Units and Nilpotents [closed]

Here's a rather different argument. First, suppose that $R$ is commutative. Suppose $u+a$ is not a unit. Then it is contained in some maximal ideal $M\subset R$. Since $a$ is nilpotent, $a\in M$ (since $R/M$ is a field, and any nilpotent element of a field is $0$). Thus $u=(u+a)-a\in M$ as well. But $u$ is a unit, so it can't be in any maximal ideal, and this is a contradiction.

If you don't know that $R$ is commutative, let $S\subseteq R$ be the subring generated by $a$, $u$, and $u^{-1}$. Then $S$ is commutative: the only thing that isn't immediate is that $u^{-1}$ commutes with $a$, and this this can be proven as follows: $$u^{-1}a=u^{-1}auu^{-1}=u^{-1}uau^{-1}=au^{-1}.$$

The argument of the first paragraph now shows that $u+a$ is a unit in $S$, and hence also in $R$.

This argument may seem horribly nonconstructive, due to the use of a maximal ideal (and hence the axiom of choice) and proof by contradiction. However, it can be made to be constructive and gives an explicit inverse for $u+a$ in terms of an inverse for $u$ and and an $n$ such that $a^n=0$.

First, we observe that all that is actually required of the ideal $M$ is that it is a proper ideal which contains $u+a$ and all nilpotent elements of $R$. So we may replace $M$ with the ideal $(u+a)+N$ where $N$ is the nilradical of $R$, and use the fact that if $I=(u+a)$ is a proper ideal in a commutative ring then $I+N$ is still a proper ideal. This is because $R/(I+N)$ is the quotient of $R/I$ by the image of $N$, which is contained in the nilradical of $R/I$. If $I$ is a proper ideal, then $R/I$ is a nonzero ring, so its nilradical is a proper ideal, so $R/(I+N)$ is a nonzero ring and $I+N$ is a proper ideal.

Next, we recast this argument as a direct proof instead of a proof by contradiction. Letting $I=(u+a)$, we observe that $I+N$ is not a proper ideal since $u=(u+a)-a\in I+N$ and $u$ is a unit. That is, a nilpotent element (namely $a$) is a unit in the ring $R/I$, which means $R/I$ is the zero ring, which means $I=R$, which means $u+a$ is a unit.

Finally, we chase through the explicit equations witnessing the statements above. Letting $v=u^{-1}$, we know that $v((u+a)-a)=1$ so $$-va=1-v(u+a),$$ witnessing that $a$ is a unit mod $u+a$ (with inverse $-v$). But $a$ is nilpotent, so $a^n=0$ for some $n$, and thus $0$ is also a unit mod $u+a$. We see this explicitly by raising our previous equation to the $n$th power: $$0=(-v)^na^n=(1-v(u+a))^n=1-nv(u+a)+\binom{n}{2}v^2(u+a)^2+\dots+(-v)^n(u+a)^n,$$ where every term after the first on the right-hand side is divisible by $u+a$. Factoring out this $u+a$, we find that $$1=(u+a)\left(nv-\binom{n}{2}v^2(u+a)+\dots-(-v)^n(u+a)^{n-1}\right)$$ and so $$-\sum_{k=1}^n \binom{n}{k}(-v)^k(u+a)^{k-1}= nv-\binom{n}{2}v^2(u+a)+\dots-(-v)^n(u+a)^{n-1}$$ is an inverse of $u+a$.

The fact that this complicated formula is hidden in the one-paragraph conceptual argument given at the start of this answer is a nice example of how powerful and convenient the abstract machinery of ring theory can be.

Let $v$ be the inverse of $u$, and suppose $a^2=0$. Note that $$(u+a)\cdot v(1-va)=(1+va)(1-va)=1-v^2a^2=1-0=1.$$ See if you can generalize this.

Note that since $u$ is a unit and

$ua = au, \tag 1$

we may write

$a = u^{-1}au, \tag 2$

and thus

$au^{-1} = u^{-1}a; \tag 3$

also, since $a$ is nilpotent there is some $0 < n \in \Bbb N$ such that

$a^n = 0, \tag 4$

and thus

$(u^{-1}a)^n = (au^{-1})^n = a^n (u^{-1})^n = (0) (u^{-1})^n = 0; \tag 5$

we observe that

$u + a = u(1 + u^{-1}a), \tag 6$

and that, by virtue of (5),

$(1 + u^{-1}a) \displaystyle \sum_0^{n - 1} (-u^{-1}a)^k = \sum_0^{n - 1} (-u^{-1}a)^k + u^{-1}a\sum_0^{n - 1} (-u^{-1}a)^k$ $= \displaystyle \sum_0^{n - 1} (-1)^k(u^{-1}a)^k + \sum_0^{n - 1} (-1)^k(u^{-1}a)^{k + 1}$ $= 1 + \displaystyle \sum_1^{n - 1} (-1)^k (u^{-1}a)^k + \sum_0^{n - 2} (-1)^k(u^{-1}a)^{k + 1} + (-1)^{n - 1}(-u^{-1}a)^n$ $= 1 + \displaystyle \sum_1^{n - 1} (-1)^k (u^{-1}a)^k + \sum_1^{n - 1} (-1)^{k - 1}(u^{-1}a)^k = 1 + \displaystyle \sum_1^{n - 1} ((-1)^k + (-1)^{k - 1})(u^{-1}a)^k = 1; \tag 7$

this shows that

$(1 + u^{-1}a)^{-1} = \displaystyle \sum_0^{n - 1} (-u^{-1}a)^k, \tag 8$

and we have demonstrated an explicit inverse for $1 + u^{-1}a$. Thus, by (6),

$(u + a)^{-1} = (u(1 + u^{-1}a))^{-1} = (1 + u^{-1}a)^{-1} u^{-1}, \tag 9$

that is, $u + a$ is a unit.

Nota Bene: The result proved above has an application to this question, which asks to show that $I - T$ is invertible for any nilpotent linear operator $T$. Taking $T = a$ and $I = u$ in the above immediately yields the existence of $(I - T)^{-1}$. End of Note.

I think that you do not need to overthink of this problem. The point is to get some idea of the multiplicative inverse.

Firstly, think about if $x$ is a nilpotent, what is the multiplicative inverse of $1+x$. Let's forget about the commutative algebra for a moment. A natural choice is $\frac{1}{1+x}$. However, you need to write $\frac{1}{1+x}$ in terms of $x$, because this is the only thing you know that belongs to the ring. This brings you the Taylor expansion of $$\frac{1}{1+x}= \sum_{k=0}^{\infty}(-1)^{k}x^{k}.$$ The point is that $x^{n}=0$ for some $n>0$, so this sequence is kind of repeating. I do not want to go to details analysis here, but this should give you an idea to try to verify that $$(1+x)\Bigg(\sum_{k=0}^{n}(-1)^{k}x^{k}\Bigg)=1,$$ and this is true.

Ok, for general case, if $x^{n}=0$ for some $n>0$ and $y$ is a unit, then $y^{-1}$ and all the power of $y^{-1}$ make sense. Now, we use the same idea. The natural choice of the inverse of $(x+y)$ is for sure $\frac{1}{x+y}$, but then again you need to write it as in terms of $x,y$ or $y^{-1}$. This again brings you the Taylor expansion. Treat $y$ as a constant, you expand in terms of $x$. You can feel free to choose the expansion around any point (even around infinty can work I believe), here I expand it around $x=0$, which is $$\frac{1}{x+y}=\sum_{k=0}^{\infty}(-1)^{k}x^{k}y^{-(k+1)}.$$ Again, the $y^{-(k+1)}$ make sense because $y$ is a unit. Hence, what you should consider to prove is $$(x+y)\Bigg(\sum_{k=0}^{n}(-1)^{k}x^{k}y^{-(k+1)}\Bigg)=1.$$ This is again easy to verify.

Given, $R$ is a commutative ring with unity. a be the unit and $b^2=0$. Let $c$ be the multiplicative inverse of $a$.

Case $1$:
If $b=0$
$a.c=1$
$\Rightarrow(a+0).c=1$
$\Rightarrow(a+b).c=1$

$\implies a+b$ is a unit.

Case $2$:
$If b≠0$
$b^2=0$
$\Rightarrow(bc)²=0$
$\Rightarrow1-(bc)²=1$
$\Rightarrow[(1+bc)(1-bc)]=1$. [ Since, $R$ is a commutative ring]
$\Rightarrow[(a+b)(c-bc²)]=1$. [ Since, $c$ is the multiplicative inverse of $a$]

$\implies a+b$ is unit.

Hence proved.

Let $R$ denote a ring, $u\in R$ denote a unit, and $w\in R$ nilpotent. Then $(u+w)(u-w)=u^2-w^2$ (by our commutativity assumption). Notice that $u^2$ will remain a unit and $w^2$ a nilpotent element. If we repeat this process (e.g. we compute $(u^2+w^2)(u^2-w^2)$), we should arrive at the equation $(u^{2^{n}}+w^{2^{n}})(u^{2^{n}}-w^{2^{n}})=(u^{2^{n+1}}-w^{2^{n+1}})$ after $n$-repetitions.

Define $k\in\mathbb{N}$ to be the least positive integer such that $w^k=0$ (such an integer will exist be definition of nilpotent elements). Then $n\geq \lceil\frac{k}{2}\rceil$ implies $w^{2^{n+1}}=0$, and thereby $(u^{2^{n}}+w^{2^{n}})(u^{2^{n}}-w^{2^{n}})$ gives us a unit. Noting that this final result is true if and only if $(u^{2^{n}}+w^{2^{n}})$ and $(u^{2^{n}}-w^{2^{n}})$ are units, induction will give us the desired result.

I think this is just rephrasing arguments above, but the perspective is a little different. Let $R$ be any ring with $u \in R$ a unit and $a\in R$ nilpotent and $ua=au$. Write $R^\times$ for the group of units in a ring $R$, so that $u \in R^\times$. Then since $R^\times$ is a group and $u+a = u(1+u^{-1}a)$ it follows that that $u+a$ is a unit if and only if $1+u^{-1}a$ is a unit. But since $ua=au$ we have $(u^{-1}a)^m = u^{-m}a^m=0$, that is, $u^{-1}a$ is nilpotent. But this follows from the following claim:

Claim: If $R$ is any ring and $n \in R$ is nilpotent, then $1+n$ is a unit.

Proof of claim: The claim is a property of a pair $(R,n)$ consisting of a ring $R$ and a nilpotent element $n \in R$. Now clearly the claim is true of the pair $(R,n)$ if and only if it is true for the pair $(\mathbb Z[n],n)$ where $\mathbb Z[n]\subseteq R$ is the subring generated by $n\in R$. Moreover, $\mathbb Z[n] \cong \mathbb Z[t]/\langle t^m \rangle$ where $m=\min\{k \in\mathbb Z_{>0}: a^k=0\}$, hence the claim is true for $(\mathbb Z[n],n)$ if and only if it is true for the pair $(\mathbb Z[t]/\langle t^m \rangle,t+\langle t^m \rangle)$.

Thus if we let $t_m = t+\langle t^m \rangle$, we are reduced to showing that $1+t_m$ is a unit in $\mathbb Z[t]/\langle t^m\rangle$ for every $m \geq 1$. But if $P = \mathbb Z[\![t]\!]$ is the ring of formal power series in $t$, then $\pi_m \colon P \to \mathbb Z[t]/\langle t^m \rangle$ given by $\pi_m(t) = t_m$ defines a surjective ring homomorphism from $S$ to $\mathbb Z[t]/\langle t^m \rangle$, hence it suffices to show that $1+t$ is a unit in $S$, but this follows from the identity $(1+t)^{-1} = \sum_{k=0}^\infty (-1)^kt^k$.

Since $a$ is nilpotent, it belongs to the Jacobson radical $R(J)$. (To see this, note that $R(J)$ is the intersection of all maximal ideals, which are also all prime.)

Now suppose for contradiction that $u + a$ is not a unit. Then it belongs to some maximal ideal $m$, but $a \in R(J) \subseteq m$. This implies that $u \in m$, which is absurd, since it would make $m$ equal to the whole ring, contradicting the fact that it is a proper ideal.

Since $u$ is a unit we have $u'$ in $R$ such that $uu'=1$. Now note that if $a$ is a nilpotent then $au$ also a nilpotent element in $R$. Let $$a^nu'^n =0$$

Now $$(1-au')(1+au')=1-a²u'²$$ $$(1-a²u'²)(1+a²u'²)=1-a⁴u⁴$$ After nth step$$1-a^{2^n}u^{2^n}=1$$

Since we start with $1+au'$ it's a unit in $R$ so that $uu'+au' =u'(u+a)$ and so that $u+a$.