Artin's Algebra, maximal ideals of $\mathbb{C}[x, y]/(y^2+x^3-17)$

Question

I am trying to solve the following problem from Artin's Algebra.

Let $I$ be the principal ideal of $\mathbb{C}[x, y]$ generated by the polynomial $y^2 + x^3 - 17$. Which of the following sets generate maximal ideals in the quotient ring $R = \mathbb{C}[x, y]/I$? $(x-1, y-4), (x+1, y+4), (x^3 - 17, y^2)$.

For the latter ideal I was able to use the Correspondence Theorem and the identity $(A/I)/(B/I) \simeq A/B$ to write it as $\mathbb{C}[x, y]/(x^3 - 17, y^2)$, and then show that this is not a field. But for the first two neither of them contain $(y^2 + x^3 - 17)$, so the Correspondence Theorem doesn't apply.

Answer 1

I wonder if you might have a typo in your post, and your second ideal should read $\langle x-1, y+4 \rangle$. Indeed, $\langle x-1, y-4 \rangle$ and $\langle x-1, y+4 \rangle$ both contain $\langle y^{2}+x^{3}-17 \rangle$.

The general principle here is both simple and powerful. Consider the (unique) $\mathbb{C}$-algebra homomorphism $\varphi \colon \mathbb{C}[x, y] \to \mathbb{C}$ which sends $x$ to $1$ and $y$ to $4$. It is straightforward to see that $\langle x-1, y-4 \rangle$ is contained in the kernel of this homomorphism. I encourage you to show that in fact, $\langle x-1, y-4 \rangle$ is exactly the kernel of $\varphi$; as a hint, you might consider using Euclidean division, which is possible since the leading coefficients of $x-1$ and $y-4$ are units. (You may have already seen something like this as a theorem in Artin.)

But now you can see that $\varphi$ sends $y^{2}+x^{3}-17$ to $(4)^{2}+(1)^{3}-17 = 0$, so $y^{2}+x^{3}-17$ belongs to $\langle x-1, y-4 \rangle$. A very similar argument shows that $y^{2}+x^{3}-17 \in \langle x-1, y+4 \rangle$ - again, I encourage you to write this out for yourself.

In the interest of explicitness, incidentally, it's not too hard to express $y^{2}+x^{3}-17$ as a $\mathbb{C}[x, y]$-linear combination of $x-1$ and $y-4$. Indeed, by just playing around and trying to naively cancel the highest order terms, I found the decomposition:

$$y^{2}+x^{3}-17 = (x-1)^3+3(x-1)^2+3(x-1)+(y-4)^2+8(y-4)$$

You might see for yourself how to modify this decomposition with $y+4$ in place of $y-4$ (flip a few signs). Or, if you prefer cute tricks, you might like the following more:

$$y^{2}+x^{3}-17 = x^{3}-1 + y^{2}-16 = (x-1)(x^{2}+x+1) + (y+4)(y-4)$$

which demonstrates simultaneously why $y^{2}+x^{3}-17$ belongs to both $\langle x-1, y-4 \rangle$ and $\langle x-1, y+4 \rangle$.