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I saw the following statement "If polynomial ring $k[X_1,\cdots,X_n]$ is field then this implies that $n=0$" but I can't understand this statement. I want to completed proof of this statement.

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Consider whether the nonzero element $X_1$ has a multiplicative inverse. –  Shaun Ault Sep 1 '11 at 12:45
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Maybe it's easier if you think of the contrapositive: if $n\ge1$, then $k[X_1,\dots,X_n]$ is not a field (because, for example, the element $X_1$ has no multiplicative inverse). –  Gerry Myerson Sep 1 '11 at 13:04
    
Just another reason: if $n \geq 1$, then the ring $k[X_1, \dots, X_n]$ has at least $\vert k \vert^n$ maximal ideals (one for each point of $\mathbb{A}^n_k$), so it is not a local ring. –  Andrea Sep 1 '11 at 13:32

1 Answer 1

up vote 6 down vote accepted

HINT $\rm\qquad\rm X_1 \ f\:(X_1,\ldots,X_n) = 1 \ $ in $\rm\: k[X]\:\ \Rightarrow\: \ 0 = 1 \ $ in $\rm\: k \ \ $ by evaluating at $\rm\ X_i = 0\:. $

NOTE $\:\ $ This has a very instructive universal interpretation: if $\rm\: x\:$ is a unit in $\rm\: R[x]\:$ then so too is every $\rm\: R$-algebra element $\rm\: r\:,\:$ as follows by evaluating $\ \rm x \ f(x) = 1 \ $ at $\rm\ x = r\:.\:$ Therefore to present a counterexample it suffices to exhibit any nonunit in any $\rm R$-algebra. $\:$ A natural choice is the nonunit $\;\rm 0\in R\:,\:$ which yields the above simple proof.

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