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  "body": "\n## Definition\n\nLet $p$ be a prime, let $n \\ge 1$, and let $f \\in \\F_p[x]$ be a monic [[irreducible polynomial|00X2A9G9W3]] of [[degree|0DPF6WRZW6]] $n$.\nThe [[quotient ring|0VFW4Y69BT]] $\\F_p[x]/(f)$ of the [[polynomial ring|0Q9EZS90GY]] is a [[finite field|08RTDHX1SK]] with $p^n$ elements.\n*Finite field arithmetic* in this representation computes with the polynomials $a(x)$ of degree less than $n$ over $\\F_p$, one for each element, as follows.\n\n1. The sum $a + b$ is the coefficientwise sum modulo $p$.\n2. The product $a \\cdot b$ is the remainder of the polynomial product $a(x) b(x)$ under [[division|D6D0SR1FK3]] by $f$.\n3. The inverse of $a \\ne 0$ is the polynomial $u$ of degree less than $n$ with $u a + v f = 1$ for some $v \\in \\F_p[x]$, found by the extended [[Euclidean algorithm|0TRNT4X7J0]].\n\n## Remarks\n\nTake $p = 2$ and $f = x^3 + x + 1$, which has no root in $\\F_2$ and so is irreducible.\nThen $(x^2 + 1)(x + 1) = x^3 + x^2 + x + 1$, and its remainder modulo $f$ is $x^2$.\nThe inverse of $x$ is $x^2 + 1$, since $x(x^2 + 1) = x^3 + x = f + 1$.\nThe polynomial $x^2 + 1$ over $\\F_2$ cannot serve as $f$, since $x^2 + 1 = (x + 1)^2$, and then $x + 1$ has no inverse.\nOver $\\F_2$, an element is stored as the bit string of its coefficients, and addition is the bitwise exclusive or.\nThe inverse can also be computed as $a^{p^n - 2}$, since $a^{p^n - 1} = 1$ for $a \\ne 0$.\n",
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