It is enough to prove that every prime ideal $\mathfrak{q}$ of $S$ is the intersection of the maximal ideals containing it. Replace $R\subseteq S$ byThe new extension is integral, both rings are domains, and the base remains a Jacobson ring.

Let $0\ne s\in S$. Choose an integral equation of least degreeAs above, $a_0\ne 0$. Since the zero ideal of $R$ is the intersection of its maximal ideals, choose a maximal ideal $\mathfrak{m}$ with $a_0\notin \mathfrak{m}$. By the Lying-over theorem, some maximal ideal $\mathfrak{n}$ of $S$ contracts to $\mathfrak{m}$. If $s\in \mathfrak{n}$, the integral equation would imply $a_0\in \mathfrak{n}\cap R=\mathfrak{m}$, a contradiction. Thus every nonzero $s$ is omitted by some maximal ideal, so their intersection is zero. Therefore $S$ is Jacobson, proving Integral extension of a Jacobson ring.

Pass to the integral extensionThe ring $B/I$ is an integral domain, and the ideal $J/I$ has zero contraction to the base. If $J/I$ contained a nonzero element $b$, choose an integral equation for $b$ of least degree:Its constant term is nonzero, since otherwise the domain property would let us cancel $b$ and obtain an equation of lower degree. Butis a nonzero element of $(J/I)\cap (A/(I\cap A))$, a contradiction. Therefore $J/I=0$ and $I=J$. This proves ideal contraction rigidity under an integral extension.

For an algebraically closed field $k$, the Weak Hilbert Nullstellensatz says that every maximal ideal of $k[x_1,\dots ,x_n]$ isfor a unique $a\in k^n$. Equivalently, every proper ideal has a common zero. The Strong Hilbert Nullstellensatz says that for every ideal $I$,

To deduce the strong form, let $f$ vanish on $V(I)$ and introduce a variable $y$. The equations in $I$ together with $1-yf$ have no common zero: a common zero would satisfy both $f=0$ and $yf=1$. The weak theorem therefore givesSubstitute $y=f^{-1}$ in the localization $k[x_1,\dots ,x_n,f^{-1}]$. The last term vanishes, and clearing a power of $f$ yields $f^N\in I$. Thus $f\in \sqrt{I}$. The reverse inclusion is immediate, completing the Rabinowitsch trick proof.

The Krull intersection theorem givesbecause $(R,\mathfrak{m})$ is a Noetherian local ring and $x\in \mathfrak{m}$. Consequently every nonzero $a\in R$ has a largest $x$-adic order: one can write

Suppose nonzero elements $a,b$ satisfy $ab=0$. Write $a=x^r{a}^{\prime }$ and $b=x^s{b}^{\prime }$ with $a^{\prime },b^{\prime }\notin (x)$. Since $x$ is a non-zero-divisor, cancellation of $x^{r+s}$ gives $a^{\prime }{b}^{\prime }=0$. Reducing modulo $(x)$ now gives a product of two nonzero elements equal to zero in $R/(x)$, contradicting that this quotient is an integral domain. Hence $R$ is an integral domain.

Choose $a\in K\setminus F$. The multiplication homomorphismhasThis element is nonzero. Since $1,a$ are linearly independent over $F$, there is an $F$-linear functional $\lambda :K\to F$ with $\lambda (1)=0$ and $\lambda (a)=1$. Applying $\lambda \otimes {\mathrm{id}}_K$ sends the displayed tensor to $1$.

False. LetThe element $1/y$ does not belong to $A$: assigning degrees $\mathrm{deg}x=\mathrm{deg}y=1$ and $\mathrm{deg}(x/y)=0$ shows that every element of $A$ has nonnegative total degree, whereas $1/y$ has degree $-1$.

If $A=T^{-1}R$ for some multiplicative subset $T$, the membership $x/y\in T^{-1}R$ would give $x/y=a/t$ for some $a\in R$ and $t\in T$. Since $R$ is a unique factorization domain and $x,y$ are coprime, the equation $xt=ay$ implies $y\mid t$. Write $t=yu$. As $t$ is invertible in $T^{-1}R$, so iscontrary to $1/y\notin A$. Thus a subring of a localization of a ring need not itself be a localization of the original ring.

Take $V=U={\mathbb{R}}^3$ and identify $V{\otimes }_{\mathbb{R}}U$ with the space of $3\times 3$ real matrices, where pure tensors correspond to matrices of rank at most one. Letand let $f$ be the quotient map. It is not injective because its kernel is the nonzero line $\mathbb{R}{I}_3$.

If $f(v_1\otimes u_1)=f(v_2\otimes u_2)$, thenfor some $\lambda \in \mathbb{R}$. The left side has matrix rank at most two. If $\lambda \ne 0$, the right side has rank three, which is impossible. Thus $\lambda =0$ and the two pure tensors were equal. Hence $f$ is injective on the set of pure tensors while failing to be injective linearly.

Yes. Let $\mathfrak{m}$ be maximal and setThen $L$ is a field generated as a $\mathbb{C}$-algebra by countably many elements. Since the polynomial ring in countably many variables has a countable monomial basis, $L$ has at most countable dimension as a $\mathbb{C}$-vector space.

Suppose $t\in L$ were transcendental over $\mathbb{C}$. The familywould be linearly independent over $\mathbb{C}$. Indeed, after multiplying a finite relation by ${\prod }_j(t-x_j)$, evaluation at $t=x_i$ forces the $i$th coefficient to vanish. This would be an uncountable linearly independent subset of the countable-dimensional vector space $L$, a contradiction.

Thus $L/\mathbb{C}$ is algebraic. Since $\mathbb{C}$ is an algebraically closed field, $L=\mathbb{C}$. If $x_i$ is the image of $T_i$, the quotient map is evaluation at $(x_1,x_2,\dots )$ and

Take $R=\mathbb{Z}$, $M=\mathbb{Q}/\mathbb{Z}$, and $k=2$. The group $M$ is nonzero and divisible. Since every element of $M$ has finite order, $M$ is the filtered union of finite cyclic groups. For every $n\ge 1$,Tensor products commute with filtered colimits, so

Now suppose a nonzero finitely generated $R$-module $M$ satisfied $M^{\otimes k}=0$. Choose a maximal ideal $\mathfrak{m}$ in the support of $M$. The localized module $M_{\mathfrak{m}}$ is nonzero and finitely generated. By Nakayama lemma,This is a nonzero vector space over the residue field $\kappa (\mathfrak{m})$, so its $k$-fold tensor power is nonzero. But it is the reduction modulo $\mathfrak{m}$ of $(M^{\otimes k}{)}_{\mathfrak{m}}$, a contradiction. Thus a nonzero tensor-nilpotent module cannot be finitely generated.

For a commutative ring $R$, the Jacobson radical isLet $A\subseteq B$ be integral. If $\mathfrak{n}$ is maximal in $B$, then its contraction $\mathfrak{n}\cap A$ is maximal in $A$. Therefore every $a\in J(A)$ belongs to every $\mathfrak{n}$, andConversely, the Lying-over theorem puts a maximal ideal $\mathfrak{n}$ of $B$ above every maximal ideal $\mathfrak{m}$ of $A$. Hence an element of $J(B)\cap A$ lies in every $\mathfrak{m}$, provingThis is the Jacobson radical under an integral extension formula.

The maximum isThe product of two fields $K\times L$ attains it: its only nonzero proper ideals are $K\times 0$ and $0\times L$, and both are maximal ideals.

For the upper bound, suppose ${\mathfrak{m}}_1$ and ${\mathfrak{m}}_2$ are distinct maximal ideals of $R$. Their intersection cannot be nonzero, since a nonzero proper ideal is maximal and cannot be properly contained in either of two distinct maximal ideals. Hence ${\mathfrak{m}}_1\cap {\mathfrak{m}}_2=0$. Also ${\mathfrak{m}}_1+{\mathfrak{m}}_2=R$, so the Chinese remainder theorem givesBoth factors are fields, and this product has exactly two maximal ideals. Thus a third maximal ideal is impossible.

An $R$-module $M$ is Noetherian when every submodule is finitely generated, equivalently when every ascending chain of submodules stabilizes.

A free module is an $R$-module with a basis: every element has a unique expression as a finite linear combination of basis elements.

A flat module $M$ is one for which the tensor functor $-{\otimes }_{R}M$ is exact. Since tensor products are always right exact, it is equivalent to require that tensoring with $M$ preserve injections.

A projective module $P$ has the lifting property: for every surjection $E\twoheadrightarrow F$ and every map $P\to F$, there is a map $P\to E$ making the resulting triangle commute. Equivalently, $P$ is a direct summand of a free module.

For a degree-three isogeny whose kernel becomes ${\mu }_3$ under the Weil pairing, the Kummer connecting map embeds the quotient by the isogeny into $K^{\times }/(K^{\times }{)}^3$. Valuations outside the primes of bad reduction restrict its image to an S-unramified power class group.

A finite-dimensional complex Lie algebra is semisimple if and only if its Killing form is nondegenerate.