# 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.

• $\mathbb Q[x,y]/(x^2+1,y-1)\cong \Bbb Q(i)$, so is maximal. Commented May 18, 2020 at 14:34
• Ok..can you give me a hint, please? Commented May 18, 2020 at 14:35
• $\mathbb Q[x,y]/(x^2+1,y-1)\cong \Bbb Q[x]/(x^2+1)\cong\Bbb Q(i)$ Commented May 18, 2020 at 14:36
• I think you are referring to a ring homomorphsim..but what should be the exact map? Commented May 18, 2020 at 14:39
• $f(x,y)\mapsto f(x,1)\mapsto f(i,1)$ Commented May 18, 2020 at 14:40

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.

• Thanks sir , very helpful answer. Can you please elaborate the part where you say $\mathbb{Q}[x,y]/ \langle x^2+1,y-1 \rangle$ is isomorphic to $R$.? Sorry for being slow !! I understand the rest part of your answer. Commented May 20, 2020 at 7:38
• Ah, I thought you might raise that. It comes from one of the isomorphism theorems, but I will write it down anyway, to leave no stone unturned. Commented May 20, 2020 at 8:44
• I have got your point after a long thinking. Thanks again. Commented May 20, 2020 at 11:28
• Great. I've just got to $50$k! Commented May 20, 2020 at 11:50
• Congrats sir ! Can you be kind enough to tell me some book's name to learn more about ring of polynomials in several variables starting from the introduction.? Commented May 20, 2020 at 11:53