Why this Artin ring is simple? Suppose we have a non-commutative left Artin ring $R$, such that the subring generated by any nonzero two-sided ideal together with the unity is $R$ itself. I read that $\mathrm{Z}(R)$, which is the center of $R$, forms an integral domain, but why is this true? And how can one proceed to deduce that $R$ is simple? I tried to use DCC on the sets of all two-sided ideals to go further, but don’t see why this implies there can only be extremal two-sided ideals in this case. Any hints on this?
 A: Let $R$ be left artinian such that, for all non-zero two-sided ideals $I$, the subring generated by 1 and $I$ is all of $R$. We will describe all such $R$, including the commutative ones.
Write $J$ for the Jacobson radical of $R$, a two-sided ideal. Suppose first that $J^2$ is non-zero. Then $R$ is generated as a ring by 1 and $J^2$, so $R/J^2$ is generated by 1, and hence is a quotient of $\mathbb Z$. In particular, $J/J^2$ is generated by the image of some integer $a$, so by Nakayama's Lemma $J=Ra$, and hence, using the nilpotency of $J$, we see that $\mathbb Z$ surjects onto $R$. Thus $R=\mathbb Z/n$ for some integer $n$.
Otherwise $J^2=0$. Assume $J\neq0$. Then $R$ is generated by 1 and $J$, so $R/J$ is generated by 1, so is a quotient of $\mathbb Z$. If the image of some $a\in\mathbb Z$ is non-zero and in $J$, then $R$ is generated by $1$ and $Ra$, so by 1, and hence is a quotient of $\mathbb Z$.
Otherwise, the image $A$ of $\mathbb Z$ intersects $J$ trivially, and $R=A\oplus J$. If $e\in A$ is a non-trivial idempotent, then $R$ is generated by 1 and $Re$, so $J=Je$. Similarly $J=J(1-e)$, so $J=0$, a contradiction. Thus $A=\mathbb Z/p$ for some prime $p$. Taking any non-zero $t\in J$ we see that $R=(\mathbb Z/p)[t]/(t^2)$.
Finally, if $J=0$, then $R$ is semisimple artinian, so a product of matrices over division rings. If $R=A\times B$ is a proper product, then $R$ is generated as a ring by 1 and either $A$ or $B$. Thus both $A$ and $B$ are quotients of $\mathbb Z$ by square-free integers. This holds for any such decomposition, so either $R=\mathbb Z/m$ with $m$ square-free, or else $R=(\mathbb Z/p)^2$ for some prime $p$. Note that up to now, all such $R$ are commutative.
The only other case is when $R$ is a matrix ring over a division ring, and so simple. In particular, every such non-commutative $R$ is simple artinian.
In summary, then only such rings are
$$ \mathbb Z/n, \quad (\mathbb Z/p)[t]/(t^2), \quad (\mathbb Z/p)^2, \quad \mathbb M_r(D) $$
with $p,n,r\in\mathbb N$ and $p$ prime, and $D$ a division ring.
A: This is what I was working on.  Let $R$ be a right ring such that $1\cdot \mathbb Z+I=R$ for every nonzero ideal $I$, such that $R$ is not commutative.

Lemma 1: under the conditions above, there is no nonzero element $x$ such that $x\mathbb Z$ is an ideal of $R$.

Proof: if $x\mathbb Z$ were a two-sided ideal, $1\cdot\mathbb Z+x\mathbb Z$ would be commutative, contradicting our standing hypotheses.

Lemma 2: under the conditions above, if $I$ is a nonzero ideal, $I^2\neq \{0\}$.

Proof: if $I^2=\{0\}$, then $R=1\cdot \mathbb Z+I$ is commutative, so the square of $I$ must be nonzero.

Proposition:  The center of $R$ is a domain.

Proof: If $xy=0$, and $x,y$ are nonzero elements of the center of $R$, then from $R=1\cdot \mathbb Z+(y)$, we learn $xR=x\mathbb Z$. Since the right hand side is manifestly a nonzero ideal, the Lemma 1 above contradicts this. Therefore there is no such pair $x,y$, and the center is a domain.

Proposition: $R$ is simple Artinian

Proof: it suffices to show that $R$ is prime, i.e. there are no nonzero ideals $A,B$ such that $AB=\{0\}$.  Suppose two such ideals existed.  Since $(BA)^2=\{0\}$, we see by Lemma 2 that $BA=\{0\}$ as well.
Now let $a$ be a nonzero element of $A$.  From $R=1\cdot \mathbb Z +B$ we learn both $aR=a\mathbb Z$ and $Ra=a\mathbb Z$. Therefore $aR=Ra=a\mathbb Z$ is a nonzero ideal of $R$, something prevented by the Lemma 1 above.
Therefore $R$ is a right Artinian prime ring, which is well-known to be a simple Artinian ring.

If one does not recall the proof that a right Artinian prime ring is simple, we could also say that $R$ is right Artinian and semiprime, hence semisimple. It would remain to show that it is simple, and this would follow from a similar argument about existence of two ideals $A,B$ such that $AB=BA=\{0\}$.
A: After having needed to spend a bit of time in order for things to come to light, may I say that this is quite a lovely problem and may I convey my regards to your professor for having included it in his textbook.
Let us assume a left Artinian, non-commutative ring $A$ such that for every nonzero bilateral (I dislike the English terminology of "two-sided") ideal we have $A=[I \cup \{1_A\}]_{\mathrm{Ann}}$, where for arbitrary subset $X \subseteq A$ I use the notation $[X]_{\mathrm{Ann}}$ to refer to the subring generated by $X$ in $A$. Let us remark that for any bilateral ideal $I \leqslant_{\mathrm{b}} A$ we have the relation $[I \cup \{1_A\}]_{\mathrm{Ann}}=I+\mathbb{Z}_A$, where I have used the abbreviation $\mathbb{Z}_A\colon=\mathbb{Z}1_A\colon=\{n1_A\}_{n \in \mathbb{Z}}$ (on the same note let us also introduce $n_A\colon=n1_A$ for arbitrary integer $n \in \mathbb{Z}$).
We seek to prove that $A$ is simple under these conditions and we shall proceed in the following sequence:


*

*Introducing the abbreviation $C\colon=\mathrm{Z}(A)$ for the centre of $A$, we first argue that $C$ is a field.


Since $A$ is non-commutative it is in particular nonzero, which means $0_A \neq 1_A$ and therefore $C$ is also nonzero. Given arbitrary ring $R$ we shall write $\mathrm{U}(R)$ for the subset of units of $R$ and similarly $R^{\times}\colon= R \setminus \{0_R\}$. For arbitrary subset $X \subseteq R$ we recall the left and right annihilators $\mathrm{Ann}_{\mathrm{s}}(X)\colon=\{t \in A \mid tX \subseteq \{0_R\}\}$ respectively $\mathrm{Ann}_{\mathrm{d}}(X)=\{t \in A \mid Xt \subseteq \{0_R\}\}$, remarking that left (right) annihilators are always left (right) ideals: $\mathrm{Ann}_{\mathrm{s}}(X) \leqslant_{\mathrm{s}} R$ respectively $\mathrm{Ann}_{\mathrm{d}}(X) \leqslant_{\mathrm{d}} R$. In the particular case when $X \subseteq \mathrm{Z}(R)$ is a central subset it is easily seen that the two lateral annihilators coincide and are both a bilateral ideal of $R$: $\mathrm{Ann}_{\mathrm{s}}(X)=\mathrm{Ann}_{\mathrm{d}}(X) \leqslant_{\mathrm{b}} R$.
We first establish the following characterisation: $z \in \mathrm{U}(C) \Leftrightarrow \mathrm{Ann}_{\mathrm{s}}(z)=\{0_A\}$ (a central element is invertible if and only if it is cancellable). Indeed, the implication $z \in \mathrm{U}(C) \Rightarrow \mathrm{Ann}_{\mathrm{s}}(z)=\{0_A\}$ is a particular case of the more general implication $x \in \mathrm{U}(A) \Rightarrow \mathrm{Ann}_{\mathrm{s}}(x)=\mathrm{Ann}_{\mathrm{d}}(x)=\{0_A\}$, relation which is valid in any ring without imposing any supplementary hypotheses.
As for the converse implication, we turn to the left Artinian nature of $A$ by considering the decreasing sequence of principal left (they are actually bilateral) ideals $\left(Az^n\right)_{n \in \mathbb{Z}}$ and inferring the existence of $k \in \mathbb{N}$ such that $Az^k=Az^{k+1}$, which further entails the existence of $x \in A$ such that $z^k=z^{k+1}x$ or equivalently $(1_A-zx)z^k=0_A$ and thus $1_A-zx \in \mathrm{Ann}_{\mathrm{s}}\left(z^k\right)$. In general, for arbitrary $y \in A$ it is elementary that $\mathrm{Ann}_{\mathrm{s}}(y)=\{0_A\}$ entails $\mathrm{Ann}_{\mathrm{s}}\left(y^n\right)=\{0_A\}$ for any $n \in \mathbb{N}$ (if $y$ has a zero left annihilator then the group endomorphism of $(A, +)$ given by right multiplication with $y$ is injective, property which is preserved by all the powers of this endomorphism in the ring $\mathrm{End}_{\mathbf{Gr}}((A, +))$, powers which are none other than the endomorphisms given by right multiplication with $y^n$ for generic $n \in \mathbb{N}$). We thus infer that $zx=xz=1_A$, which means that $z \in \mathrm{U}(A) \cap \mathrm{Z}(A)=\mathrm{U}(C)$ (recall that in general for any monoid $M$ the equality $\mathrm{U}(\mathrm{Z}(M))=\mathrm{U}(M) \cap \mathrm{Z}(M)$ holds, since indeed any central element of $M$ which is a unit of $M$ will necessarily have a central inverse, rendering it into a unit of the centre).
Having obtained the above characterisation, let us further establish the relation $z=0_A \Leftrightarrow z \in C \wedge \mathrm{Ann}_{\mathrm{s}}(z) \neq \{0_A\}$. The $\Rightarrow$ direction is clear, so we focus on the converse, considering an arbitrary central element $z \in C$ whose (left...) annihilator $I\colon=\mathrm{Ann}_{\mathrm{s}}(z)$ is nonzero. As we have already remarked, $I \leqslant_{\mathrm{b}} A$ is in this case a bilateral ideal and as it is nonzero we gather that $A=I+\mathbb{Z}_A$ by virtue of our very particular assumption. This immediately entails the description $Az=\left(I+\mathbb{Z}_A\right)z=Iz+\mathbb{Z}z=\mathbb{Z}z$, which means that any element in the principal (left...) ideal generated by $z$ is actually an integer multiple of $z$. Let us now additionally assume that $z \neq 0_A$ and derive from this a contradiction. If indeed $z \neq 0_A$ then $z \in Az \leqslant_{\mathrm{b}} A$, in other words $Az$ is a nonzero bilateral ideal (left ideals generated by central subsets are automatically bilateral). The particular assumption of our problem once again applies to the conclusion that $A=Az+\mathbb{Z}_A=\mathbb{Z}z+\mathbb{Z}_A$, which further entails the commutativity of $A$ thus contradicting the hypothesis. It follows that with necessity $z=0_A$, under the above assumptions.
Hence, an arbitrary central element $z \in C$ either has nontrivial (left...) annihilator - in which case it is $0_A$ - or a trivial one, in which case it is a unit of $C$. This establishes $C$ as a field.



*We now argue that $A$ is a simple ring, for which it will suffice to show that its Jacobson radical $R\colon=\mathrm{Rad}(A)$ is null.


Indeed, having established the fact that $R=\{0_A\}$ we recall the general property of any left Artinian ring of having a semisimple quotient by its radical. It follows from one of the Wedderburn structure theorems that $A/\{0_A\} \approx A \ (\mathbf{Ann})$ is semisimple and thus isomorphic to the direct product of a finite family $S$ - indexed by $\Lambda$ - of nonzero simple rings. This entails that the centre $C$ is isomorphic to the direct product of the family $(\mathrm{Z}(S_{\lambda}))_{\lambda \in \Lambda}$ of centres. By virtue of another one of the Wedderburn structure theorems, it can be easily inferred that the centre of a nonzero simple ring is necessarily a field. Combining this with the fact that $C$ is itself a field and hence a nonzero domain, we gather that $|\Lambda|=1$, in other words the direct product isomorphically expressing $A$ must consist of precisely one factor (the case $|\Lambda| \geqslant 2$ would contradict the integrity of $C$ whereas $\Lambda=\varnothing$ is prohibited by our previous observation that since $A$ is noncommutative it is in particular nonzero).
Let us proceed by Reductio ad Absurdum and assume that $R \neq \{0_A\}$. We first remark that under this hypothesis it is impossible that $\mathrm{char}\ C=0$. Indeed, if this were the case then $2_A \neq 0_A$. Since $R \leqslant_{\mathrm{b}} A$ is nonzero we gather that $A=R+\mathbb{Z}_A$ and as $C$ is a field we can claim that $\frac{1}{2_A} \in A$, whence there must exist $a \in R$ and $k \in \mathbb{Z}$ such that $\frac{1}{2_A}=a+k_A$. This equation leads to $a=-k_A+\frac{1}{2_A} \in C^{\times}=\mathrm{U}(C)$ and signifies that the proper ideal $R$ would contain a unit, clearly an absurd situation (the hypothesis that $C$ is of characteristic $0$ is the one which justifies the assertion that $k_A \neq \frac{1}{2_A}$, equivalent to $(2k-1)_A \neq 0_A$).
Since $C$ is a field, we must have $\mathrm{char}\ C=p$ for a certain (strictly positive) prime number $p$. This means that $A$ has a canonical (associative and unitary) $\mathbb{Z}_p$ - algebra structure, where I am using the notation $\mathbb{Z}_r\colon=\mathbb{Z}/r\mathbb{Z}$ for arbitrary $r \in \mathbb{N}$. From the particular assumption in the hypothesis we gather that given any proper nonzero bilateral ideal $I \leqslant_{\mathrm{b}} A$, we have $A=I+(\mathbb{Z}_p)1_A$, where the sum on the right-hand side is direct: indeed, $I \cap \left(\mathbb{Z}^{\times}_p\right)1_A \subseteq I \cap \mathrm{U}(A)=\varnothing$ whence it follows that $I \cap (\mathbb{Z}_p)1_A=\{0_A\}$. From this we infer that $\mathrm{codim}^{A}_{\mathbb{Z}_p}I=\mathrm{dim}_{\mathbb{Z}_p}(A/I)=1$, in other words any nonzero proper bilateral ideal of $A$ is a hyperplane in the $\mathbb{Z}_p$-vector space structure subjacent to the algebra structure introduced above.
Let $\mathscr{Id}_{\mathrm{b}}(A)$ denote the set of all bilateral ideals of $A$, $\mathscr{I}\colon=\mathscr{Id}_{\mathrm{b}}(A) \setminus \{A\}$ and $\mathscr{I}’\colon=\mathscr{I} \setminus \{\{0_A\}\}$. As $A$ is nonzero we clearly have $\mathscr{I} \neq \varnothing$. Bearing in mind the general relation $R \subseteq \displaystyle \bigcap \mathrm{Max}\left(\mathscr{I}\right)$ ($\mathscr{I}$ is considered implicitly ordered by inclusion and the syntax $\mathrm{Max}$ refers to the collection of maximal elements with respect to this ordered set, in other words the collection of all maximal bilateral ideals), we infer that $\mathscr{I}’ \neq \varnothing$ (otherwise we would have $\mathscr{I}=\{\{0_A\}\}$, $\mathrm{Max}\left(\mathscr{I}\right)=\{0_A\}$ and eventually $R \subseteq \{0_A\}$).
It is also an elementary fact that the ordered set $\mathscr{I}$ is inductive and therefore admits at least a maximal element, say $M \in \mathrm{Max}\left(\mathscr{I}\right)$. It is clear that $M \neq \{0_A\}$, for assuming the contrary would once again entail $\mathscr{I}=\{\{0_A\}\}$, situation which we have already ruled out as impossible under the given assumptions. Since $M, R \in \mathscr{I}’$ we have that $M$ and $R$ are both $\mathbb{Z}_p$ - hyperplanes of $A$, according to one of the previous observations. Since $R \subseteq M$, we thus gather that $R=M$, meaning that the Jacobson radical $R$ is a maximal bilateral ideal. Since $R$ is at the same time included in every maximal bilateral ideal, we infer that $\mathrm{Max}\left(\mathscr{I}\right)=\{R\}$.
Since any proper bilateral ideal is included in a certain maximal bilateral ideal (by virtue of the same token of inductivity of $\mathscr{I}$ ordered by inclusion), the existence of just one maximal bilateral ideal means that $A$ actually has $R$ as a maximum proper bilateral ideal, symbolically $R=\max \mathscr{I}$. Thus, if we furthermore consider an arbitrary $I \in \mathscr{I}’$ we gather that $I \subseteq R$ and $I$, $R$ are both $\mathbb{Z}_p$ - hyperplanes of $A$. Once again, this entails $I=R$ and signifies that we actually have the relation $\mathscr{I}=\{\{0_A\}, R\}$.
Let us however keep in mind another general property of left Artin rings, namely that they have nilpotent Jacobson radicals. I will use the notation $I^{.n}$ to refer to the power of exponent $n \in \mathbb{N}$ of the bilateral ideal $I$ within the multiplicative monoid $(\mathscr{Id}_{\mathrm{b}}(A), \bullet)$. In our case, since $R^{.2} \in \mathscr{I}$ we have either $R^{.2}=\{0_A\}$ or $R^{.2}=R$. The latter option would mean that $R$ is at the same time a nonzero idempotent as well as a nilpotent element, which constitutes a contradiction. This leaves the only possible option $R^{.2}=\{0_A\}$.
The above finding means in particular that $R^{(2)}=RR=\{0_A\}$ and - together with the relation $A=R+\mathbb{Z}_A$ - allows us to infer that $R \subseteq \mathrm{C}_A(R)$, in other words $R$ centralises itself. As it is also obvious that $R \subseteq \mathrm{C}_A(\mathbb{Z}_A)$, taking into account the relation $A=R+\mathbb{Z}_A$ we easily gather that $A \subseteq \mathrm{C}_A(A)=\mathrm{Z}(A)$, contradicting the restriction that $A$ is not commutative. This contradiction is caused by the assumption $R\neq \{0_A\}$ and signifies that indeed $R=\{0_A\}$, as we had set out to prove.
Our argument thus comes to an end.
