Let $K$ be the set of all polynomials with rational coefficients that are integral on the integers. Does the set of polynomials $\binom{x}{j}$ for $j=0,1,2,...$ form an integral basis for K? For example, $\frac{x^2+5x+4}{2}=1\cdot\binom{x}{2}+3\cdot\binom{x}{1}+2\cdot\binom{x}{0}$ and $\frac{2x^3-21x^2+37x+6}{3}=4\binom{x}{3}-10\binom{x}{2}+6\binom{x}{1}+2\binom{x}{0}$. In other words is every polynomial in $K$, of degree $n$, of the form $\sum_{j=0}^n a_j\binom{x}{j}$ for some integers $a_i$? Clearly those polynomials are in $K$, but how do we show that $K$ consists of precisely these polynomials?
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The degree of each $\dbinom x j$ is equal to $j$. So you have a polynomial for each degree. – player3236 Sep 13 '20 at 04:13
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See answer1 and answer2. – Bill Dubuque Sep 13 '20 at 04:45
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You can prove it by induction or you can just think of the $n$ polynomials of degree up to $n-1$ as vectors in $\mathbb R^n$ formed by the coefficients. Since the polynomials all have different degree you get a triangular matrix with nonzero entries on the diagonal, hence the determinant is nonzero.
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