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

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]\$

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]\$

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.

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)

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 $$1$$.

## Math Genius: Zero divisors in annihilator of modules

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. 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?

## Math Genius: Ring of Witt vectors over finite fields

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 $$Bbb Q_p$$.

## Math Genius: Definition of multimodules

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 $$lambda,lambda’in L$$?

## Math Genius: Descending chaincondition for cyclic ideals

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?

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.)

## 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.

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.

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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