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  title: 'Euclid's Lemma on prime numbers',
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  text: '[summary: The [4mf prime numbers] have a special property that they "can't be distributed between terms of a product": if $p$ is a prime dividing a product $ab$ of [48l integers], then $p$ wholly divides one or both of $a$ or $b$. It cannot be the case that "some but not all of $p$ divides into $a$, and the rest of $p$ divides into $b$".]\n\n[summary(Technical): Let $p$ be a [4mf prime] natural number. Then $p \\mid ab$ implies $p \\mid a$ or $p \\mid b$.]\n\nEuclid's lemma states that if $p$ is a [4mf prime number], which divides a product $ab$, then $p$ divides at least one of $a$ or $b$.\n\n# Proof\n\n## Elementary proof\n\nSuppose $p \\mid ab$ %%note:That is, $p$ divides $ab$.%%, but $p$ does not divide $a$.\nWe will show that $p \\mid b$.\n\nIndeed, $p$ does not divide $a$, so the [-5mw] of $p$ and $a$ is $1$ (exercise: do this without using integer factorisation); so by [bezouts_theorem B├ęzout's theorem] there are integers $x, y$ such that $ax+py = 1$.\n\n%%hidden(Show solution to exercise):\nWe are not allowed to use the fact that we can factorise integers, because we need "$p \\mid ab$ implies $p \\mid a$ or $p \\mid b$" as a lemma on the way towards the proof of the [-5rh] (which is the theorem that tells us we can factorise integers).\n\nRecall that the highest common factor of $a$ and $p$ is defined to be the number $c$ such that:\n\n- $c \\mid a$;\n- $c \\mid p$;\n- for any $d$ which divides $a$ and $p$, we have $d \\mid c$.\n\n[euclidean_algorithm Euclid's algorithm] tells us that $a$ and $p$ do have a (unique) highest common factor.\n\nNow, if $c \\mid p$, we have that $c = p$ or $c=1$, because $p$ is [4mf prime].\nBut $c$ is not $p$ because we also know that $c \\mid a$, and we already know $p$ does not divide $a$.\n\nHence $c = 1$.\n%%\n\nBut multiplying through by $b$, we see $abx + pby = b$.\n$p$ divides $ab$ and divides $p$, so it divides the left-hand side; hence it must divide the right-hand side too.\nThat is, $p \\mid b$.\n\n## More abstract proof\n\nThis proof uses much more theory but is correspondingly much more general, and it reveals the important feature of $\\mathbb{Z}$ here.\n\n$\\mathbb{Z}$, viewed as a [3gq ring], is a [-5r5]. ([integers_is_pid Proof.])\nThe theorem we are trying to prove is that the [5m1 irreducibles] in $\\mathbb{Z}$ are all [5m2 prime] in the sense of ring theory.\n\nBut it is generally true that in a PID, "prime" and "irreducible" coincide ([5mf proof]), so the result is immediate.\n\n# Converse is false\n\nAny composite number $pq$ (where $p, q$ are greater than $1$) divides $pq$ without dividing $p$ or $q$, so the converse is very false.\n\n# Why is this important?\n\nThis lemma is a nontrivial step on the way to proving the [-5rh]; and in fact in a certain general sense, if we can prove this lemma then we can prove the FTA.\nIt tells us about the behaviour of the primes with respect to products: we now know that the primes "cannot be split up between factors" of a product, and so they behave, in a sense, [5m1 "irreducibly"].\n\nThe lemma is also of considerable use as a tiny step in many different proofs.',
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