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Is it true that $\langle x^2+y^2-1\rangle$ is prime in $\mathbb{C}[x,y]$ ?

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    Yes, it is: there is no polynomial $f(y)\in \mathbb C[y]$ such that $f(y)^2+y^2-1=0$ and it is monic. (If don't like this use Eisenstein.) – 2017-01-22

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One way to see that this ideal is prime is to show that the quotient ring is an integral domain. By change of variables $u=x+iy, v=x-iy$ you get that $$\mathbb{C}[x,y]/\left\cong \mathbb{C}[u,v]/\left\cong \mathbb{C}[t^{\pm 1}].$$

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    I wonder how would you solve the problem over $\mathbb R$ instead of $\mathbb C$, and more generally over any field of characteristic $\ne 2$? – 2017-01-22
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    @user26857 Over $\mathbb{R}$ you can use the embedding $\mathbb{R}[x,y]/ \hookrightarrow \mathbb{C}[x,y]/$ in order to get that the quotient is an integral domain. Also, the same argument used for $\mathbb{C}$ should work for every field which contains the square root of -1, and if it does not contain, you should add $\sqrt{-1}$ as in the argument for $\mathbb{R}$ – 2017-01-22
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    @Orfir Is it obvious that $\mathbb{R}[x,y]/\left\hookrightarrow\mathbb{C}[x,y]/\left$? (Not mentioning the general case you suggested.) – 2017-01-22
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    @user26857 It is not trivial, but not that hard to prove. What you need to show is that if $f\in \mathbb{R}[x,y]$ is divisible by $x^2+y^2-1$ over $\mathbb{C}$, then this is also true over $\mathbb{R}$. This is true since if $f=h(x,y)(x^2+y^2+1)$, then $Re(f)=Re(h)(x^2+y^2-1)$. – 2017-01-22
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    Right. Btw, this property holds in general: if $f,g\in K[X_1,\dots,X_n]$ and $K\subset L$ is a field extension such that $f\mid g$ in $L[X_1,\dots,X_n]$ then $f\mid g$ in $K[X_1,\dots,X_n]$. – 2017-01-22
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Yes. Since the conic $x^2+y^2=1$ is an irreducible curve in $\mathbb{C}^2$.

More than irreducible, its projective closure is smooth. One can simply check the vanishing of the partial derivatives.

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    And how exactly do you prove that the curve is irreducible? – 2017-01-22
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    How to prove the irreducibility ? – 2017-01-22
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    Are you serioussly suggesting that someone who cannot prove that polynomial irreducible use the fact that smooth curves in projective space are irreducible to do it? – 2017-01-22
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    @ Mariano Suárez-Álvarez. The answer meant to show an alternative way of looking at the problem. Many of the OP’s questions have the (alg. geometry)/(comm. algebra) tags. So I would not assume that he is not capable of understanding this possible approach. Even if that were the case, the answer can very well work as a motivation/invitation to explore related notions. I really don't see how this can be harmful. – 2017-01-22