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I'm studying Sharp's Steps in Commutative Algebra, and I need a hint how to proceed with this exercise in the page 26:

First of all, I didn't understand even the notation, what did the author mean by $(X_1-\alpha_1,...,X_n-\alpha_n)$?

Thanks a lot.

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  • $\begingroup$ Definition 2.15 in my copy describes that notation as an alternative to $(X_1-a_1)K+\dots+(X_n-a_n)K$ where $K$ is the given polynomial ring. $\endgroup$ – Karl Kronenfeld Jun 8 '13 at 2:21
  • $\begingroup$ For the proof you can change coordinates so that you can assume WLOG that $\alpha_1 = ... = \alpha_n = 0$. $\endgroup$ – Qiaochu Yuan Jun 8 '13 at 2:29
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    $\begingroup$ I believe $(X_1-\alpha_1,\dotsc,X_n-\alpha_n)$ is standard notation for the ideal generated by the $X_i-\alpha_i$, which is the set user1 describes. You might also see it written $\langle X_1-\alpha_1,\dotsc,X_n-\alpha_n\rangle$. $\endgroup$ – Josh Chen Jun 8 '13 at 2:36
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Basically the problem can be translated as "Show that if a polynomial in $n$ variables has $(\alpha_1,\ldots,\alpha_n)$ as a root, then it is of the form $f_1\cdot(x_1-\alpha_1)+\ldots +f_n\cdot (x_n-\alpha_n)$ for some polynomials $f_1,\ldots,f_n$.

Can you do this for a polynomial in 1 or 2 variables?

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  • $\begingroup$ I solved for a polynomial of 1 variable, I'm having troubles to prove this statement for more than 1 variable: $p\in Kerf\implies p\in (X_1-\alpha_1,...,X_n-\alpha_n)$, where f is the evaluation homomorphism. I've already done the converse. $\endgroup$ – user42912 Jun 8 '13 at 3:08
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Hint: By the division algorithm, $\, f = (x_1\! -\! a_1) q_1 + r_1,\ $ for $\ r_1\in R[x_2,\ldots,x_n].\,$ By induction, $\,r_1 = (x_2\!-\!a_2)q_2 +\cdots + (x_n\!-\!a_n)q_n + r_n,\,$ for $\,r_n\in R.\ $ Evaluating at $\,x_i = a_i\,$ yields $\,f(a_i) = 0\iff r_n = 0\iff f\in (x_1\!-\!a_1,\ldots,x_n\!-\!a_n)$

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  • $\begingroup$ f is the evaluation homomorphism? $\endgroup$ – user42912 Jun 8 '13 at 11:31

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