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In a book on algebra I'm currently working with a proof that uses the binomial theorem for $(x+y)^m$ where $x,y$ are elements of some arbitrary field $k$. This looks strange to me, so I did some research on the Internet, but all pages which state (and prove) this theorem never specify what $x$ and $y$ are. A binomial coefficient is defined by faculties and fractions of integers. This is not evaluable in a general field that does not extend the integers like $\mathbb Q$.

We can use the map $$\mathbb Z \to k,\quad n\mapsto \sum_1^n 1$$ to get a representation of the integers in $k$, but this is only a group, and it is not clear (to me), why the inverses of these internal representations in $k$ are internal integers as well (so that we could define the binomial coefficient via internal integers).

A further question would be, whether we can weaken the requirements from general fields to (not necessarily general) rings. Except for the definition of binomial coefficients, we don't need inverses in the statement, so maybe this is possible as well in some cases (of course this works for all rings that we obtain via a forgetful functor from a field, but I mean rings that are not fields).

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  • $\begingroup$ What do you mean by "inverses of this internal representations in k are internal integers as well"? In particular, what are "internal integers"?) FWIW, I don't see any obstacle to the usual proof working just as well for any unital commutative ring, but perhaps I am overlooking something. $\endgroup$ – Travis Willse Apr 27 '15 at 11:28
  • $\begingroup$ with internal integers I mean the subgroup of the additive group of $k$ that I obtain as the image of the given map. Does this make clear what I mean with inverses of internal integers are internal integers? $\endgroup$ – Sebastian Bechtel Apr 27 '15 at 11:32
  • $\begingroup$ Yes, thank you. In the case $\text{char} k$ is prime, this additive subgroup (in fact, it is a field) is called the prime subfield; I don't know whether it has a particular name when $\text{char} k = 0$. (And this is a nice question, by the way, (+1).) $\endgroup$ – Travis Willse Apr 27 '15 at 11:47
  • $\begingroup$ faculties $\mapsto$ factorials $\:$ ? $\;\;\;\;$ $\endgroup$ – user57159 Apr 28 '15 at 4:10
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I think you're looking at $$ (x+y)^n = \sum_{i=0}^n \binom ni x^i y^{n-i} $$ and thinking that the multiplication of $\binom ni$ with the other factors is the multiplication operation of the field. It's not, really; it's the "scalar" multiplication $\mathbb Z\times k\to k$ that does repeated addition. (Groups are $\mathbb Z$-modules, you see.) The binomial coefficient is really an integer.

Indeed, to prove the binomial theorem, you'd multiply out $(x+y)^n$ and then gather like terms. It's that gathering that produces the binomial coefficients — you count how many terms you have of each type. That's plain old counting, so it yields plain old integers, not elements of the field.

The binomial theorem holds, in this form, in any commutative ring. (We need multiplication to be commutative for the "like terms" to actually be like each other.) (Well, as user26857 pointed out, it's enough that $x$ and $y$ commute with each other, and the scenario where $x$ and $y$ commute but the multiplication is not commutative in general does come up pretty often.)

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  • $\begingroup$ Great answer! I would never have thought that I can just treat the sum as a term in a $\mathbb Z$-module... $\endgroup$ – Sebastian Bechtel Apr 27 '15 at 11:47
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    $\begingroup$ @SebastianBechtel: Yes that is the crux; the (real) integers are merely acting on the ring elements, just like powers of ring elements are using (real) integers rather than anything from the ring itself. $\endgroup$ – user21820 Apr 27 '15 at 12:06
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    $\begingroup$ @user26857 How about this compromise: the ring generated by $x$ and $y$ has to be commutative. $\endgroup$ – user21467 Apr 28 '15 at 0:52
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This is a good question. You can prove the binomial theorem (where the choose functions are interpreted as elements of the "copy" of $\mathbb{Z}$ in the ring by exactly the map you described) in any commutative ring. Induct on $n$, the case of $n = 1$ being clear. $$ (x+y)^n = (x+y)(x+y)^{n-1} = (x+y)\sum_{k=0}^{n-1}\binom{n-1}{k} x^ky^{n-1-k} $$ Expanding, the coefficient of $x^jy^{n-j}$ is $$ \binom{n-1}{j-1} + \binom{n-1}{j} $$ and we win since we are reduced to the usual statement in the integers that $$ \binom{n}{j} = \binom{n-1}{j-1} + \binom{n-1}{j} $$ (which you can prove bijectively, but which you should already accept since you're willing to accept the binomial theorem over $\mathbb{Z}$.)

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    $\begingroup$ NB you had a typo in the summation index. I fixed that for you. $\endgroup$ – AlexR Apr 27 '15 at 11:38
  • $\begingroup$ It seems to me that one needs the commutative ring to be unital as well, so that $n \mapsto \sum_{k = 1}^n 1$ is well-defined. (Of course, some people include that in the definition of ring anyway.) $\endgroup$ – Travis Willse Apr 27 '15 at 11:40
  • $\begingroup$ Unfortunately your answer doesn't fill the gap in my understanding. Basically it just sketches the normal proof of the binomial theorem. But the point is, why do the rules for binomial coefficients hold in this setup? I think Steven's post addresses this. $\endgroup$ – Sebastian Bechtel Apr 27 '15 at 11:43
  • $\begingroup$ @AlexR very belated thanks. $\endgroup$ – hunter Sep 2 '16 at 9:32
  • $\begingroup$ @Travis there seems to be a proof here for rings without unit. $\endgroup$ – Natan Yellin Nov 15 '17 at 14:49

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