Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mathbb{Z}[x]/\langle x,2 \rangle$. We know $\langle x,2 \rangle = \{f \in \mathbb{Z}[x] : f(0)$ is even integer$\}$

Also $$\mathbb{Z}[x]/\langle x,2 \rangle = \{f(x) + \langle x,2 \rangle\} = \{a + \langle x,2 \rangle\}$$ since we can write $$f(x) = a_0 + \cdots+ a_nx^n = a_0 + x(a_1 + \cdots + a_nx^{n-1}) = a + 2k + x(a_1 +\cdots + a_mx^{n-1}) = a + 2k + xg(x)$$ and since $\langle x,2 \rangle$ is an ideal, it absorbs $2k + xg(x)$. So, my approach would be to show that $\mathbb{Z}[x]/\langle x,2 \rangle$ is an integral domain. This would give that $\langle x,2 \rangle$ is prime and maximal. Am I correct so far?

share|improve this question
4  
If $A$ is a ring, by which I mean a commutative ring with identity, then an ideal $I$ of $A$ is prime if and only if $A/I$ is a domain, and $I$ is maximal if and only if $A/I$ is a field. Since fields are domains, every prime ideal is maximal, but the converse is not generally true. So, in your case, proving that $\mathbf{Z}[x]/\langle x,2\rangle$ is a domain implies that $\langle x,2\rangle$ is prime, but not maximal. You would need to prove that $\mathbf{Z}[x]/\langle x,2\rangle$ is a field. –  Keenan Kidwell Nov 16 '12 at 2:07
2  
To do this, try to figure out what field it might be, and write down a surjective homomorphism from $\mathbf{Z}[x]$ to this field with kernel $\langle x,2\rangle$. –  Keenan Kidwell Nov 16 '12 at 2:09
    
The simplification I made is correct? For the Factor ring –  Marvin Gaye Nov 16 '12 at 2:11
3  
You haven't said which field, so I'm not sure you are reasoning correctly. If you can show the quotient ring is an integral domain with 2 elements, you can fall back on a theorem that says a finite integral domain is a field. Alternatively, you can simply show directly that your ideal is maximal, by showing that if you try to make a strictly bigger ideal you get the whole ring. –  Gerry Myerson Nov 16 '12 at 3:04
1  
@KeenanKidwell Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Sep 14 '13 at 10:47

2 Answers 2

up vote 3 down vote accepted

MSE--sometimes I just can't stop typing!

My way of looking at this is very similar to Citizen's, maybe redundantly so; like I said, sometimes I can't stop typing, so please bear with me . . .

Let us denote the ideal $\langle 2, x \rangle$ of $\mathbf Z [x]$ by $J$; thus $J = \langle 2, x \rangle \subset \mathbf Z [x]$.

First off, notice that the ideal $J = \langle 2, x \rangle$ of $\mathbf Z [x]$ consists of all polynomials of the form $2g(x) + xh(x)$ with $g(x), h(x) \in \mathbf Z [x]$. Next, observe that we can always express a polynomial of the form $2g(x) + xh(x)$ as $xp(x) + 2m$ for some $p(x) \in \mathbf Z [x]$ and $m \in \mathbf Z$. Then two polynomials $q_1(x), q_2(x) \in \mathbf Z [x]$ represent the same element of $\mathbf Z [x]/J$ precisely when $q_1(x) - q_2(x) \in J$, that is when $q_1(x) - q_2(x) = xp(x) + 2m$ with $p(x)$, $m$ as above. But since $xp(x)$ is an arbitrary polynomial with constant term $0$, it follows that $q_1(x) + J = q_2(x) + J$ exactly when the constant terms of $q_1(x)$, $q_2(x)$, which are $q_1(0)$, $q_2(0)$ respectively, differ by an even integer, i.e. $q_1(0) - q_2(0)$ is a mutiple of $2$. Thus $q_1(0)$ and $q_2(0)$ are either both even, or both odd. From this it follows that there are precisely two equivalence classes in $\mathbf Z [x]/J$, one containing all polynomials $q(x) \in \mathbf Z [x]$ with $q(0)$ even, which is in fact $J$ itself, and the other consisting of those $q(x) \in \mathbf Z [x]$ with $q(0)$ odd, which is $1 + J$. The natural projection homomorphism $\pi: \mathbf Z [x] \to \mathbf Z [x]/J$ thus effectively maps $\mathbf Z [x]$ onto a ring with precisely two elements, and since the product $q_1(x)q_2(x)$ of two polynomials $q_1(x), q_2(x) \in \mathbf Z [x]$ with $q_1(0), q_2(0)$ odd again has $q_1(0)q_2(0)$ odd, we have $(1 + J)^2 = (1 + J) \ne 0$ in $Z [x]/J$, which shows that multiplication in the quotient ring $Z [x]/J$ is precisely the same (up to isomorphism) as that in the two-element field $\Bbb F_2$. Thus we see that $Z [x]/J$ is in fact a field itself, whence $J = \langle 2, x \rangle$ is a maximal ideal in $\mathbf Z [x]$.

Now in fact the maximality of $J$ can actually be seen directly once we have established that the ideal $\langle 2, x \rangle$ consists of precisely those polynomials of the form $xp(x) + 2m$, for any ideal $J'$ with $J \subset J'$, $J' \ne J$, must contain some $q(x) \in \mathbf Z [x]$, $q(x) \notin J$. But we have shown that any such $q(x)$ must have $q(0)$ odd; but then we would have $1 \in J'$, whence $J' = \mathbf Z [x]$, showing $J$ is maximal in $\mathbf Z [x]$.

It goes on (still can't stop typing!): apparently for any $n \in \mathbf Z$ we have $\mathbf Z [x]/\langle n, x \rangle \simeq \mathbf Z_n$; this follows since, as in the above, $J = \langle n, x \rangle$ consists of those polynomials of the form $ng(x) + xh(x) = xp(x) + mn$; now $q_1(x) - q_2(x) \in J$ means $q_1(0) - q_2(0) = mn$ for some $m \in \mathbf Z$; thus $q_1(0) \equiv q_2(0)\mod n$; as in the case $n = 2$, $q_1(x) + J = q_2(x) + J$ if and only if $q_1(0), q_2(0)$ are in the same residue class modulo $n$; furthermore the natural projection $\pi:\mathbf Z[x] \to \mathbf Z [x] / J$ is easily seen to map the polynomial $q_1(x)q_2(x)$ to $q_1(0)q_2(0) + J$. Thus there are precisely $n$ elements in $\mathbf Z [x] / J \simeq \mathbf Z_n$.

For primes $p$, we can also argue the maximality of $J = \langle p, x \rangle$ via the observation that the elements of $J$ are of the form $xq(x) + mp$; then if $r(x) \notin J$ it must be of the form $r(x) = xs(x) + t$ where $(t, p) = 1$; but then any ideal $K$ containing $J$ and $r(x)$ must contain $1$, since there exist integers $a, b$ with $at + bp = 1$; as such $K = \mathbf Z [x]$ and thus $J$ is maximal.

If, in the above, $n$ is taken to be a prime, the we have $\mathbf Z [x] / J \simeq \mathbf Z_n = \Bbb F_n$, a field, and the ideal $J = \langle x, n \rangle$ is maximal in $\mathbf Z [x]$. When $n$ is composite, however, $\mathbf Z_n$ is not a field and $J$ will not be maximal. If $n = rs$ with $r, s > 1$, then we have $\langle n, x \rangle \subset \langle r, x \rangle \subset \mathbf Z [x]$, with no equalities between these sets.

OK, fingers tired; brain dead; can stop typing now.

I really hope this one helps someone, somewhere.

I say, "Cheers"; exhausted, I proclaim

Fiat Lux!

share|improve this answer

On the urging of @Julian Kuelshammer, I'm making an answer out of my comments. Since maximal ideals are prime it suffices to prove that $(x,2)$ is maximal, and for this it is enough to prove that $R:=\mathbf{Z}[x]/(x,2)$ is a field. What field should it be? Well, the image of $2$ in this ring is $0$ since $2\in(x,2)$, so if this is a field, it has to be of characteristic $2$. In fact the canonical map $\mathbf{Z}\rightarrow R$ has kernel containing $2\mathbf{Z}$, and in fact this must be the kernel, since $2\mathbf{Z}$ is a maximal ideal of $\mathbf{Z}$, so we get an induced ring map $\mathbf{F}_2=\mathbf{Z}/2\mathbf{Z}\rightarrow R$. It is injective since $2\mathbf{Z}=\ker(\mathbf{Z}\rightarrow R)$, or if you prefer, because $\mathbf{F}_2$ is a field. We claim that it is also surjective. Well, given $f(X)=a_nX^n+\cdots+a_1X+a_0\in\mathbf{Z}[X]$, we have $f(X)\equiv a_0\pmod{X}$. But the image of $a_0+2\mathbf{Z}\in \mathbf{F}_2$ under our map $\mathbf{F}_2\rightarrow R$ is precisely $a_0+(x,2)=f(X)+(x,2)$. So indeed the map is surjective, and therefore an isomorphism.

share|improve this answer
    
so educate me, already! What is the definition of the canonical map $\mathbf Z \to R$? –  Robert Lewis Sep 15 '13 at 4:33
3  
Every ring $R$ with identity $1_R$ admits a unique map ring homomorphism from $\mathbf{Z}$. It is given by $n\mapsto n\cdot 1_R$ (the multiplication being the abelian group structure on the underlying additive group of $R$). –  Keenan Kidwell Sep 15 '13 at 16:40
    
Hey, Keenan, you give good education! Of course I was already aware of such homomorphisms, but hadn't heard of them referred to as "canonical". Thanks, Bob Lewis. –  Robert Lewis Sep 15 '13 at 16:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.