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I want to solve this problem, but I have no idea how I can start:

If $K$ is a field, $(a_1,...,a_n) \in K^n,$ and $I$ the ideal $I=\langle x_1-a_1,...,x_n-a_n\rangle$, then how can we prove that $I$ is a maximal ideal?

One example: Is $\langle x^2+1 \rangle$ a maximal ideal of $ \mathbb{R}[x]$?


marked as duplicate by user26857, Grigory M, Simon S, Hagen von Eitzen, Micah Dec 17 '14 at 19:51

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  • $\begingroup$ Do you know that an ideal is maximal iff the quotient by the ideal is a field? $\endgroup$ – Tobias Kildetoft Dec 16 '14 at 12:27
  • $\begingroup$ A maximal ideal in which ring? Certainly not in $K[x_1,\ldots,x_n,x_{n+1}]$ $\endgroup$ – Hagen von Eitzen Dec 17 '14 at 19:34

For your first question, suppose that $J\supset I$ is a larger ideal. You want to show that in fact $J=K[x_1,\ldots,x_n]$. Now let $f\in J\setminus I$. Apply the division algorithm to $f$ with respect to $x_1-a_1,\ldots,x-a_n$ and let $r$ be the remainder. Why must $r\in K$? Why must $r\neq 0$? So $J$ contains a nonzero element of $K$. Why then must $J= K[x_1,\ldots,x_n]$?

For your second question, consider $\mathbb{R}[x]/\langle x^2+1\rangle.$ Can you recognize this as a certain field?

  • $\begingroup$ If r=0 then I=K[x_1,...,x_n], right? and r must be in K, because if $r \notin K$ then J/I=J , right? $\endgroup$ – user40491 Dec 16 '14 at 13:27
  • $\begingroup$ If $r=0$, then $f\in I$ but we chose $f$ specifically to not be in $I$. We have $r\in K$ by the division algorithm: The remainder must not be divisible by the leading term of any of the polynomials we are dividing by, which means $r$ has degree $0$. $\endgroup$ – Casteels Dec 16 '14 at 13:36
  • $\begingroup$ thank you for your useful answers, for the second question, since for every polynomial in R[x], if we use the division with rest algorithm, then the rest is always is $\mathbb{R}[x]$ consequently $\langle x^2 + 1 \rangle$ is maximal ideal.? $\endgroup$ – user40491 Dec 16 '14 at 14:01
  • $\begingroup$ I guess for my hint you need the result that Tobias mentioned in his comment. $\endgroup$ – Casteels Dec 16 '14 at 14:34

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