From what I know, $\mathbb{R}^2$ is a group under addition, defined as $(a, b) + (c, d) = (a+b,c+d)$. However, this answer on another question seems to suggest that $\mathbb{R}^2$ is actually a ring with multiplication defined as $(a, b)\cdot (c, d) = (ac, bd)$. I thought that we usually only define multiplication over the group $\mathbb{R}^2$ as $(a, b)\cdot (c,d) = (ac - bd, ad+bc)$, and in result end up making it a field called $\mathbb{C}$?
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6You can put multiple different algebraic structures on the same set. That's why a group/ring is a set together with specified operations, not just a set by itself. – Randall Jun 01 '20 at 13:56
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Is $\mathbb{R}^2$ not an algebraic structure -- a group? @Randall – Bad at Mathematics Jun 01 '20 at 13:59
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As you described it, $\mathbb{R}^2$ is a ring: commutative, with identity, but not a field nor integral domain. – Randall Jun 01 '20 at 14:01
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@ICanKindOfCode $\Bbb R^2$ does have a "default" additive structure, so it is indeed typically considered a group. However, it does not have a "default" multiplicative structure, so it is not usually considered a ring. – Ben Grossmann Jun 01 '20 at 14:02
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Interesting question. – justadzr Jun 01 '20 at 14:05
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@Randall so if it's a ring, then it does have a default multiplicative structure? – Bad at Mathematics Jun 01 '20 at 14:10
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There is no default. That's the whole point. – Randall Jun 01 '20 at 14:11
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Then what exactly is $\mathbb{R}^2$ if it does not have a default multiplicative structure defined? It should be a group and not a ring, right? – Bad at Mathematics Jun 01 '20 at 14:13
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It is just the set of all ordered pairs (until you clarify in your writing what you want the symbol to mean). – Randall Jun 01 '20 at 14:13
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As the answer you have linked indicates, the key here is that the exact meaning of $\Bbb R^2$ (or equivalently $\Bbb R \times \Bbb R$) depends in the context that we are working in.
In settings where $\Bbb R$ and/or $\Bbb C$ are the only rings/fields being discussed (typically in problems of an area whose name includes the word "analysis"), $\Bbb R^2$ typically refers to the abelian group/vector space over the set $\Bbb R^2$. In other words, no multiplication between elements is defined or considered.
However, in settings where $\Bbb R$ and/or $\Bbb C$ are considered to be one ring among many (typically in problems of an area whose name includes the word "algebra(ic)"), $\Bbb R^2$ refers to the multiplication associated with the product $\Bbb R \times \Bbb R$ of rings. That is, $\Bbb R^2$ a ring with multiplication defined by $(a,b)\cdot(c,d) = (ac,bd)$.
The symbol $\Bbb R^2$ is never used to refer to $\Bbb R^2$ with the complex-number multiplication $(a,b)\cdot (c,d) = (ac - bd,ad + bc)$, except perhaps for pedagogical reasons. Where the set $\Bbb R^2$ is given this multiplication rule, the symbol $\Bbb C$ is used instead.
Ben Grossmann
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So in conclusion I have to define any sort of algebraic structure on it, and $\mathbb{R}^2$ does not have structure by itself(but in different contexts it can be interpreted to have different structures). – Bad at Mathematics Jun 01 '20 at 14:17
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1@ICanKindOfCode I would say that the following is perhaps more accurate: in most contexts, $\Bbb R^2$ does not come with any sort of multiplicative structure, but it almost always comes with the usual notion of $+$. The only time where $\Bbb R^2$ does have a multiplicative structure that does not explicitly need to be explained is when we are already discussing the product $R \times S$ of rings (which comes with a multiplication over $R \times S$), and $\Bbb R^2$ arises as one such product. – Ben Grossmann Jun 01 '20 at 14:22
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You have that from the same abelian group (in this case $\mathbb{R}^2$) you can obtain different non-isomorphic rings if you consider it with different multiplications.
In this case $\mathbb{C}$ and $\mathbb{R}^2$ are both rings but with different multiplications.
In the first case, as Randall commented, you have that the group you are considering became a field. In the second case it's not a field (for example $(1,0)$ doesn't admit any inverse).
So, even if the abelian group is the same, you can construct different rings on it.
J. W. Tanner
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Sant97
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Wait, so when i say "$\mathbb{R}^2$", are we talking about the abelian group or the ring? can it mean both? – Bad at Mathematics Jun 01 '20 at 14:07
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If you use the term $\mathbb{R}^2$, you will be commonly understood to be referring to an abelian group, or an $\mathbb{R}$-vector space. – rogerl Jun 01 '20 at 14:09
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So my doubt in this was just a problem of nomenclature, and the other multiplicative structure I showed(componentwise) can also form a ring from the abelian group $\mathbb{R}^2$, but then $\mathbb{R}^2$ by itself is not a ring because we only define an additive structure to it? @rogerl – Bad at Mathematics Jun 01 '20 at 14:11
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It is your job to specify the ring structure, since as this answer points out, it is not unique. – Randall Jun 01 '20 at 14:12
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$\mathbb{R}^2$ by itself is not anything other than a set of points. The point is that the symbol is by general agreement used to represent the abelian group. If you wish it to represent something else, then as @Randall has pointed out, you must explicitly specify what. – rogerl Jun 01 '20 at 14:14