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Let $R$ be a commutative ring. We say that an ideal ${\mathcal I}$ in $R$ is generated by elements $\iota_1, \ldots \iota_k$ in $R$, written:

${\mathcal I}=(\iota_1,\ldots,\iota_k)\quad or \quad{\mathcal ​I}=R\iota_1+R\iota_2+\ldots+R\iota_k$

...if ${\mathcal I}$ consists precisely of all linear combinations of the form $r_1\iota_1+\ldots+r_k\iota_k$ with $r_1, \ldots, r_k\in R$.

Let $k$ be a field and let $R=k[x_1,\ldots,x_n]$ be the ring of polynomials in $n$ variables with coefficients in $k$.

If ${\mathcal S}$ is a finite subset of $R$, let $V({\mathcal S})$ be the points $a=(a_1,\ldots,a_n)\in k^n$ such that $P(a)=0$ for all $P\in {\mathcal S}$.

It is easy to check that $V({\mathcal S})=V({\mathcal I})$ where ${\mathcal I}$ is the ideal generated by the elements of ${\mathcal S}$.

Conversely, if $X$ is a subset of $k^n$, let ${\mathcal I}(X)$ be the ideal in $R$ consisting of the polynomials in $R$ vanishing at all the points of $X$.

Which of the following are correct?

Select ALL that apply.


If $X\subseteq Y\subseteq k^n$, then ${\mathcal I}{\mathcal (X)}\subseteq {\mathcal I}(Y)$.


If ${\mathcal S}\subseteq{\mathcal T}\subseteq R$, then $V({\mathcal S})\subseteq V({\mathcal T})$.


If ${\mathcal S}\subseteq R$ and $X\subseteq k^n$, then ${\mathcal S}\subseteq {\mathcal I}(V({\mathcal S}))$ and $X\subseteq V({\mathcal I}(X))$.


When $k$ has infinite cardinality, there is no subset $X$ of $k^n$ with ${\mathcal I}(X)=\{0\}$.


If ${\mathcal S}\subseteq R$, and $X\subseteq k^n$, then $V({\mathcal I}(V({\mathcal S})))=V({\mathcal S})$ and ${\mathcal I}(V({\mathcal I}(X)))={\mathcal I}(X)$.

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