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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

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    $\begingroup$ Then let $R = \mathbb{Z}[x]$ and look at ring homomorphisms from $R[y]/(y^2 + x^3 - 1)$. $\endgroup$
    – user14972
    Oct 5, 2017 at 7:35
  • $\begingroup$ 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? $\endgroup$
    – user14972
    Oct 5, 2017 at 7:36
  • $\begingroup$ 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. $\endgroup$
    – 2ndYearFreshman
    Oct 5, 2017 at 8:02

1 Answer 1

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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.

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  • $\begingroup$ I found the answer very quick with your explanation,its 11! Thank you for breathing some wisdom into my brain, $\endgroup$
    – 2ndYearFreshman
    Oct 5, 2017 at 8:00
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    $\begingroup$ No problem. Thinking in terms of generators and kernels is often very effective when dealing with relatively simple quotients of polynomial rings. $\endgroup$
    – Arthur
    Oct 5, 2017 at 8:02
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    $\begingroup$ (The reason being that polynomial rings are just free generators, and quotients are intricately linked to kernels, as mentioned above.) $\endgroup$
    – Arthur
    Oct 5, 2017 at 8:04
  • $\begingroup$ 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 $\mathbb{Z}_7$? $\endgroup$
    – kombucza
    Jan 10 at 13:14
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    $\begingroup$ @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 $\Bbb Z[x, y]/(x^3+y^2-1)\to \Bbb Z_7$. $\endgroup$
    – Arthur
    Jan 10 at 14:20

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