# Counting ring homomorphisms from $\mathbb{Z}[x,y]/(x^3+y^2-1)$ to $\mathbb{Z_7}$

How many ring homomorphisms there is between $\mathbb{Z}[x,y]/(x^3+y^2-1)$ and $\mathbb{Z_7}$? Here $\mathbb{Z_7}$ denote ring of integers mod 7.

I don't know how to approach this problem,so far I've only worked with one variable ring so I can't tell the properties of $\mathbb{Z}[x,y]/(x^3+y^2-1)$. Thanks in advance

• Then let $R = \mathbb{Z}[x]$ and look at ring homomorphisms from $R[y]/(y^2 + x^3 - 1)$.
– user14972
Oct 5, 2017 at 7:35
• But... is the two variable case really different from the one variable case in any essential way? What methods would you normally use that you think don't apply?
– user14972
Oct 5, 2017 at 7:36
• Your first comment is indeed insightful,I was stuck in concepts such as prime ideals and properties of kernel and forgot to look at the quotient as you did,thank you. Oct 5, 2017 at 8:02

Note that any homomorphism $\mathbb{Z}[x,y]/(x^3+y^2-1) \to \Bbb Z_7$ induces by composition a unique canonical homomorphism $\Bbb Z[x, y]\to \Bbb Z_7$, so we can start by looking at those, because that's a lot easier.

Let's say we have a homomorphism $f$. The ring $\mathbb{Z}[x,y]$ has three generators, $1, x$ and $y$. The homomorphism has to send $1$ to $1$, which leaves in total $49$ possibilities for $f(x)$ and $f(y)$.

That would be the final answer if we were interested in maps from $\Bbb Z[x, y]$ to $\Bbb Z_7$. However, we are interested in maps from $\mathbb{Z}[x,y]/(x^3+y^2-1)$ to $\Bbb Z_7$. By the universal property of quotient rings, this is equivalent to counting the homomorphisms $\Bbb Z[x, y]\to \Bbb Z_7$ whose kernel contains the ideal $(x^3 + y^2 -1)$.

That specifically means that we want $f(x)$ and $f(y)$ to satisfy the relation $f(x)^3 + f(y)^2 - 1 = 0$, which limits the possibilities greatly. For instance, if $f(x) = 0$, then we must have $f(y)^2 = 1$, which has two solutions: $1$ and $6$. Thus there are two possible homomorphisms with $f(x) = 0$. Do this for the $6$ remaining possible $f(x)$, and you should have your answer.

• I found the answer very quick with your explanation,its 11! Thank you for breathing some wisdom into my brain, Oct 5, 2017 at 8:00
• No problem. Thinking in terms of generators and kernels is often very effective when dealing with relatively simple quotients of polynomial rings. Oct 5, 2017 at 8:02
• (The reason being that polynomial rings are just free generators, and quotients are intricately linked to kernels, as mentioned above.) Oct 5, 2017 at 8:04
• Why in this homomorphism $\mathbb{Z}[x,y] \rightarrow \mathbb{Z}_7$ we can map $x \mapsto 0$ or $y \mapsto 0$ if $0$ is not a generator of $\mathbb{Z}_7$? Jan 10 at 13:14
• @kombucza Because you're forgetting the constant polynomial $1$. We must have $1_{\Bbb Z[x, y]}\mapsto 1_{\Bbb Z_7}$ by definition of unital ring homomorphism. Also, there was no requirement that our final homomorphism is surjective in the first place, so it's not really a concern. Also, $x\mapsto0, y\mapsto 0$ turns out to not work when out main goal is to look at homomorphsims $\Bbb Z[x, y]/(x^3+y^2-1)\to \Bbb Z_7$. Jan 10 at 14:20