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Let be $k$ a field and $k[x_1,x_2,...,x_n]$ its polynomial ring in $n$ variables. Let be $I$ the ideal generated by $x_1-c_1,x_2-c_2,...,x_n-c_n$, where $c_1,...,c_n$ are elements of $k$. I want to show that $I$ is a maximal ideal.

I think it's useful to argue by induction on $n$, using the isomorphism between $k[x_1,x_2,...,x_n]/(x_1,...,x_{n-1})$ and $k[x_n]$, and that quotient of ring by a maximal ideal is a field. Moreover the base case $n=1$ is easy to verify.

In addition, I think there is the more general statement: Let be $p_1(x),...,p_n(x)$ irreducible in $k[x]$, then the ideal generated by $p_1(x_1),...,p_n(x_n)$ is maximal in $k[x_1,...,x_n]$. Maybe it's easier to prove that, since we don't get messed up in details as in the above specific case.

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No induction. Just compute the quotient. –  Martin Brandenburg Mar 6 at 0:16
    
I have difficulties with that, since I don't see clearly, which elements are in a. For example, is $x_1$, $(x_1)^m$ in a where $m$ is a natural number? Is the quotient ring by a isomorphic to k? –  bjn Mar 6 at 0:23

4 Answers 4

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There is no need to do induction. Instead, try to prove that the ring homomorphism $$\phi: k \rightarrow \frac{k[x_1,\dots,x_n]}{(x_1-c_1,\dots,x_n-c_n)}$$ given by $\phi(\alpha) = \bar{\alpha}$ is an isomorphism.

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You can easy answer to the 1st request using the isomorphism

$$R/I \simeq (R/J) / (I/J)$$ that holds if $I$ and $J$ are ideals such that $J \subseteq I$.

Then you obtain

$$\mathbb{k}[x_1,\dots, x_n]/(x_1-c_1, \dots, x_n-c_n) \simeq $$ $$\simeq \mathbb{k}[x_1,\dots, x_n]/(x_n-c_n)/(x_1-c_1, \dots x_{n-1}-c_{n-1})^* \simeq$$ $$ \simeq \mathbb{k}[x_1,\dots, x_{n-1}]/(x_1-c_1, \dots x_{n-1}-c_{n-1}) \simeq$$ $$\simeq \mathbb{k}[x_1,\dots, x_{n-2}]/(x_1-c_1, \dots x_{n-2}-c_{n-2})\simeq \dots \simeq \mathbb{k}[x_1]/(x_1-c_1) \simeq \mathbb{k}$$

Now we come to the general statement: It is FALSE!

Let consider the ring $\mathbb{R}[x,y]$ and the ideal $I=(x^2+1,y^2+1)$

The quotient $\mathbb{R}[x,y]/I$ is isomorphic to $\mathbb{C}[x]/(x^2+1)\simeq\mathbb{C^2}$ that is not a domain! (Then the ideal is not only non maximat but is also not prime!)

Your 2nd statement is TRUE if the field is algebrically closed. In fact in this case the only irreducible polynomials are of degree 1 and we are in the case above.

(This is one of the Hilbert's Nullstellensatz forms that says that if $\mathbb{k}=\bar{\mathbb{k}}$ then the maximal ideals of $\mathbb{k}[x_1,\dots, x_n]$ are all of the form $(x_1-c_1, \dots, x_n-c_n)$)

*In this passage and in the following I'm writing $((x_1-c_1, \dots x_{n-1}-c_{n-1}))$ for "the class of the ideal $(x_1-c_1, \dots x_{n-1}-c_{n-1})$ in the quotient $R/(x_n-c_n)$". I'm sorry for the abuse of notation!

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Your general statement is unfortunately false:
Consider the irreducible polynomial $p_1(x)=p_2(x)=x^2+1\in \mathbb R[x]$.
Then the ideal $\langle x^2+1, y^2+1\rangle \subset \mathbb R[x,y]$ generated by $p_1(x)$ and $p_2(y)$ is not maximal.
Indeed $$\mathbb R[x,y]/\langle x^2+1, y^2+1\rangle\cong (\mathbb R[x]/\langle x^2+1\rangle)[y] /\langle y^2+1\rangle \cong \mathbb C[y]/\langle y^2+1\rangle\cong \mathbb C\times \mathbb C$$ is not a field.

Edit
The above calculation is strongly linked to the formula $A/I\otimes_R B/J\cong A\otimes_R B/(I^e+J^e)$ for the tensor product of quotient $R$-algebras, in which the notation is more or less self-explanatory once you know that $(...)^e$ means extension of ideals.

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Hint: Recall that an ideal is maximal if quotienting by it gives a field.

Now consider the homomorphism $\phi :k[x_1,\dots,x_n]\rightarrow k$ given by $\phi(f(x_1,\dots x_n)) = f(c_1,\dots,c_n)$.

For the generalisation, do you know much about what happens when you form the quotient $\frac{k[X]}{f(X)}$ for a given irreducible polynomial $f(X)$?

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