# Can you talk about (the rest of the) field axioms when the operations are not closed? [closed]

Note: Updated based on this.

In my course, my instructor posed the following exercise:

Let $$S$$ be the subset of $$\mathbb R^n$$, $$S=\{(a_1,a_2,a_3...a_n) | a_2 = \pm a_1, a_3=...=a_n=0 \}$$. Define addition and multiplication as follows: For $$(a_1,a_2, ..., a_n),(b_1,b_2, ..., b_n) \in S$$, define $$(a_1,a_2, ..., a_n)+(b_1,b_2, ..., b_n)=(a_1+b_1,a_2+b_2, ..., a_n+b_n)$$ and $$(a_1,a_2, ..., a_n)(b_1,b_2, ..., b_n)=(a_1 \times b_1,a_2 \times b_2, ..., a_n \times b_n)$$, where $$+$$ and $$\times$$ are the usual addition and multiplication in $$\mathbb R$$. Which axioms in the definition of a field are satisfied by $$S$$? Is $$S$$ a field and why?

Now, I notice that the given operation $$+: S^2 \to \mathbb R^n$$ is not closed i.e. its image is not a subset of $$S$$. Now, I notice that the given operation $$+: S^2 \to \mathbb R^n$$ is not closed i.e. its image is not a subset of $$S$$. In particular it really just says 'define' instead of like giving explicitly the domain and range as $$+: S^2 \to \mathbb R^n$$. I think this is much like: https://math.stackexchange.com/questions/856226/how-do-you-prove-the-domain-of-a-function

(It) is part of the assignment (as to whether or not the operations are closed). If an operation is not closed, which of the rest of the axioms are satisfied?

Question 1: So basically this is a trick for students like someone studying this for the 1st time may think of 'define' to really mean $$+: S^2 \to S$$ instead of $$+: S^2 \to \mathbb R^n$$ and thus not realise the operation is not closed ?

Question 2:

If not, then what?

If so, then is this a fair trick? I mean is the question umambiguous, 'reasonable' and 'of considerable pedagogical value' (terms used in comments below) ? Or is this question perhaps unclear and an unfair way to trick students?

Cross-posted on maths se: https://math.stackexchange.com/questions/4256472/can-you-talk-about-the-rest-of-the-field-axioms-when-the-operations-are-not-cl --> I justify the cross-post in that maths education se might respond more to the trick of questions while maths se might respond more to the talking about field axioms when assumptions are violated.

• Because this is the Mathematics Educators site, I will suggest focusing on the pedagogical reasons for asking a homework question like this: What do you want students to learn from the problem? Based on that, is posing this homework problem really the best way to help them learn the concept/skill you have in mind? In general, I'd say that as soon as you're even asking something like, "Is this question fair or unambiguous or not weird?".... then the answer is pretty clearly, "Yeah, don't use this homework problem as is." Sep 21 at 17:47
• @BCLC: To be perfectly honest, I cannot understand what it is you're asking here. When you say, "Consider a homework question that goes...", that sounds to me like you're the one choosing to assign this problem to students. (Maybe you've created it yourself, or maybe you found it in a book, whatever it is...) If that is indeed the case, then only you can answer, "What is the pedagogical purpose?" I am not in your class! And, by the way, if that is not the case, and you're not thinking of using this problem with students... then why are we having this discussion?! Sep 21 at 18:01
• You should indicate whether or not this is graded homework that can affect their course grade, rather than one of possibly many problems you suggest they do (but don't take up and grade). My feeling is that this would work better AND take up less of your time if you used this as a classroom discussion example. Pose this as a lecture question and let the students hash it out for a couple (or more) minutes. Sep 21 at 18:45
• @BCLC The question seems perfectly reasonable. Addition and multiplication are being defined on $\mathbb{R}^n$ to make it a ring. The question is whether this subset of $\mathbb{R}^n$ is a field. The answer is no because the subset is not closed under the operations. Sep 22 at 10:04
• @BCLC The question is of considerable pedagogical value. The example is closed under multiplication, it is closed under taking additive inverses, every nonzero element has a multiplicative inverse which lives in the set. It just isn't closed under addition. This drives home the point that these items are independent, and require independent arguments. Sep 22 at 10:13

This is a little long for a comment so it will have to be an answer. First, you should talk to your professor; I'm pretty sure that they will be able to definitively clear your issue up with a small amount of back-and-forth discussion.

You seem to have trouble starting around your encounter with the word "define". I'm not really sure what we are supposed to get out of your link. (Is that a game? an anime? it certainly isn't a universal enough piece of culture to pass unexplained.) At an informal level, when we say "define..." it merely sets up some common terminology. In this case, an extension of $$+$$ and $$\times$$ to vectors. Someone else could propose a different $$+$$ and $$\times$$ on the same set at a later time, but this is the one you will be working with in this problem. Maybe it satisfies the field axioms and maybe it doesn't; calling them $$+$$ and $$\times$$ doesn't mean much.

Secondly you get tripped up around the words "binary operation". A binary operation on $$\mathcal{S}$$ is simply a function $$\mathcal{S}\times \mathcal{S} \to \mathcal{S}$$. i.e. it takes two elements of $$\mathcal{S}$$ to one element of $$\mathcal{S}$$. Aside from verifying that the operations are well-defined and have the proper (co-)domains, there is nothing to do.

But, trust me, your professor wanted you to do something so much simpler. You have a direct sum of commutative rings and hence a commutative ring. What else do you need for a field?

• I see now that your 1st comment adds those important details. Yes, the $+$ operation is not closed. e.g $(1,1)+(1,-1)=(2,0)\not\in S$. This was a pretty important omission. It still remains binary; it is just that the codomain is too big.
• @MarkS. maybe. i was thinking more of when people say $f$ meromorphic on $G$. i know that's more about the domain of $f$ than the range/codomain of $f$ but still however annoying i find it, i can't deny it's a thing people do.