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Suppose every irreducible element in a domain $D$ is prime.

I'm trying to prove this implication:

In a integral domain $D$, if $a=c_1c_2...c_n$ and $a=d_1d_2...d_m$ ($c_i,d_i$ irreducible), then $n=m$ and up to order $c_i$ and $d_i$ are associates for every $i$.

My Solution

For each $i$, $c_i$ divides $d_1...d_m$, since $c_i$ is irreducible, hence prime, $c_i$ has to divide some $d_j$, $j\le m$, but $c_i$ and $d_i$ are irreducible, so $c_i$ and $d_j$ are associate.

Is my solution is correct so far?

how can I prove $n=m$?

Thanks in advance

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  • $\begingroup$ The way you had formulated the first few lines made your question seem confused, so I changed them to what I thought you meant. If I changed your meaning, feel free to revert it. $\endgroup$
    – rschwieb
    Commented May 30, 2013 at 13:06
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    $\begingroup$ You've jumped right into deducing things without having a clear strategy, I think :) That's fine, you will need these thoughts shortly anyway. You could go about this a couple ways. You could strive to prove the statement directly by induction as vadim123 has recommended, or you could try an argument by contradiction by supposing $n<m$. With either of these setups, you would use your step above and vadim's cancellation suggestion to proceed. $\endgroup$
    – rschwieb
    Commented May 30, 2013 at 13:10

1 Answer 1

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You're doing great, now cancel and continue. (all integral domains are cancellative)

More precisely: prove your claim by induction on $\min(m,n)$.

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    $\begingroup$ Tip: when you cancel the associate primes, you may be left with a spurious unit factor. That can be absorbed into one of the other primes. $\endgroup$
    – Key Ideas
    Commented May 30, 2013 at 13:37

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