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Let $A=k[x_1,\ldots, x_n]$ and $I=(f_1, \ldots, f_n)\subset A$ with $f_i\in k[x_i]$ irreducible polynomials. Is it true that $I$ is a maximal ideal in $A$?

$I$ is a maximal ideal $\iff$ $1\in (I, g)$ for every $g\in A\setminus I$ $\iff$ $1\in (I, g)$ for every $g\in A$ with $\deg g<\deg f_i$ $\forall i$ (with the the lexicographical order). Any other ideas?

user26857
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Of course not! $A/I\simeq k[x_1]/(f_1)\otimes_k\cdots\otimes_kk[x_n]/(f_n)$, and a tensor product of fields is not necessarily a field. For a concrete (counter)example see this answer.

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user26857
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  • Do you know where I can find a proof of this isomorphism in your answer? – Vik78 Apr 15 '16 at 01:03
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    Why don't you just look at an example: Let $k=\mathbb{Q}$ and take $f=x^2-2, g=y^2-2$, Both are irreducible but the ideal $I=(f,g)\subset k[x,y]$ is not maximal, since $x^2-y^2\in I$, but $x-y, x+y$ are not in $I$. – Mohan Apr 15 '16 at 01:09
  • @Mohan Is there something wrong with understanding what's $A/I$ in terms of tensor product of fields? – user26857 Apr 15 '16 at 05:34
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    Not at all. I mentioned my example just as a starting point, to have an explicit example, if the OP was not yet comfortable with tensor products. – Mohan Apr 15 '16 at 13:23