Let $R:=\Bbb Z_5[X,Y]$ and $I:=\langle Y \rangle \trianglelefteq R.$

1) Prove that $I$ is prime but not maximal ideal.

2) Find all maximal ideals of $R$, which contain $I$.

Answer. 1) If we take te evaluation epimorphism $$\epsilon:R\longrightarrow \Bbb Z_5[X], \ f(X,Y)\longmapsto \epsilon (f(X,Y)):= f(X,0)$$ we deduce that $I=\ker \epsilon$ and thus from 1st Isomorphism Theorem for Rings, $$\frac{\Bbb Z_5[X,Y]}{\langle Y \rangle} \cong \Bbb Z_5 [X],$$ where the isomorphism is $$\theta : \frac{\Bbb Z_5[X,Y]}{\langle Y \rangle}\longrightarrow \Bbb Z_5[X],\ f(X,Y)+\langle Y \rangle \longmapsto \theta(f(X,Y)+\langle Y \rangle):=\epsilon (f(X,Y))=f(X,0). $$

So, $\Bbb Z_5[X]$ is an integral domain but not field $\iff \frac{\Bbb Z_5[X,Y]}{\langle Y \rangle}$ is an integral domain but not field $\iff \langle Y \rangle$ is prime but not maximal ideal of $R$.

2) As for the second statement, we know that if we have an ideal $I\trianglelefteq R$, then the mapping \begin{align*} \phi:\{J\trianglelefteq R:J\supseteq I\} & \longrightarrow \{K\trianglelefteq R/I\} \\ J & \longmapsto \phi(J):=J/I \end{align*} is a bijection and one can prove that an isomorphic image of maximal ideal is maximal ideal.

So, let's take the bijection \begin{align*} \phi:\{J\trianglelefteq \Bbb Z_5[X,Y]:J\supseteq I\} & \longrightarrow \{K\trianglelefteq \Bbb Z_5[X,Y]/\langle Y \rangle\} \\ J&\longmapsto \phi(J):=J/\langle Y \rangle. \end{align*}

We know that the maximal ideals of $\Bbb Z_5[X]$ have the form $\langle p(X) \rangle \trianglelefteq \Bbb Z_5[X]$, for some irreducible polynomial $p(X)\in \Bbb Z_5[X]$.

Update: Taking into account the comment, I change a little bit my thoughts:

Since $\frac{\Bbb Z_5[X,Y]}{\langle Y \rangle} \cong \Bbb Z_5 [X],$ $\langle p(X) \rangle$ is maximal in $\Bbb Z_5[X]$ iff $ \theta^{-1}(p(X))$ is maximal in $\frac{\Bbb Z_5[X,Y]}{\langle Y \rangle}$ iff $\phi^{-1}(\theta^{-1}(p(X))) $ is maximal in $R$ and contains $ I$.

So we have to compute the above ideals, but first observe that $f(X,Y)+\langle Y \rangle = f(X)+\langle Y \rangle$ (*), because $a(X,Y)Y^i\in \langle Y \rangle,\ \forall i$, so every expression with $Y$ is disappeared.

Now, \begin{alignat*}{2} \theta^{-1}(p(X))\quad = \quad & \{ f(X,Y)+\langle Y \rangle \in \frac{\Bbb Z_5[X,Y]}{\langle Y \rangle}:\theta(f(X,Y)+\langle Y \rangle)\in \langle p(X) \rangle \} \\ \quad = \quad & \{ f(X)+\langle Y \rangle \in \frac{\Bbb Z_5[X,Y]}{\langle Y \rangle}:f(X,0)\in \langle p(X) \rangle \} \\ \quad = \quad & \{ f(X)+\langle Y \rangle \in \frac{\Bbb Z_5[X,Y]}{\langle Y \rangle}:f(X)= p(X)h(X),\ h(X)\in \Bbb Z_5[X] \} \\ \quad = \quad & \{ p(X)h(X)+\langle Y \rangle \in \frac{\Bbb Z_5[X,Y]}{\langle Y \rangle}: h(X)\in \Bbb Z_5[X] \} \\ \quad = \quad & \{ p(X)h(X,Y)+\langle Y \rangle \in \frac{\Bbb Z_5[X,Y]}{\langle Y \rangle}: h(X,Y)\in \Bbb Z_5[X,Y] \} \\ \quad = \quad & \langle p(X) \rangle / \langle Y \rangle \end{alignat*} and the last equality holds because of (*).

But I don't like this result, because then $\langle p(X) \rangle \supseteq \langle Y \rangle\iff p(X)|Y$ and this can not happen.

What do I miss?

Are these above correct?

Of course any other easier way is welcome.


  • 2
    $\begingroup$ Looks good. I would go a little simpler on 1). Namely, since $Y$ is irr. then $(Y)$ is prime, and it is contained in $(X,Y)$ for example. $\endgroup$
    – Ruben
    Jun 21, 2019 at 14:58
  • 1
    $\begingroup$ By $\mathbb{Z}_5$ you mean the ring of 5-adic integers? Or do you mean $\mathbb{Z}/5\mathbb{Z}$? $\endgroup$ Jun 21, 2019 at 14:58
  • 1
    $\begingroup$ @RubenduBurck Thanks for your comment. $Y$ is irreducible in which ring and why? $\endgroup$
    – Chris
    Jun 21, 2019 at 15:00
  • 1
    $\begingroup$ @RubenduBurck Well, $Y$ is irreducible in $(\Bbb Z_5[X])[Y]$. Can we conclude that it is irreducible in $\Bbb Z_5[X,Y]$? $\endgroup$
    – Chris
    Jun 21, 2019 at 15:12
  • 2
    $\begingroup$ @Chris It is much simpler: write $Y=A\cdot B$, conclude on the degrees of $A,B$, conclude that $Y$ is irreducible. $\endgroup$
    – lisyarus
    Jun 21, 2019 at 16:22

1 Answer 1


@Chris, Right, so I think that the problem we are having is that we are getting confused as to which elements belong to which rings. First, remember that $\theta^{-1}(\langle p(X) \rangle$ is an element of $\frac{\mathbb{Z}_5[X,Y]}{\langle Y \rangle}$ not $\mathbb{Z}_5[X,Y]$. Thus, it really does not make sense to write $\frac{\langle p(X) \rangle}{\langle Y \rangle}$ (even though it is true that the $Y$ does 'disappear' in this ring). What you are forgetting is that when you take $\theta^{-1}(\langle p(X) \rangle)$, this does in fact contain $\langle Y \rangle$. Specifically, $\theta^{-1}(\langle p(X) \rangle)$ consists of all elements of $f(X,Y) + \langle Y \rangle \in \frac{\mathbb{Z}_5[X,Y]}{\langle Y \rangle}$ such that $\theta(f(X,Y) + \langle Y \rangle) \in \langle p(X) \rangle$. But, by your definition of your function $\theta$ this means $\epsilon(f(X,Y)) \in \langle p(X) \rangle$ which in turn means that $f(X,0) \in \langle p(X) \rangle $.

Now, think about what functions in $\mathbb{Z}_5[X,Y]$ have this property. Certainly every function in $\langle p(X,0) \rangle$ has this property (this is an ideal in $\mathbb{Z}_5[X,Y]$). But, as I mentioned in my comments, the ideals you are seeking need to contain $\langle Y \rangle$ as well. Thus, consider the ideal $\langle p(X,0), Y \rangle$ (again in $\mathbb{Z}_5[X,Y]$) which does contain $\langle Y \rangle $. Check that for any $h(X,Y) \in \langle p(X,0), Y \rangle$ that $\epsilon (h(X,Y)) \in \langle p(X) \rangle$. Hence, $\theta^{-1}(\langle p(X) \rangle) = \frac{\langle p(X,0), Y \rangle}{\langle Y \rangle}$. Let me know if this helps at all.

  • $\begingroup$ Sorry for the delay. This was actually the detail which makes the difference. Thank you very much! $\endgroup$
    – Chris
    Jun 24, 2019 at 23:21

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