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I am trying to prove that $f(gh)=(fg)h$ for $f,g,h \in R[\mathbb{N}^n]$. Context to the exercise:

We define the polynomial ring $R[X_1,...,X_n]$ in $n$ variables as $$ R=[X_1,...,X_n] = R[\mathbb{N}^n] = \\\{ f:\mathbb{N}^n \to R | f(v)=0,|v| >> 0 \} $$ where $v=(v_1,...,v_n)\in \mathbb{N}$ and $|v| = v_1+...+v_n.$ A polynomial $f \in R[X_1,...,X_n]$ is the same as a function $f:\mathbb{N}^n \to R$ that is non-zero for only finitely many $v \in \mathbb{N}^n$. We let $X^v \in R\left[\mathbb{N}^n\right]$ denote the function given by $$ X^v(w)= \begin{cases}1 & \text { if } v=w, \\\\ 0 & \text { if } v \neq w .\end{cases} $$ With this notation, every polynomial $f \in R\left[\mathbb{N}^n\right]$ can be written as a (finite) sum $$ f=\sum_{v \in \mathbb{N}^n} a_v X^v, $$ where $a_v \in R$ (an element $r \in R$ is identified with the function mapping the zero vector to $r$ and everything else to $0 \in R$ ). If $f, g \in R\left[\mathbb{N}^n\right]$ we define $f+g$ by $(f+g)(v)=f(v)+g(v)$ and $f g$ by the (finite) sum $$ (f g)(v)=\sum_{v_1+v_2=v} f\left(v_1\right) g\left(v_2\right) $$ where $v_1,v_2 \in \mathbb{N}^n$.

MNS1517
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  • In the definition of $R[\mathbb{N}^n]$, you surely mean to write ${ f:\mathbb{N}^n \to R \mid f(v) = 0 \rm{\ for\ } \lvert v \rvert \gg 0 }$. – Andreas Caranti Nov 29 '22 at 07:41

2 Answers2

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One should be able to prove associativity by figuring out the coefficient of $x_1^{m_1} \cdots x_n^{m_n}$ in a general product of two polynomials, then analysing that for the product of three functions. For

$$f(x) = \sum_{v\in {\bf N}^n} a_{v} x^v,$$

where if $v = (m_1,\ldots,m_n)$ and $x=(x_1,\ldots,x_n)$, then $x^v$ denotes ${x_1}^{m_1}\cdots{x_n}^{m_n}$. (Of course the $a_v$ are zero for cofinitely many $v$.) Let

$$g(x) = \sum_{v\in {\bf N}^n} b_{v} x^v,$$

Then $fg$ is given by

$$(fg)(x) = \sum_{v\in {\bf N}^n} \sum_{u+w =v} a_ub_w x^v.$$

So if we introduce a third polynomial

$$h(x) = \sum_{v\in {\bf N}^n} c_{v} x^v,$$

we have

$$(gh)(x) = \sum_{v\in {\bf N}^n} \sum_{u+w =v} b_uc_w x^v.$$

And from here it is easy to see that both $(fg)h$ and $f(gh)$ are equal to

$$\sum_{v\in {\bf N}^n} \sum_{t+u+w =v} a_tb_uc_w x^v,$$

by the associativity of addition of natural numbers, and the associativity of multiplication in a ring.

marcelgoh
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  • Thanks for the answer! Can you elaborate on the last part in which the result follows from associativity of addition and multiplication of natural numbers. Is it to say that I can change the subscripts of the elements a,b,c in R? – MNS1517 Nov 25 '22 at 18:17
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    I could be wrong, but I believe you will need associativity of addition so that $(t+u)+w = t+(u+w)$ for the integer vectors, and you'll need associativity of multiplication to delete brackets in $(a_tb_u)c_w = a_t(b_uc_w)$. – marcelgoh Nov 25 '22 at 18:20
  • I am unfortunately not sure I follow. a_t, b_u, c_w are elements in our ring R, so why does associativity of multiplication of the natural numbers make us able to delete the brackets? – MNS1517 Nov 25 '22 at 18:40
  • Hi, sorry, that was a mistake. I'll fix it. Associativity of multiplication will come from the ring $R$, but we still need associativity of addition from ${\bf N}$. Incidentally, there is ${\bf R}$ and normal $R$ in your question. Are both of these meant to be the same or does one denote the real numbers and the other denote an arbitrary ring? – marcelgoh Nov 25 '22 at 18:43
  • my apologies, all the R's is my question are suppose to refer to the same ring R. – MNS1517 Nov 25 '22 at 18:47
  • Is this an arbitrary ring? If so, I would avoid using the boldface ${\bf R}$, since that is usually used to denote the real numbers. – marcelgoh Nov 25 '22 at 18:48
  • Yes, R is an arbitrary commutative ring. – MNS1517 Nov 25 '22 at 18:50
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    Okay, I think my answer is valid across arbitrary rings. I see that you are a new contributor so a reminder that if my post answered your question, you can click the checkmark beside it to accept it! – marcelgoh Nov 25 '22 at 18:54
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Since it seems like you are learning university level algebra, this simpler but more theoretical proof could be instructive.

Note that the map $(fg)h - f(gh)$ is an $R$-module map $$M: R[X_1, \ldots, X_n]^{\otimes 3} \to R[X_1, \ldots, X_n]$$ since it is obtained as compositions and subtractions of the multiplication, which is a $R$-module map. The key observation is that the triple tensor product is generated by tensor products of monomials $X^{\alpha}\otimes X^{\beta}\otimes X^{\gamma}$. If we manage to show that $M$ is zero in generators, we show associativity.

Product of monomials is particularly easy: you can show that the summation $X^{\alpha}X^{\beta}(\nu)$ is zero unless $\nu=\alpha+\beta$, and in this case the result is one, since only one summand is non-zero. We deduce that $X^{\alpha}X^{\beta} = X^{\alpha+\beta}$, which yields

$$M(X^{\alpha}\otimes X^{\beta}\otimes X^{\gamma}) = (X^{\alpha}X^{\beta})X^{\gamma} - X^{\alpha}(X^{\beta}X^{\gamma} ) = X^{(\alpha+\beta)+\gamma} - X^{\alpha+(\beta+\gamma)} = 0$$

Voilà!

Andrea Marino
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