0

The following fact is totally obvious, but I cannot find a way to prove it.

Let $R$ be a UFD and $a \in R$ be non zero and non invertible. Factor it as a product of irreducible elements: $a=p_1 \cdot ... \cdot p_n. $ Let $b \in R$ be a proper divisor of $a$, that is to say, $b$ is not a unit nor associate to $a$. Write also $b$ as a product of irreducible elements: $b=q_1\cdot ... \cdot q_m$. Then $m <n$.

Any suggestions to prove this?

carciofo21
  • 569
  • 3
  • 13
  • 3
    Hint: write $a=bc$ and factor $c$ into irreducibles. Compare the factorizations of $a$ and $bc$ you get this way. – Wojowu Jan 29 '20 at 14:19
  • Well in this case, $c$ must be a proper divisor itself, hence non zero and non invertible and we can factor it as $c=r_l \cdot ... \cdot r_l$ we have $p_1 \cdot ... \cdot p_n = q_1 \cdot ... \cdot q_m \cdot r_1 \cdot ... \cdot r_l$ and by essential uniqueness of factorization $n=m+l$. From here the thesis is trivial. Does this work? – carciofo21 Jan 29 '20 at 14:38
  • Yep, that works. – Wojowu Jan 29 '20 at 14:38

2 Answers2

1

Hint:

The shortest way uses induction on the number $m$ of prime factors of $b$:

  • If $m=1$, as anyway $n\ge 1$, there's nothing to prove.
  • Suppose the assertion is true an element with $m$ prime factors and consider the case $b=q1\,q_2\dotsm q_m\,q_{m+1}$. The last prime $q_{m+1}$ divides $b$, hence $a=p_1\,p_2\dots p_n$. By Euclid's lemma, $q_{m+1}$ is associated to one of the $p_i$s, say $p_n$.

Set $a'=\frac a{p_n}$, $b'=\frac{b}{q_{m+1}}$ and check $b'$ divides $a'$.

Bernard
  • 175,478
1

Hint: $\,b\mid a\,\Rightarrow\, a = bc.\,$ By hypothesis $\,c\,$ is nonunit so has at last $\rm\color{#c00}{one}$ prime factor. Counting primes in the unique prime factorization of $\,\color{#90f}a=\color{#0a0}b\,\color{#c00}c\,$ yields that $\, \color{#90f}n \ge \color{#0a0}m+\color{#c00}1 > m$.

Remark $ $ More generally in a UFD: $\rm \ a\mid b_1\cdots b_n \iff a = a_1\cdots a_n,\,$ $\rm\, a_1\mid b_1,\ldots, a_n\mid b_n.\,$ By induction we can extend this to Riesz refinement matrices which show how any two factorizations of an integer have a common refinement, e.g. if we have two factorizations $\rm\: a_1 a_2 = n = b_1 b_2 b_3\:$ then Schreier refinement implies that we can build the following refinement matrix, where the column labels are the product of the elements in the column, and the row labels are the products of the elements in the row

$$\begin{array}{c|ccc} &\rm b_1 &\rm b_2 &\rm b_3 \\ \hline \rm a_1 &\rm c_{1 1} &\rm c_{1 2} &\rm c_{1 3}\\ \rm a_2 &\rm c_{2 1} &\rm c_{2 2} &\rm c_{2 3}\\ \end{array}$$

This implies the following common refinement of the two factorizations

$$ \begin{align}\rm a_1 a_2 &\rm =(c_{1 1} c_{1 2} c_{1 3}) (c_{2 1} c_{2 2} c_{2 3})\\ \rm = b_1 b_2 b_3 &\rm =(c_{1 1} c_{2 1}) (c_{1 2} c_{2 2}) (c_{1 3} c_{2 3})\end{align}$$

Bill Dubuque
  • 272,048