What is meaning of $(2,3)$? where ( ) denotes a Principal Ideal

Question

first instincts is that $(x,y)=(x) \cup (y) $ or $(x,y)=(x) \cap (y) $

but from class notes there is 1 line where putting in 2,3 $$(2,3)=3\mathbb{Z}+ 2\mathbb{Z} $$ It comes from extension fields, right?. Does it have a name?

Is there an alternate set notation for $(2,3)$?

would it be $(2,3)=\{ 2k_1+3k_2:k_1,k_2 \in \mathbb{Z}\} $

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First instincts is to say $$(2,3)=3\mathbb{Z}+ 2\mathbb{Z}=5\mathbb{Z} $$ But $$ \begin{aligned} (2)=&\{ 0,\pm 2,\pm 4,\pm 6,\pm 8,\pm 10, \dots\} = \{ 2k: k \in \mathbb{Z} \} \\ (3)=&\{0, \pm 3, \pm 6, \pm 9,\pm12,\pm15 \dots \} =\{ 3k:k\in \mathbb{Z} \} \end{aligned}$$ and $$(2)+(3)=\{0,\pm 1,\pm 2,\pm 3,\pm 4,... \}=\mathbb{Z}=(1) \neq 5 $$

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Found a relatable question once writing it here, confirming that is is not (5)

I know it seems like a softball question but trying to get my foundations right.

Appreciate any constructive input, thanks.

Answer 1

In general, given a ring $R$ and a subset $S \subseteq R$, the ideal generated by $S$ is denoted $(S)$, is defined to be the set of finite $R$-linear combinations of elements of $S$, that is $$(S) = \left\{ \sum_{i=1}^n r_is_i : n \in \mathbb{N},\ r_i \in R,\ s_i \in S \right\}$$ When $S = \{ s_1, s_2, \dots, s_k \}$ is finite, we just write $(s_1, s_2, \dots, s_k)$ instead of $(\{s_1, s_2, \dots, s_k \})$.

So for example if $R = \mathbb{Z}$ and $S = \{ 2, 3 \}$, then $$(S) = (2,3) = \{ 2a+3b : a,b \in \mathbb{Z} \} = \mathbb{Z}$$ Such ideals need not be principal; in fact, principal ideals are precisely those of the form $(s)$, where $s \in R$. For example, $$(2,X) = \{ 2a+bX : a,b \in \mathbb{Z} \}$$ is not a principal ideal of the polynomial ring $\mathbb{Z}[X]$.

Answer 2

First of all, the parentheses denote an ideal (not necessarily principal). For instance, in the ring $\Bbb{Z}$, $$ (3) = \{3k \mid k \in \Bbb{Z}\} = 3\Bbb{Z}. $$ This ideal is principal since it's generated by a single element. You can look at ideals generated by more than one element, though. For instance, in $\Bbb{Z}$, $$ (6, 15) = \{6i + 15j \mid i,j \in \Bbb{Z}\}. $$ It turns out that in the ring $\Bbb{Z}$, every ideal is principal (but this is not the case for every ring). In other words, we can find some $n \in \Bbb{Z}$ such that $$ (6, 15) = n\Bbb{Z}. $$ What is this $n$? Consider the greatest common divisor: $$ n = \gcd(6, 15) = 3. $$ Notice that since $3$ divides $6$, we have $6 \in (3)$. Analogously, $15 \in (3)$. So we can conclude that $3$ divides any linear combination of $6$ and $15$, i.e., $(6, 15) \subseteq (3)$.

But what about the reverse inclusion? Can we express $3$ as a linear combination of $6$ and $15$? In other words, are there integers $x$ and $y$ such that $$ 6x + 15y = 3? $$ Yes! One way is with $(x, y) = (-2, 1)$. This shows that $3 \in (6, 15)$; therefore, $(3) \subseteq (6, 15)$. Thus, $$ (6, 15) = (3) $$ and, generally, if $a, b \in \Bbb{Z}$, $$ (a, b) = (d), \quad\text{where } d = \gcd(a, b). $$

Answer 3

It is indeed the set of linear combinations of 2 and 3, as you wrote: $(2,3)=\{ 2k_1+3k_2:k_1,k_2 \in \mathbb{Z}\} $ You can simplify that though, as that's also known as any multiple of the gcd of 2 and 3, and since 2 and 3 are relatively prime, we get $(2,3)=(1)=\mathbb Z$.

Answer 4

The notation $(x_1,x_2,\dots, x_k)$ in a ring $R$ is used to denote the ideal generated by the elements $x_i\in R$. Linear algebraically this is the $R$-linear span of the elements $\{x_1, \dots, x_k\}$. Similarly, as a module, this denotes the direct sum $Rx_1\oplus R x_2\oplus \cdots \oplus Rx_k$