The ideals $\langle y-x-1\rangle$ and $\langle x-2,y-3\rangle$ in $\mathbb C[x,y]$ are prime

Question

The following is a quote from Wolfram MathWorld article about prime ideals.

A maximal ideal is always a prime ideal, but some prime ideals are not maximal. In the integers, $\{0\}$ is a prime ideal, as it is in any integral domain. Note that this is the exception to the statement that all prime ideals in the integers are generated by prime numbers. While this might seem silly to allow this case, in some rings the structure of the prime ideals, the Zariski topology, is more interesting. For instance, in polynomials in two variables with complex coefficients $\mathbb C[x,y]$, the ideals $$\langle 0\rangle \subset \langle y-x-1\rangle \subset \langle x-2,y-3\rangle$$ are all prime.

How can we prove that $\langle y-x-1\rangle$ and $\langle x-2,y-3\rangle$ are prime ideals in $\mathbb C[x,y]$?

Perhaps showing that quotient rings are $\mathbb C[x,y]/\langle y-x-1\rangle\cong \mathbb C[x]$ and $\mathbb C[x,y]/\langle x-2,y-3\rangle\cong\mathbb C$ would be way to go? (This would mean the quotient ring is an integral domain and therefore the ideal is prime.)

Answer 1

Let me try to expand anon's comment to an answer.

Let us start with this simple observation:

Let $R$ be a commutative ring. If $f(x)\in R[x]$ and $a\in x$ then there exist uniquely determined $q(x)\in R[x]$ and $b\in R$ such that $$f(x)=q(x)(x-a)+b.$$ Moreover, we have $b=f(a)$.

This follows from the fact that we can divide by monic polynomials in $R[x]$ for any commutative ring $R$. Using this observation we can also show that the map $f(x)\mapsto f(a)$ is a homomorphism from $R[x]$ to $R$ and the kernel is precisely the ideal $\langle x-a\rangle$.

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Since we have $\mathbb C[x,y]=\mathbb C[x][y]$ we can use the above to get a homomorphism $$\varphi \colon \mathbb C[x,y] \to \mathbb C[x] \qquad \varphi \colon f(x,y) \mapsto f(x,x+1).$$ Kernel of this homomorphism is $\langle y-x-1 \rangle$. This shows that $$\mathbb C[x,y]/\langle y-x-1 \rangle \cong C[y].$$ So the quotient ring is an integrity domain and the ideal is a prime ideal.

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Similarly, we can get, using the above observation, the homomorphisms $$\psi_1 \colon \mathbb C[x][y] \to \mathbb C[x] \qquad \psi_1 \colon f(x,y) \mapsto f(x,3)$$ and $$\psi_2 \colon \mathbb C[x] \to \mathbb C \qquad \psi_2 \colon g(x) \mapsto g(2).$$ By composing these two homomorphisms we get $\psi=\psi_2\circ\psi_1$ from $\mathbb C[x,y]$ to $\mathbb C$ given by $f(x,y)=f(2,3)$.

Again we have that $\operatorname{Ker} \psi_2 = \langle x-2 \rangle$.

The polynomials in $\operatorname{Ker} \psi$ are precisely the polynomials such that $\psi_1(f)\in \operatorname{Ker} \psi_2$. So $f\in \operatorname{Ker} \psi$ if and only if $$f(x,y) = q(x,y)(y-3) + h(x)(x-2).$$ Every such polynomial belongs to $\langle x-2,y-3 \rangle$. And since both $x-2$ and $y-3$ belong to the kernel of $\psi$, we get $$\operatorname{Ker} \psi = \langle x-2,y-3 \rangle.$$

From this we get $\mathbb C[x,y]/\langle x-2,y-3 \rangle\cong\mathbb C$. Again, since the quotient ring is an integrity domain, we get that the ideal is prime.