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.)

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    $\begingroup$ Yes to your questions at the very end. They are kernels of evaluation maps. $\endgroup$ – anon Apr 24 '16 at 5:22
  • $\begingroup$ @user26857 Thanks for the edit. As you can see, I quoted the text from the link, so I written in in the same way. (As it was an exact quote.) What somewhat surprise me was the removal of (commutative-algebra) tag. According to the tag-excerpt this tag is for: "Questions about commutative rings, their ideals, and their modules." (But you are certainly more active in questions from this area as I am. So you probably know the norms of this math.SE subcommunity better.) $\endgroup$ – Martin Sleziak Apr 24 '16 at 6:42
  • $\begingroup$ To see that $<y - x- 1>$ is a prime ideal, you can also note that, like every other degree $1$ polynomial, it is irreducible, and that irreducible elements are prime in, for example, UFDs (polynomial rings over $\mathbb{C}$ are certainly unique factorization domains). $\endgroup$ – Badam Baplan Aug 11 '18 at 15:50
  • $\begingroup$ @BadamBaplan Maybe you could expand your comment a bit and post it as an answer. $\endgroup$ – Martin Sleziak Aug 11 '18 at 17:09

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 $\langle x-a\rangle$.

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.

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.


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