# On the maximal ideals of $\Bbb Z_5[X,Y]$ which contain $\langle Y \rangle$

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.

Thanks

• 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. Jun 21, 2019 at 14:58
• By $\mathbb{Z}_5$ you mean the ring of 5-adic integers? Or do you mean $\mathbb{Z}/5\mathbb{Z}$? Jun 21, 2019 at 14:58
• @RubenduBurck Thanks for your comment. $Y$ is irreducible in which ring and why? Jun 21, 2019 at 15:00
• @RubenduBurck Well, $Y$ is irreducible in $(\Bbb Z_5[X])[Y]$. Can we conclude that it is irreducible in $\Bbb Z_5[X,Y]$? Jun 21, 2019 at 15:12
• @Chris It is much simpler: write $Y=A\cdot B$, conclude on the degrees of $A,B$, conclude that $Y$ is irreducible. Jun 21, 2019 at 16:22

@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.