The ideal $\langle x^2+1, y-1 \rangle$ in $\mathbb{Q}[x,y]$ Check whether $\langle x^2+1, y-1 \rangle$ is prime/maximal ideal in $\mathbb{Q}[x,y]$
I can see that $x^2+1$ and $y-1$ are irreducible and that $\mathbb{Q}[x,y]$ is not a PID since $\mathbb{Q}[x]$ is not a feild but I don't know if these are useful at all to determine the answer.
I have limited knowlege in ring of polynomials with several variables. Please help me in this problem. Thanks for your help. 
 A: The key to doing these kind of problems is seeing if we can bring it to analysis of domains extending $\mathbb Q$. Once we do that, it becomes easier to observe if such a thing is a domain or not.
The way to bring in such analyses in this case is to use the evaluation map to eliminate variable-by-variable.
In our case, $S= \mathbb Q[x,y]/\langle x^2+1,y-1\rangle$ is isomorphic to $R= (\mathbb Q[x]/\langle x^2+1\rangle)[y]/\langle y-1\rangle$ using one of the isomorphism theorems (well, I've done that below). Now this is isomorphic to $\mathbb Q[x]/\langle x^2+1\rangle$ because of the following : any element of $R$ is a polynomial of the form $a_0+a_1y + a_2y^2 + ... + a_ny^n$ for some $a_0,...,a_n \in \frac{\mathbb Q[x]}{\langle x^2+1\rangle}$. But then each of $y-1,y^2-1,y^3-1,...,y^n-1$ is a multiple of $y-1$, so in $R$ this element is equivalent to $a_0+a_1+a_2+...+a_n$, which is an element of $\frac{\mathbb Q[x]}{\langle x^2+1\rangle}$. Finally, the map is obvious, sending every equivalence class to this element. One can easily check that it is an isomorphism.
Now, $x^2+1$ is irreducible over $\mathbb Q$. We conclude using a common theorem that this is a field, and therefore the original ideal is maximal. Every maximal ideal is prime, hence it is prime as well.

To see the isomorphism between $\mathbb Q[x,y]/\langle x^2+1,y-1\rangle$ and $(\mathbb Q[x]/\langle x^2+1\rangle)[y]/\langle y-1\rangle$, first we make the "in words" interpretation of both these domains. This allows us to construct the isomorphism "in words". We then make things formal, like all mathematicians do.
Right : what is $\mathbb Q[x,y]/\langle x^2+1,y-1\rangle$? We first have $\mathbb Q[x,y]$, which is the set of all polynomials with rational coefficients in two variables $x,y$. Then, we "quotient" by $x^2+1$ and $y-1$ . This means that polynomials that differ by a (rational polynomial) multiple of either $x^2+1$ or $y-1$ will be "related", and we take the set of all equivalence classes under this relation.
What is the second domain, $(\mathbb Q[x]/\langle x^2+1\rangle)[y]/\langle y-1\rangle$? This is the set of all polynomials in $y$ with coefficients in $\mathbb Q[x]/\langle x^2+1\rangle$, after which we identify the polynomials which differ by a (polynomial in $\mathbb Q[x]/\langle x^2+1 \rangle$) multiple
of $y-1$.
Now,the idea is to use the fact that you can first handle $x^2+1$ and second handle $y-1$. For this, it is important to realize that $\mathbb Q[x,y]$ is nothing but $(\mathbb Q[x])[y]$, the set of polynomials in $y$ with coefficients in $\mathbb Q[x]$. 
Example : $$x^2+xy+3y^2 +2y^4x^8 + 5 = (5+x^2) + (x)y + (3)y^2 + (2x^8)y^4$$
So now, if you have to remove multiples of $x^2+1$, you do this from the coefficients of these $y^i$. After doing this, you will be left with a polynomial in $y$, whose coefficients will be in $\mathbb Q[x]/\langle x^2+1 \rangle$.
Therefore, an isomorphism can be described as follows : take a member of $\mathbb Q[x,y]/\langle x^2+1,y-1\rangle$ and gather terms containing the same powers of $y$ together as like terms. The resulting polynomial can be interpreted as a polynomial in $y$ with coefficients in $\mathbb Q[x]/\langle x^2+1 \rangle$, which has been reduced with respect to $y-1$ because the original polynomial already was.

Formally, if an element of $S$ is $[\sum_{i,j=0}^n a_{ij}x^iy^j]_{\langle x^2+1,y-1\rangle}$ (used to denote equivalence class modulo the ideal on the bottom), then the isomorphism to $R$ takes this to $\sum_{j=0}^n [[\sum_{i=0^n} a_{ij}x^i]_{\langle x^2+1\rangle} y^j]_{\langle y-1 \rangle}$. It is obvious that this works.
