Prove by definition that $(x,2)\subset\mathbb Z[x]$ is a maximal ideal

Question

When the polynomial ring $\mathbb{Z}[x]$ is quotiented by the ideal $(2,x)$ we get a field as $\mathbb{Z}[x]/(x,2)\cong\mathbb{Z}/(2)\cong\mathbb{Z}_{2}$ which is a field.

But I want to prove it with the definition of maximal ideal. So let $I$=$(2,x)$ be not maximal. If possible, there is ideal $J$ such that $I\subsetneq J\subsetneq \mathbb{Z}[x]$ . If only $\mathbb{Z}[x]$ were a PID, I could say $J$ should have one element dividing both $2$ and $x$ but their gcd being $1$ , $1\in J$ i.e. $J$=$\mathbb{Z}[x]$. But since $\mathbb{Z}[x]$ is not a PID this argument won't work , $J$ can be generated by two different elements . So how do I contradict the existence of $J$ here?

Thanks for any help.

Answer 1

Try it like this. Let $p(x)= a_0 + a_1x + a_2x^2 + \dots + a_nx^n$ (with each $a_k \in \mathbb{Z}$) be any element that is in $J$ but not in $I$. Now note that $a_1x + a_2x^2 + \dots + a_nx^n$ is already in $I$ (because it is a multiple of $x$), hence it must be true that $a_0 \in J$ but $a_0 \notin I$. Now $a_0$ can't be an even integer -- if it were, it would be in $I$ -- so it must be an odd integer. From this, it is easy to conclude that $1 \in J$, and you are done.

Answer 2

One might argue in this fashion, instead: Suppose that $J$ is an ideal such that $I\subsetneq J\subseteq\Bbb Z[x],$ and take any $y\in J\setminus I.$ By definition of $I,$ we know that $x$ doesn't divide $y,$ and so if $y=px+n$ for some $p\in\Bbb Z[x]$ and some $n\in\Bbb Z,$ then $n\ne 0.$ In fact, $n$ cannot be any even integer. (Why?) Thus, if $y=px+n$ for some $p\in\Bbb Z[x]$ and some $n\in\Bbb Z,$ then there is some $k\in\Bbb Z$ such that $y=px+2k+1.$ Finally, we note that $px+2k\in I,$ whence $px+2k\in J,$ and so $1=y-(px+2k)\in J,$ and so $J=\Bbb Z[x].$

All that remains is to justify the steps above, and to demonstrate that every element of $\Bbb Zx$ can be written in the form $px+n$ for some $p\in\Bbb Z[x]$ and some $n\in\Bbb Z,$ which I leave to you.

Answer 3

You shouldn't need to worry about PIDs or factorization or anything like that.

The proof that $I\subset A$ is maximal if and only if $A/I$ is a field is a short one, and therefore it is easy to convert your proof that $\mathbb Z[x]/(x,2)$ is a field into a direct proof that $(x,2)$ is a maximal ideal in $\mathbb Z[x]$. Let me show you how that's done.

Recall the proof that $A/I$ is a field $\Rightarrow$ $I$ is a maximal ideal in $A$: Suppose $I\subsetneq J\subset A$. Then there is $x\in J\setminus I$. Now the image $x+I$ of $x$ in $A/I$ is non-zero, so has a multiplicative inverse; i.e., there is $y\in A$ such that $xy+I=1+I$; i.e, $1-xy\in I$. In particular, $1-xy\in J$; since $xy\in J$, it must be the case that $1\in J$; i.e., $J=A$. So $I$ is maximal.

Now the key fact that we used here is the following: if $A/I$ is a field and $x\in A\setminus I$, then there is $y\in A$ such that $1-xy\in I$. Having shown that, it was not difficult to prove that $I$ had to be a maximal ideal.

So how do we show that in this case? Well, it's fairly easy to mirror your proof that $\mathbb Z[x]/(x,2)\cong \mathbb Z/2$ here. Suppose $f\in\mathbb Z[x]\setminus (x, 2)$. $(x, 2)$ is just the set of polynomials whose constant term is even, so $f$ must be a polynomial whose constant term is odd. Can we find a polynomial $g(x)\in\mathbb Z[x]$ such that $1-fg\in(x, 2)$? Easy! Just take $g(x)=1$. In this case, $1-f\in(x,2)$. Indeed, any polynomial with odd constant term would have done here - and that's because the quotient field $\mathbb Z[x]/(x,2)$ only has two elements: $0$ (corresponding to elements of $(x,2)$) and $1$ (corresponding to elements of $\mathbb Z[x]\setminus (x,2)$). To show that this is a field, we need only show that $1$ has a multiplicative inverse; in this case, the inverse is $1$.

Our completed proof then becomes:

Proof. Suppose $J$ is an ideal of $\mathbb Z[x]$ such that $(x,2)\subsetneq J$. Then there exists $f(x)\in J\setminus(x,2)$. Then $f$ has odd constant term, so $1-f$ has even constant term and is therefore contained in $(x,2)$. Therefore, $1-f\in J$. But $f\in J$, and so $1\in J$, and therefore $J=\mathbb Z[x]$.

It follows that $I$ is maximal. $\Box$.

Answer 4

We prove that if $I\subsetneq J$ is an ideal, then $J= \mathbb Z[X]$.

Let $P(x) \in J \backslash I$ be any polynomial. Then you can do long division by $X$ and get $$P(x)=XQ(X)+m$$

Now, $XQ(X) \in I \subset J$ which implies that $m \in J$.

If $m$ is even, then $XQ(X) \in I, m \in I$ would imply that $P(X) \in I$.

Therefore $m=2k+1$ for some $k$. This shows that $$1=m-2 k \in J-J \subset J$$

As $1 \in J$ we get $J=\mathbb Z[X]$.