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  "body": "\n## Proof\n\nLet $K$ be a [[field|0QA8CHH8MG]]. For a nonzero $h \\in K[x]$, let $Z(h)$ be the set of [[roots|00EHX6FWC0]] of $h$ in $K$. For $d \\in K$, let $m_h(d)$ be the [[multiplicity|03XVP8Q4GY]] of $d$ as a root of $h$. We prove by induction on $n \\ge 0$ this claim: every nonzero $f \\in K[x]$ of [[degree|0DPF6WRZW6]] $n$ has $Z(f)$ finite and $\\sum_{d \\in Z(f)} m_f(d) \\le n$.\n\n1. The field $K$ is an [[integral domain|09S3PFBEPM]]: $1 \\ne 0$, and if $a b = 0$ with $a \\neq 0$, then $b = a^{-1} a b = 0$.\n2. So $K[x]$ is an integral domain by [[the theorem on polynomial rings|0B5Z1EC66J]], and $\\deg(g h) = \\deg g + \\deg h$ by the [[degree of a product|0T233AZ1GE]]. Nonzero factors [[cancel|0GJM24EXP9]] in $K$ and in $K[x]$.\n3. Let $h \\neq 0$ and $d \\in Z(h)$. By the [[factor theorem|0K3MWA7DMC]], $x - d$ divides $h$, so $m_h(d) \\ge 1$. Conversely, if $m_h(d) \\ge 1$, then $x - d$ divides $h$ and $d \\in Z(h)$.\n4. By step 3, the claim gives $\\abs{Z(f)} \\le \\sum_{d \\in Z(f)} m_f(d) \\le n$. So the claim implies both parts of the statement.\n5. Let $f \\neq 0$ have degree $n$. If $f$ has no root, then $Z(f)$ is empty and the sum is $0 \\le n$. This covers $n = 0$: a nonzero constant $a$ has the value $a \\neq 0$ everywhere.\n6. Now let $n \\ge 1$, assume the claim for degree $n - 1$, and let $c$ be a root of $f$. By the factor theorem, $f = (x - c) g$ with $g \\in K[x]$.\n7. Since $f \\neq 0$, also $g \\neq 0$. By step 2, $n = 1 + \\deg g$, so $\\deg g = n - 1$.\n8. For $d \\in K$, the [[evaluation homomorphism|0125QVNCFQ]] gives $f(d) = (d - c) g(d)$. By step 1, $f(d) = 0$ exactly when $d = c$ or $g(d) = 0$. So $Z(f) = Z(g) \\cup \\set{c}$.\n9. By the induction hypothesis, $Z(g)$ is finite, so $Z(f)$ is finite.\n10. Let $d \\neq c$. We show by induction on $m \\ge 0$ that for every $g_0 \\in K[x]$, if $(x - d)^m$ divides $(x - c) g_0$, then $(x - d)^m$ divides $g_0$. The case $m = 0$ is trivial.\n11. Let $m \\ge 1$ and $(x - c) g_0 = (x - d)^m k$ with $k \\in K[x]$. Evaluating at $d$ gives $(d - c) g_0(d) = 0$. Since $d - c \\neq 0$, step 1 gives $g_0(d) = 0$.\n12. By the factor theorem, $g_0 = (x - d) g_1$ with $g_1 \\in K[x]$. Then $(x - d)(x - c) g_1 = (x - d)(x - d)^{m-1} k$.\n13. Cancelling the nonzero factor $x - d$ gives $(x - c) g_1 = (x - d)^{m-1} k$. By the induction hypothesis of step 10, $(x - d)^{m-1}$ divides $g_1$, so $(x - d)^m$ divides $g_0$.\n14. If $(x - d)^m$ divides $g$, then it divides $f = (x - c) g$. With step 10 for $g_0 = g$, this gives $m_f(d) = m_g(d)$ for each $d \\neq c$.\n15. Let $m \\ge 1$. If $(x - c)^m k = (x - c) g$, then cancelling $x - c$ gives $g = (x - c)^{m-1} k$. Conversely, $g = (x - c)^{m-1} k$ gives $f = (x - c)^m k$.\n16. So $(x - c)^m$ divides $f$ exactly when $(x - c)^{m-1}$ divides $g$, for each $m \\ge 1$. Taking the largest such $m$ gives $m_f(c) = 1 + m_g(c)$.\n17. If $c \\notin Z(g)$, then $m_g(c) = 0$ by step 3. So in all cases $m_g(c) + \\sum_{d \\in Z(g), d \\neq c} m_g(d) = \\sum_{d \\in Z(g)} m_g(d)$.\n18. By steps 8, 14, 16 and 17, $\\sum_{d \\in Z(f)} m_f(d) = 1 + m_g(c) + \\sum_{d \\in Z(g), d \\neq c} m_g(d) = 1 + \\sum_{d \\in Z(g)} m_g(d)$.\n19. By the induction hypothesis for $g$, this is at most $1 + (n - 1) = n$. This proves the claim, and with step 4 the statement.\n\n## Idea\n\nEach root $c$ splits off a linear factor: $f = (x - c) g$, and $g$ has degree one less. The roots of $f$ are $c$ and the roots of $g$, because a field has no zero divisors. The factor $x - c$ adds exactly one to the multiplicity of $c$ and leaves the other multiplicities unchanged, so induction on the degree gives the bound.\n",
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