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Background

Example: If $k$ is a field, the ideal $(x)$ is maximal in $k[x]$. In $k[x,y]$, the ideal $(x)$ is not maximal, the ideal $(x,y)$ is maximal. We have $(x)\subseteq (x,y)\subseteq k[x,y]$.

Questions

In the above exampe, where it states why $(x)$ is not maximal in $k[x,y]$, I just want to make sure I understand the notation properly. In the cases of both $(x)$ and $(x,y)$ both being elements of $k[x,y]$, they are defined respectively as $(x)=\{x\cdot f(x,y)\mid f(x,y)=\sum_{i,j}a_{i,j}x^i,y^j\in k[x,y]\}$ and $(x,y)=\{x\cdot g_1(x,y)+y\cdot g_2(x,y)\mid g_1(x,y)=\sum_{i,j}a_{i,j}x^i,y^j\in k[x,y], g_2(x,y)=\sum_{i,j}a_{i,j}x^i,y^j\in k[x,y]\}$. Also both $(x)$ and $(x,y)$ being subset of $k[x,y]$ doesn't explicitly mean $(x)=\{x\cdot f(x,0)\mid f(x,0)=\sum_{i,j}a_{i,j}x^i,0^j\in k[x,y]\}$ and $(x,y)=\{x\cdot g_1(x,0)+y\cdot g_2(0,y)\mid g_1(x,0y)=\sum_{i,j}a_{i,j}x^i,0^j\in k[x,y], g_2(0,y)=\sum_{i,j}a_{i,j}0^i,y^j\in k[x,y]\}$. It seems that $(x)\subseteq k[x]$ and $(x)\subseteq k[x,y]$ along with $(x,y)\subseteq k[x,y]$ could mean different things.

Anne Bauval
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Seth
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  • I think your trouble is this: $(x)$ and $(x)$ look identical, but they mean different things when one is an ideal in $k[x]$ and the other is an ideal in $k[x, y]$. This is true. You need to remember the ring in which you are working to correctly interpret this notation. (Alternatively, you can already consider $x$ to be different from $x$ if they are indeterminate in different polynomial rings.) – Eike Schulte Aug 04 '24 at 08:04
  • @EikeSchulte yes, that is what is causing me confusion. $(x)$ means different things depending on which polynomial ring one is talking about. In the case of $(x)\subseteq k[x,y]$ and similarily for $(x,y)$, is it $x\cdot f(x,0)\in k[x,y]$, $f(x,0), g(0,y)$ are both elements of $k[x,y]$. For $(x)\subseteq k[x,y]$, is it $x\cdot f(x,y)\in k[x,y]$ or $x\cdot f(x,0)\in k[x,y]$. Simiilarly for the case of $(x,y)$, is it $x\cdot f(x,0)+y\cdot g(0,y)$ or is it $x\cdot f(x,y)+ y\cdot g(x,y)$ where $f(x,0), g(0,y), f(x,y), g(x,y)$ are all elements of $k[x,y]$? – Seth Aug 04 '24 at 08:20
  • You always multiply by all elements of the ring in question. All the variants where you have $f(x,0)$ or $g(0,y)$ are wrong. (You get these sets if you first compute the ideal in $k[x]$ and then embed the result in $k[x,y]$, but if this is meant, it should usually be made more explicit; they’re not ideals of $k[x,y]$, though.) – Eike Schulte Aug 04 '24 at 09:38
  • @EikeSchulte so basically i multiply by $(x)$ by $f(x,y)$ and similarily for $(x,y)$, it is $(x,y)={x\cdot f(x,y)+y\cdot g(x,y)\mid f(x,y), g(x,y) \in k[x,y]}$$ – Seth Aug 04 '24 at 10:21
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    That's correct. – Eike Schulte Aug 04 '24 at 11:33

1 Answers1

2

This notation for ideals has nothing specifically to do with polynomial rings. In any commutative ring $A$ with $a_1, \ldots a_n \in A$, the ideal $(a_1,\ldots,a_n)$ in $A$ means the set of all $A$-linear combinations of $a_1, \ldots, a_n$: $$ (a_1,\ldots,a_n) = \{a_1b_1+\cdots + a_nb_n : b_i \in A\}. $$ This is the smallest ideal in $A$ that contains $a_1, \ldots, a_n$. This notation does not indicate what the ring $A$ is, so you simply need to pay attention to what ring you are in. When $A = \mathbf Z$, $(2)$ is a proper ideal, but when $A = \mathbf Q$, $(2) = (1) = A$ since $2$ is invertible in $\mathbf Q$.

KCd
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  • I edited my post. I added the reasons why I am confused about theh notation. Does your comment still applies. Sorry I hope I am not being difficult, I feel there is an ambigiuity about what $()$ means in $[]$ versus $k[x,y]$. . – Seth Aug 04 '24 at 08:27
  • An ideal $(a)$ in a commutative ring $A$ means all multiples of $a$ in $A$: it is the set of all $ab$ as $b$ runs over the elements of $A$. As long as you know what $A$ is, I see no ambiguity in the notation: $(x)$ in $k[x]$ or in $k[x.y]$ is all multiples of $x$ in $k[x]$ or $k[x,y]$. What is ambiguous? – KCd Aug 04 '24 at 13:04
  • thank you for the clarification. Your answer helped me with sorting out my confusion. – Seth Aug 04 '24 at 13:25