Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 101 / 3

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.

The assertion is 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.

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 $1$.

Thus the unital ring homomorphism $\mu$ has a nonzero kernel. A unital homomorphism from a field is injective, so $K{\otimes }_{F}K$ is not a field.

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.