TITLE: Checking that some ideal is maximal in the multivariate polynomial ring
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: 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.