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Let $f_i\in\mathbb{C}[X_i]$ be a monic polynomial of $X_i$, and we consider : $$ I=(f_1,f_2,\dots,f_n) \in \mathbb{C}[X_1,X_2,\dots,X_n] $$ the ideal formed by $f_i$. I think that $I$ will be maximal iff each $f_i$ is of degree $1$, but when this ideal will be prime and when it will be trivial ?

user26857
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Hamza
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    Your intuition works well when you have a polynomial ring over an algebraically closed field like $\mathbf{C}$. Over the real numbers, for example when $n = 1$, the ideal generated by an irreducible polynomial of degree 2 is maximal. – Alex Macedo Dec 22 '16 at 01:54
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    yes you are right i will fix it – Hamza Dec 22 '16 at 03:36
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    $\mathbb{C}[X_1,X_2,\dots,X_n]/(f_1,f_2,\dots,f_n)\simeq\mathbb C[X_1]/(f_1)\otimes_{\mathbb C}\cdots\otimes_{\mathbb C}\mathbb C[X_n]/(f_n)$. Now you can refer to this answer in order to find out when your quotient is an integral domain. You can also use this isomorphism for finding the answer to the other questions. – user26857 Dec 22 '16 at 11:40
  • can I get the answer more directly please, because i have any formation on this level of commutative algebra or in algebraic geometry. – Hamza Dec 22 '16 at 11:47
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    I think a lot of people (aside from @user26857) are missing the part where you say that $f_i$ depends on $X_i$ only and not the other $X_j$. – Daniel McLaury Dec 22 '16 at 17:46

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Over $\mathbf{C}$ or any other algebraically closed feel $K$, you proceed as follows. Suppose $I$ is a maximal ideal of $K[x_1,\dots , x_n]$. Then $K[x_1,\dots,x_n]/I$ is a field finitely generated as an algebra over $K$. By Zariski's lemma, it is a finite (hence algebraic) extension of $K$.

The hypothesis $K$ algebraically closed now implies that $K[x_1, \dots, x_n]/I$ is itself the copy of $K$ in this quotient. Therefore, if the class of $x_i$ in the quotient equals the class of $\alpha_i \in K$, then $x_i - \alpha_i$ must be in the ideal $I$.

Since the ideal $(x_1 - \alpha_1,\dots , x_n - \alpha_n)$ is maximal, you conclude that $I = (x_1 - \alpha_1,\dots , x_n - \alpha_n)$.

(Can you see what goes wrong when $K$ is not algebraically closed?)

(And I recommend that you use the case $n = 2$ as an exercise to investigate what prime but not maximal ideals of $\mathbf{C}[x, y]$ look like)

Alex Macedo
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