I want to prove that $\bar{K}[V]/M_p \simeq \bar{K}$ where $K$ is a field, $\bar{K}$ is its algebraic closure and $$\bar{K}[V]=\bar{K}[x_1,...,x_n]/I_V,$$ where $I_V$ is the ideal attached to a variety, so it is prime.

Now, on the other hand:

$$M_p=\{f \in \bar{K}[V] : f(p)=0\}.$$

So we have to find a bijective function from this two fields (may be there are better ways to do this, but I am not quite sure if in field theory we can find that machinery), but my intuition says that in $\bar{K}[V]/M_p$ we don't have any vanishing polynomial at $V$ so I can't figure out how to find this bijection.

Can someone help me with this problem?

Thanks a lot in advance.

  • 1
    $\begingroup$ I don't understand your question. Firstly $\tilde{K}[x_1,\dotsc,x_n]$ is not a field. Secondly you define $\tilde{K}[V]$ as a quotient of $\tilde{K}[x_1,\dotsc,x_n]$, and then take its quotient by another ideal of $\tilde{K}[x_1,\dotsc,x_n]$ which doesn't make sense. $\endgroup$ – Josh Chen Feb 12 '16 at 2:36
  • $\begingroup$ Well the ideal is prime :),can you clarify please ? :) $\endgroup$ – user162343 Feb 12 '16 at 2:37
  • $\begingroup$ I think you may have some typos in your question then :) See my first comment. $\endgroup$ – Josh Chen Feb 12 '16 at 2:38
  • $\begingroup$ let me check :) so I can be sure just a second $\endgroup$ – user162343 Feb 12 '16 at 2:39
  • $\begingroup$ I've edited :) is it clear now? $\endgroup$ – user162343 Feb 12 '16 at 2:40

If $P=(a_1,\dots,a_n)$, then $M_P=(x_1-a_1,\dots,x_n-a_n)/I(V)$; see Proving that kernels of evaluation maps are generated by the $x_i - a_i$ (the proof given there works for any field).

Then $\overline K[V]/M_P\simeq\overline K[x_1,\dots,x_n]/(x_1-a_1,\dots,x_n-a_n)\simeq\overline K$; for the last isomorphism see Maximal ideals in $K[X_1,\dots,X_n]$.


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