansion of analytic functions like $\sin$ or $\cos$? If so, aren't the expansion polynomial functions, rather than rational functions? $\endgroup$ Commented May 13, 2021 at 1:51
  • $\begingroup$ Well polynomials are a special kind of rational functions, so I'm not lying. Indeed I was thinking of Taylor series i.e. approximations of analytic functions through polynomials. I was deliberately vague because on closer look, here's a distinct difference in the analogy, as e.g. when completing $\mathbb Q$ to $\mathbb R$, the subset $\mathbb Z$ is closed and hence already complete; while if we e.g. complete $\mathbb R(x)$ to $\mathbb R((x))$, the polynomials $\mathbb R[x]$ are not closed yet but get completed to $\mathbb R[[x]]$. $\endgroup$ Commented May 13, 2021 at 5:19
  • $\begingroup$ I've changed the example to include $\tan$ as an example of functions with poles, which around such a pole get approximated by true Laurent polynomials, i.e. rational functions which are not just polynomials. $\endgroup$ Commented May 13, 2021 at 5:26

That is a good analogy. It turns out that both $\mathbb Z$ and $\mathbb{C}[x]$ are unique factorization domains. In the case of $\mathbb{C}[x]$, this fact, together with the fundamental theorem of Algebra, means what you wrote: every $p(x)\in\mathbb{C}[x]$ can be written as the product of a non-zero complex number and first degree polynomials. The same thing applies to any algebraically closed field, such as the field of algebraic numbers.

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    $\begingroup$ On the other hand, every polynomial ring over a field is a UFD, so the fundamental theorem of algebra appears to be a bit more specific. $\endgroup$ Commented Sep 28, 2018 at 8:23

This analogy can be formalise in ring theory. The sets $\mathbb C[X]$ and $\mathbb Z$ are both rings, and more precisely, noetherian rings.

In noetherian rings, you have a theorem that stated that every ideal can be decomposed as the intersection of finitely many primary ideals.

Which is exactly the meaning of those two theorems.

In general: if you like this kinds of analogies, study group theory, ring theory, and really any kind of algebraic structures.

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    $\begingroup$ This is incorrect interpretation of fundamental theorem of algebra - its content is not that we have representation of ideals as intersections (which is true for all fields in place of C) but rather that the nonzero prime ideals are generated by elements of degree 1. $\endgroup$
    – Wojowu
    Commented Sep 28, 2018 at 10:33
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    $\begingroup$ Also, this incorrectly interprets fundamental theorem of arithmetic - the condition on ideals really says nothing about factorizations into elements and then it's only an existential statement. It's the uniqueness which is the profound part of FTA, and noetherianness doesn't imply it. $\endgroup$
    – Wojowu
    Commented Sep 28, 2018 at 11:58

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