Math Genius: Ring homomorphism may not preserve $1$.

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Suppose $R$ is a ring with unity and $f:Rto R’$ is a ring homomorphism. Then, $f(R)$ must have an identity $1_{f(R)}.$ But $1_{f(R)}$ may not be the identity of $R’$. Even $R’$ may not contain any identity. Suppose $R’$ contains $1_{R’};$ even still, the ring homomorphism may not necessarily map $1_R$ to $1_{R’}.$

Why does this occur? I am unable to find exactly why it is so. Can this be explained by semigroup homomorphisms or monoid homomorphisms, i.e., maps from semigroup $S_1$ to $S_2$ such that $f(a cdot b)=f(a) cdot f(b)$?

Today’s mathematicians always suppose a ring homomorphism to preserve identity, i.e. $phi:Rto S$ supposed to have $f(1_R)=1_S$ (of course we suppose $R$ and $S$ unital).

But there is some cases which the property $f(1_R)=1_S$ can be concluded.

For example,

suppose $phi:Rto S$ is onto and $R$ is unital (has identity element), then $phi(1_R)$ is the identity of $S$.

For any $sin S$, there exists an $rin R$ with $phi(r)=s$ and thus $$phi(1_R)s=phi(1_r)phi(r)=phi(1_Rr)=phi(r)=s$$ and similarly, $sphi(1_R)=s$, which shows $phi(1_R)$ is the identity of $S$.

For another one,

Suppose $S$ is an integral domain and $phi:Rto S$ is a ring homomorphism. Then $phi(1_R)=1_S$ or $phi(1_R)=0$.

We have $$phi(1_R)=phi(1_R1_R)=phi(1_R)phi(1_R)implies phi(1_R)(1_S-phi(1_R))=0$$
thus either $phi(1_R)=0$ or $1_S=phi(1_R)$.

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Math Genius: Determining Whether an Ideal is Principal in $mathbb{Z}[1/5]$

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I am considering the ring $R=mathbb{Z}[frac{1}{5}]={p(frac{1}{5})mid p(x) in mathbb{Z}[x]}$ and I want to determine whether the ideals $I=left langle 2,frac{1}{5}right rangle$ and $J=left langle 2,frac{3}{5}right rangle$ are principal ideals.

So far, I think that $I=left langle frac{1}{5}right rangle$ and is hence principal. Clearly, we have $left langle frac{1}{5}right rangle subset I$. To show the other direction, observe that if $alpha=2p(frac{1}{5})+frac{1}{5}q(frac{1}{5})in I$ for some $p,q in mathbb{Z}[x]$, we can write $alpha = frac{1}{5}left(10p(frac{1}{5})+q(frac{1}{5})right)in left langle frac{1}{5}right rangle$. Hence $I subset left langle frac{1}{5} right rangle implies I =left langle frac{1}{5} right rangle$.

For the second case, I believe that $J=left langle 2, frac{3}{5} right rangle$ is not principal in $R$. To prove this, I have tried assuming that $J=left langle tleft( frac{1}{5} right) right rangle$ for some $t(x)inmathbb{Z}[x]$. Then clearly, we must have that $tleft( frac{1}{5} right)$ divides $2$ and $frac{3}{5}$. But I am struggling to place any restrictions on $t$: for instance, if we assume that $frac{3}{5}=tleft(frac{1}{5}right)rleft(frac{1}{5}right)$ with $r in mathbb{Z}[x]$, then I don’t believe I can say much about the degree of $t$ and $r$, since it is possible the coefficients will cancel. As an example, I can write $frac{3}{5}=(3cdot 5)frac{1}{5^2}=(3cdot 5^2)frac{1}{5^3}$ and so on… As a result, I am at a loss for how to show that $J$ is or is not principal in $R$.

I imagine that I am making things too complicated, or have not understood the definition of $R$.

Two facts can be used to deal with both of these.

(1) If an ideal $I$ of a ring $R$ contains a unit, then $I = langle 1 rangle = R$.

(2) Given $r in R$ and a unit $u in R$, then $langle r rangle = langle ur rangle$, i.e., associates (elements that differ multiplicatively by a unit) generate the same ideal.

Returning to your problem, since $1/5$ is a unit then (1) immediately shows that $I = langle 1 rangle$. Again, since $1/5$ is a unit then (2) implies $langle 2, 3/5 rangle = langle 2, 3 rangle$. Since $1 = 3 – 2 in langle 2, 3 rangle$, this shows that $J = langle 1 rangle$ as well.

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Math Genius: Determining Whether an Ideal is Principal in $mathbb{Z}[1/5]$

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I am considering the ring $R=mathbb{Z}[frac{1}{5}]={p(frac{1}{5})mid p(x) in mathbb{Z}[x]}$ and I want to determine whether the ideals $I=left langle 2,frac{1}{5}right rangle$ and $J=left langle 2,frac{3}{5}right rangle$ are principal ideals.

So far, I think that $I=left langle frac{1}{5}right rangle$ and is hence principal. Clearly, we have $left langle frac{1}{5}right rangle subset I$. To show the other direction, observe that if $alpha=2p(frac{1}{5})+frac{1}{5}q(frac{1}{5})in I$ for some $p,q in mathbb{Z}[x]$, we can write $alpha = frac{1}{5}left(10p(frac{1}{5})+q(frac{1}{5})right)in left langle frac{1}{5}right rangle$. Hence $I subset left langle frac{1}{5} right rangle implies I =left langle frac{1}{5} right rangle$.

For the second case, I believe that $J=left langle 2, frac{3}{5} right rangle$ is not principal in $R$. To prove this, I have tried assuming that $J=left langle tleft( frac{1}{5} right) right rangle$ for some $t(x)inmathbb{Z}[x]$. Then clearly, we must have that $tleft( frac{1}{5} right)$ divides $2$ and $frac{3}{5}$. But I am struggling to place any restrictions on $t$: for instance, if we assume that $frac{3}{5}=tleft(frac{1}{5}right)rleft(frac{1}{5}right)$ with $r in mathbb{Z}[x]$, then I don’t believe I can say much about the degree of $t$ and $r$, since it is possible the coefficients will cancel. As an example, I can write $frac{3}{5}=(3cdot 5)frac{1}{5^2}=(3cdot 5^2)frac{1}{5^3}$ and so on… As a result, I am at a loss for how to show that $J$ is or is not principal in $R$.

I imagine that I am making things too complicated, or have not understood the definition of $R$.

Two facts can be used to deal with both of these.

(1) If an ideal $I$ of a ring $R$ contains a unit, then $I = langle 1 rangle = R$.

(2) Given $r in R$ and a unit $u in R$, then $langle r rangle = langle ur rangle$, i.e., associates (elements that differ multiplicatively by a unit) generate the same ideal.

Returning to your problem, since $1/5$ is a unit then (1) immediately shows that $I = langle 1 rangle$. Again, since $1/5$ is a unit then (2) implies $langle 2, 3/5 rangle = langle 2, 3 rangle$. Since $1 = 3 – 2 in langle 2, 3 rangle$, this shows that $J = langle 1 rangle$ as well.

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Math Genius: Example of a polynomial with a lesser degree than the minimal monic polynomial.

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Let $A$ and $B$ ($Asubset B$) be commutative rings with unity. Let $ain B$ be integral over $A$. Therefore, there exists a monic polynomial $pin A[x]$, such that $p(a)=0$.

By well-ordering principle, there exists a monic polynomial $p_0 in A[x]$, such that $p_0(a) = 0$ and
$$text{degree}(p_0) := min{text{degree}(g); gin A[x], g text{is monic and }g(a)=0}. $$

Question: Does anyone know an example of commutative rings $A$ and $B$ (with unity), such that there is a polynomial $gin A[x]$ (not necessarily monic) such that $g(a)=0$ and $text{degree}(g)<text{degree}(p_0)?$

If $A$ is a field then it is obvious that it is impossible. But in the general case, I am not being able to find a counterexample. Can anyone help me?

One can choose $A = Q[u]$ and $B = Q[u,v]/(uv, v^2+1)$.

The equation of integrality of $v$ over $A$ is $T^2+1 in A[T]$. If there exists a monic polynomial $f(T) in A[T]$ of degree $1$ such that $f(v) = 0$, it must be of the form $T-v$. Since $v notin A$, this is impossible. So, $deg p_0 = 2$.

For $g(T)$, one can take $g(T) = uT in A[T]$ which is of degree $1$.

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Math Genius: Zero divisors in annihilator of modules

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I faced a theorem

Let $R$ be a noetherian local ring, $M$ a finitely generated module of finite projective dimension.
Then if the annihilator of $M$ is not trivial, then it contains a nonzero zero divisor in $R$.

Or more precisely, in this form.

enter image description here

This is a part of Vasconcelos’ paper Ideals Generated by R-Sequences, he referred paper of Auslander and Buchsbaum Homological dimension in local ring. But there is no Proporsition 6.2 in the paper referred.

My question is, how to prove this proposition?

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Math Genius: Ring of Witt vectors over finite fields

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I have just started studying Witt vectors and I have questions about the following identity $$W_n(mathbb{F}_p)cong mathbb{Z}/p^nmathbb{Z}$$

  1. I would like proving this by finding an explicit map $phi_n: W_n(mathbb{F}_p)to mathbb{Z}/p^nmathbb{Z}$, but couldn’t come up with a reasonable map. Although I have found the isomorphism $$phi: W(mathbb{F}_p)to mathbb{Z}_p$$ via $(a_0,a_1,…)mapsto chi(a_0)+chi(a_1)p… $, where $chi$ is the Teichmüller character.

    Can I obtain $phi_n$ by composing $phi$ with $pr_n:mathbb{Z}_p to mathbb{Z}/p^nmathbb{Z}$ and identifying $W_n(mathbb{F}_p)$ with $(a_0,…,a_{n-1},0,0,…)in W(mathbb{F}_p)$? Is there a nice explicit version of this map?

  2. More generally I am interested in the case $A=mathbb{F}_q$. I know that $W(mathbb{F}_q)cong mathbb{Z}_p[mu_{q-1}]$ should hold.
    Is it possible to argue that this is an isomorphism because $mathbb{Z}_p[mu_{q-1}]$ and $W(mathbb{F}_q)$ are both strict $p$-ring with residue field $mathbb{F}_q$ and as such canonically isomorphic?

Ad 1) First of all I’m pretty sure that yes, $phi_n = pr_ncircphi$, although usually the index is off by one (so I would have expected $W_{n-1}(Bbb F_p) simeq Bbb Z/p^n$).

To make that map a bit more explicit, I remember a MathOverflow post which might be helpful here. It motivates the Witt polynomials (noted $W(X_0, …, X_{n-1})$ there), which, if I’m not mistaken, are “more or less” your map $phi_n$ (annoying index shift again, oh well …). “More or less” because you have to choose some lifting
$$Bbb F_p rightarrow Bbb Z/p^{n}: quad bar a mapsto a.$$ But well, if you just go all the way up to $Bbb Z$ and work with the old school representatives ${0, …, p-1}$, — and also remember that in this very special case, the Frobenius $(cdot)^{,p}$ is just the identity on $Bbb F_p$ –, you can explicitly write the map you call $phi_n$ as

$$(bar a_0, bar a_1, …, bar a_{n-1}) mapsto (a_0^{p^{n-1}} +p cdot a_1^{p^{n-2}} + … + ,p^{n-1} cdot a_{n-1}) +p^nBbb Z$$

To make it totally explicit, say $p=5, n=3$, and say your element in $W_3(Bbb F_5)$ is $(bar 1, bar 4, bar2)$, it gets mapped to $$(1^{5^2}+5cdot 4^5 + 5^2cdot 2) +5^3Bbb Z = 5171 + 5^3Bbb Z = 46 + 5^3Bbb Z$$

— and note that crucial and fun fact that the choice of the lifting does not change the result; if e.g. instead you lift $bar 1$ to $6$, $bar 4$ to $9$, and $bar 2$ to $22$, you still get
$$(6^{5^2}+5cdot 9^5 + 5^2cdot 22) +5^3Bbb Z = 28430288029929997171 + 5^3Bbb Z= 46 + 5^3Bbb Z$$

Added: Let’s see the connection to the first part of Jyrki Lahtonen’s comment: one can get those Teichmüller-like representatives of $Bbb F_p$ in $Bbb Z/p^n$ by raising any set of representatives to the $p^{n-1}$-th power; continuing the example above we get $chi(1) =1$ and

$$chi(2) = 2^{5^2} + 5^3Bbb Z = 33554432 + 5^3Bbb Z = 57 + 5^3Bbb Z$$
$$chi(3) = 3^{5^2} + 5^3Bbb Z = 847288609443 + 5^3Bbb Z = 68 + 5^3Bbb Z$$
$$chi(4) = 4^{5^2} + 5^3Bbb Z = 1125899906842624 + 5^3Bbb Z = 124 + 5^3Bbb Z = -1 + 5^3Bbb Z.$$

(Of course we could have noticed $chi(p-1) = -1$ easier.) Notice how they are multiplicative, indeed they are just the set of standard Teichmüller representatives $mu_{p-1}(Bbb Z_p)$ modulo $p^n$, and instead of the earlier computation you can write

$$phi_n(bar 1, bar 4, bar2) = chi(1) + 5cdot chi(4) + 5^2cdotchi(2) +5^3Bbb Z = 1+5cdot (-1) + 25cdot 57 +5^3Bbb Z = 46 +5^3Bbb Z.$$

Of course that’s all equivalent, but this way you outsource some work into a once-and-for-all-computation of the Teichmüller representatives. Also, beware that as soon as one tries to do the same with $Bbb F_q$ instead of $Bbb F_p$, things get more subtle, as one has to put in appropriate $p^i$-th roots at appropriate places.

Ad 2), I think it is totally possible to argue this way, although the proofs of that theorem which I have seen (which would be the one of Bourbaki in Commutative Algebra 9, and of Serre in Local Fields) actually go through the Witt vector machinery and thus might exhibit a bit more structure. I just want to point out that, if $q=p^k$, then another description of $Bbb Z_p[mu_{q-1}]$ is: the ring of integers (a.k.a valuation ring) of the unique unramified degree $k$ extension of $Bbb Q_p$.

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Math Genius: Definition of multimodules

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Let $A,B$ be rings and $h:Arightarrowtext{End}(E)$, $k:Brightarrowtext{End}(E)$ two ring homomorphisms. If $h_alphacirc k_beta=k_betacirc h_alpha$ for all $(alpha,beta)in Atimes B$, we say that the two (left or right–I have used left) module structures defined on $E$ by $h$ and $k$ are compatible.

Let $(A_lambda)_{lambdain L}$,$(B_mu)_{muin M}$ be two families
of rings. An $((A_lambda)_{lambdain L},(B_mu)_{muin
M})$
-multimodule is a set $E$ with, for each $lambdain L$, a left
$A_lambda$module structure and, for each $muin M$, a right
$B_mu$-module structure, all these module structures being compatible
with one another
.

Does the bolded portion of this definition imply compatibility amongst the $A_lambda$ (resp. $B_mu$) as well? For example does it imply that $A_lambda$ and $A_{lambda’}$ are compatible for $lambda,lambda’in L$?

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Math Genius: Descending chaincondition for cyclic ideals

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in generell I want to show that if a Ring $R$ satisfies the descending chaincondition for cyclic ideals, so every chain of cyclic ideals $(r_1)supset (r_2)supset dots$ in $R$ becomes stationary, then every prime ideal is maximal.

If we take a prime ideal $mathcal{p} subset R$, then $R/mathcal{p}$ is an integral domain.
For $mathcal{p}$ to be maximal, $R/mathcal{p}$ has to be a field.

Now my question: Can we show that $R/mathcal{p}$ satisfies the descending chaincondition for cyclic ideals, if $R$ does?

In this case it would be rather easy to show that every element in $R/mathcal{p}$ has an inverse and therefore $R/mathcal{p}$ would be a field.

If that’s not the case, then what approach could I try instead?

Thanks in advance!

Rather than worrying about the chain condition in the quotient, one could just note that for all $ain R$, there exist positive integers $m,n$ with $m>n$ and an element $xin R$ such that $a^n=a^{m}x$. This is a simple consequence of applying the descending chain condition to the chain $aRsupseteq a^2Rsupseteq a^3Rsupseteqldots$. (*)

Because you can do this in $R$, you can do it in any quotient of $R$.

Of course, in the quotients that are domains, $a$ is zero or cancels down to $1=a^{m-n}x$, proving $a$ is a unit.

(*) Rings satisfying this condition for all elements are called strongly $pi$-regular. Rings satisfying the descending chain condition on principal right ideals are called left perfect rings. (Yes, I do mean left. It is a very curious crossover characterization.)

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Math Genius: For $A$, a commutative ring with identity, show $J(A)={xin A:xy-1 in A^times, forall y in A}$, $J(A)$ being the Jacobson radical.

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For $A$, a commutative ring with identity, show $J(A)={xin A:xy-1 in A^times, forall y in A}$, $J(A)$ being the Jacobson radical. I should add here that the Jacobson radical is the intersection of all the maximal ideals of $A$.

Firstly, $J(A)$ is itself an ideal (proof omitted).

Now, assume $x in J(A)$, and consider the natural map $phi:A to A/J(A)$, where $x$ is mapped to the zero element in the quotient ring. Now, supposed $y$ is any element from $A$. Then
$$phi(xy-1)=phi(x)phi(y)-1=-1=phi(-1)$$
So, $xy-1=-1$ in $A$, so $xy-1$ is a unit in $A$.

Now, for the other direction ( this is where I am having trouble ). Lets, assume that $ xin A:xy-1 in A^times, forall y in A$. Now supposed that $x notin J(A)$ and consider the same map as above. Then we must have that,
$$phi((xy-1)(xy-1)^{-1})=(phi(x)phi(y)-phi(1))phi(xy-1)^{-1}=1$$
$$iff phi(x)phi(y)phi(xy-1)^{-1}-phi(xy-1)^{-1}=1$$
This needs to be true for all $y$, so I take $y=1$, so the expression reduces to
$$iff phi(x)phi(x-1)^{-1}-phi(x-1)^{-1}=1$$
I think this should lead to a contradiction, since we can see that $xne1$, or that would imply $1=0$. But I cannot see how to conclude that this always is a contradiction? Am I on the right track here?

Update

After reviewing and thinking a little more, I don’t seem to be using any specific properties of the Jacobson radical. I am thinking I must need to use the property of $J(A)$ being a maximal ideal itself in a commutative ring, so $J(A)$ would be a prime ideal. This would imply that $A/J(A)$ is a field. Still now quite sure how to use this.

Let $x in J(A)$. Suppose, for contradiction, that $1 – ax$ is not a unit for some $a in A$. Then $(1-ax)$ is a proper ideal (since it does not contain $1$), so it is contained in some maximal ideal $mathfrak{m}$ of $A$. But since $x in J(A)$, we have that $x in mathfrak{m}$ so $ax in mathfrak{m}$, and hence $1 = (1 – ax) + ax in mathfrak{m}$, which is a contradiction.

Now suppose that $x not in J(A)$. Then there is some maximal ideal $mathfrak{m}$ such that $x not in mathfrak{m}$, so $mathfrak{m} + (x)$ is an ideal strictly containing $mathfrak{m}$, so by maximality of $mathfrak{m}$, we have $mathfrak{m} + (x) = A$. Thus, there exist $m in mathfrak{m}$ and $a in A$ such that $m + ax = 1$, so $1 – ax = m in mathfrak{m}$, which means that $1 – ax$ is not a unit.

You don’t want to look at $A/J(A)$, you want to look at $A/mathfrak{m}$ for each maximal ideal $mathfrak{m}$. For the first part: $phi(xy – 1) = phi(-1)$ does not mean that $xy – 1 = -1$. Instead, assume that $xy – 1$ is not invertible, then $xy – 1$ will be contained in a maximal ideal $mathfrak{m}$.

Conversely, suppose that $x$ does not belong to some maximal ideal $mathfrak{m}$, then $x$ is a unit in $A/mathfrak{m}$.

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Math Genius: For $A$, a commutative ring with identity, show $J(A)={xin A:xy-1 in A^times, forall y in A}$, $J(A)$ being the Jacobson radical.

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For $A$, a commutative ring with identity, show $J(A)={xin A:xy-1 in A^times, forall y in A}$, $J(A)$ being the Jacobson radical. I should add here that the Jacobson radical is the intersection of all the maximal ideals of $A$.

Firstly, $J(A)$ is itself an ideal (proof omitted).

Now, assume $x in J(A)$, and consider the natural map $phi:A to A/J(A)$, where $x$ is mapped to the zero element in the quotient ring. Now, supposed $y$ is any element from $A$. Then
$$phi(xy-1)=phi(x)phi(y)-1=-1=phi(-1)$$
So, $xy-1=-1$ in $A$, so $xy-1$ is a unit in $A$.

Now, for the other direction ( this is where I am having trouble ). Lets, assume that $ xin A:xy-1 in A^times, forall y in A$. Now supposed that $x notin J(A)$ and consider the same map as above. Then we must have that,
$$phi((xy-1)(xy-1)^{-1})=(phi(x)phi(y)-phi(1))phi(xy-1)^{-1}=1$$
$$iff phi(x)phi(y)phi(xy-1)^{-1}-phi(xy-1)^{-1}=1$$
This needs to be true for all $y$, so I take $y=1$, so the expression reduces to
$$iff phi(x)phi(x-1)^{-1}-phi(x-1)^{-1}=1$$
I think this should lead to a contradiction, since we can see that $xne1$, or that would imply $1=0$. But I cannot see how to conclude that this always is a contradiction? Am I on the right track here?

Update

After reviewing and thinking a little more, I don’t seem to be using any specific properties of the Jacobson radical. I am thinking I must need to use the property of $J(A)$ being a maximal ideal itself in a commutative ring, so $J(A)$ would be a prime ideal. This would imply that $A/J(A)$ is a field. Still now quite sure how to use this.

Let $x in J(A)$. Suppose, for contradiction, that $1 – ax$ is not a unit for some $a in A$. Then $(1-ax)$ is a proper ideal (since it does not contain $1$), so it is contained in some maximal ideal $mathfrak{m}$ of $A$. But since $x in J(A)$, we have that $x in mathfrak{m}$ so $ax in mathfrak{m}$, and hence $1 = (1 – ax) + ax in mathfrak{m}$, which is a contradiction.

Now suppose that $x not in J(A)$. Then there is some maximal ideal $mathfrak{m}$ such that $x not in mathfrak{m}$, so $mathfrak{m} + (x)$ is an ideal strictly containing $mathfrak{m}$, so by maximality of $mathfrak{m}$, we have $mathfrak{m} + (x) = A$. Thus, there exist $m in mathfrak{m}$ and $a in A$ such that $m + ax = 1$, so $1 – ax = m in mathfrak{m}$, which means that $1 – ax$ is not a unit.

You don’t want to look at $A/J(A)$, you want to look at $A/mathfrak{m}$ for each maximal ideal $mathfrak{m}$. For the first part: $phi(xy – 1) = phi(-1)$ does not mean that $xy – 1 = -1$. Instead, assume that $xy – 1$ is not invertible, then $xy – 1$ will be contained in a maximal ideal $mathfrak{m}$.

Conversely, suppose that $x$ does not belong to some maximal ideal $mathfrak{m}$, then $x$ is a unit in $A/mathfrak{m}$.

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