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  "body": "\n## Proof\n\n1. Let $L/K$ be a [[finite extension|0NRYPM9Y27]], with [[degree|0GS74FGFEK]] $n = [L : K]$. Let $a \\in L$. We find a nonzero $f \\in K[x]$ with $f(a) = 0$.\n2. Case 1: $a^i = a^j$ for some $0 \\le i \u003c j \\le n$. Then $f = x^j - x^i$ is nonzero and $f(a) = 0$.\n3. Case 2: the elements $1, a, a^2, \\dots, a^n$ are pairwise distinct. They form a set $S$ of $n + 1$ elements of $L$.\n4. Suppose that $S$ were [[linearly independent|0EVHJFTG9Z]] over $K$. By [[extension to a basis|0M4SXE5WW3]], $S$ lies in a basis of $L$ over $K$ with at least $n + 1$ elements.\n5. By the [[dimension theorem|0WK0Y8RH1M]], every basis of $L$ over $K$ has $[L : K] = n$ elements, by the definition of [[dimension|05G2G0P224]]. This contradicts Step 4.\n6. So there are $c_0, \\dots, c_n \\in K$, not all zero, with $c_0 + c_1 a + \\dots + c_n a^n = 0$.\n7. Then $f = c_0 + c_1 x + \\dots + c_n x^n$ is a nonzero polynomial in $K[x]$ with $f(a) = 0$.\n8. In both cases $a$ is [[algebraic|0NPW1DWECM]] over $K$. Since $a \\in L$ was arbitrary, $L/K$ is [[algebraic|09EYMKMSB9]].\n\n## Idea\n\nIn a vector space of dimension $n$, any $n + 1$ vectors are linearly dependent. Applied to the powers $1, a, \\dots, a^n$, a linear dependence is exactly a nonzero polynomial with root $a$.\n",
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