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Surprisingly, we got only one question for our 2-hour exam and I think nobody solved it. Here is the problem:

Assuming that $K$ is a field, show that $S$ is stable under addition, multiplication and division, where $S$ is defined as follow: $$S=\left\{\sum_{i=1}^{n}{x_i}^2 \mid n\in \Bbb N ,\ x_i\in K\right\}.$$

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    I guess this is the intended set? – 2017-01-28
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    Additian and multiplication are trivial. As $0\in S$, what does it mean for $S$ to be stable under division? – 2017-01-28
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    just to correct addition and multiplication are not trivial – 2017-01-28
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    @Servaes it is likely meant that if $x,y\in S$ and $y\neq0$ then $xy^{-1}\in S$. Since $S$ is stable under multiplication and $1\in S$ this is equivalent to $y^{-1}\in S$ whenever $y\in S, y\neq0$. – 2017-01-28
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    Addition: Trivial. Any sum of two finite sums of squares is a finite sum of squares. Multiplication: Trivial. Any product of two squares is a square and the distributive property shows that a product of two finite sums has finitely many terms. – 2017-01-28
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    Btw to whoever downvoted: Please leave a comment about why you feel it is justified. – 2017-01-28
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    I downvoted because the question and subsequent comment show no effort whatsoever. – 2017-01-28
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    @Servaes my efforts are irrelevants , I still think about the problem .. when I got something I will post it don't worry – 2017-01-28
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    I disagree. See also [this meta post](http://meta.math.stackexchange.com/questions/8998/closing-questions-that-show-no-effort-official-policy) and its top voted answer, and the answers' top voted comment. But feel free to disagree, it's just a downvote. – 2017-01-28
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    @Seginus: You did not even take care to properly punctuate and capitalize the question. That's just disrespectful. – 2017-01-28

3 Answers 3

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Addition: Trivial. Any sum of two finite sums of squares is a finite sum of squares.

Multiplication: Trivial. Any product of two squares is a square and the distributive property shows that a product of two finite sums has finitely many terms.

Division: Suppose that $a=\sum_{i=1}^n x_i^2$ is not zero. We would like to show that $a^{-1}\in S$ (using s.harp's comment). If $n=1$, then the inverse of $a$ is trivial. The first interesting case is when $n=2$ Suppose that $a=x_1^2+x_2^2$.

Now, $$ \frac{1}{x_1^2+x_2^2}=\frac{x_1^2+x_2^2}{(x_1^2+x_2^2)^2}=\left(\frac{x_1}{x_1^2+x_2^2}\right)^2+\left(\frac{x_2}{x_1^2+x_2^2}\right)^2 $$ The denominators are OK because this is a field. All higher $n$'s are generalizations of this case.

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    Very nice observation! – 2017-01-28
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    We posted at the same time. +1 – 2017-01-28
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    how can we generalise this ? – 2017-01-28
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    @Seginus The obvious generalization where you replace the square by any other power should also work. For other generalizations, perhaps it would be good to open a new question so that it gets more attention. – 2017-01-28
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    @Segmus In fact it generalizes further than powers, e.g. see the remark in my answer. – 2017-01-29
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$$\frac{1}{x_1^2+x_2^2+x_3^2}=\frac{x_1^2}{(x_1^2+x_2^2+x_3^2)^2}+\frac{x_2^2}{(x_1^2+x_2^2+x_3^2)^2}+\frac{x_3^2}{(x_1^2+x_2^2+x_3^2)^2}$$

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    Posting at the same time +1 – 2017-01-28
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$\begin{align}{\bf Hint} \ \ a\neq 0\ \Rightarrow&\ \ \color{#c00}{a^{\large -2}} =\, (a^{\large -1})^{\large 2} \in S\ \ \text{by $\,S\,$ contains all squares}\\ {\rm so}\ \ \ a\in S\ \Rightarrow&\ \ a^{\large -1} =\, a\cdot \color{#c00}{a^{\large-2}}\in S\ \ \text{by $\,S\,$ is closed under multipication} \end{align}$

Remark $ $ You asked how it generalizes. The above idea works not only for squares, but for any positive power, i.e. we can obtain $\,a^{-1}\,$ from any of its positive powers via $\,a^{\large -1} = a^{\large n-1}(a^{\large -1})^{\large n}.$ Also the only property of squares (or powers) employed in proving multiplicative closure is that they are closed under multiplication. These observations lead to the following generalization.

Theorem $ $ Suppose $K$ is a field, $M$ is a subset of $K$ and $S$ is the set of all sums of elements of $M.$

$(1)\ \ S$ is closed under multiplication if $M$ is closed under multiplication.

$(2)\ \ S$ is closed under division (by all nonzero elements) if $S$ is both closed under multiplication, and every $\,a\in K$ has some positive power $\,a^{\large n}\in S.$

Proof $\ (1)\,$ is an immediate consequence of the distributive law. $\ (2)\ $ follows as above, namely if $\,0\neq s\in S\,$ then by hypothesis $\,(s^{\large -1})^{\large n}\in S\,$ for some $\,n\geq 1\,$ thus $\,s^{\large n-1}(s^{\large -1})^{\large n} = s^{\large -1}\in S\,$ because $S$ is closed under multiplication. Hence $\,a\in S\,\Rightarrow\,as^{-1} = a/s\in S,\,$ so $S$ is closed under division.

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    I found this question 9 hours late (from the Hot List). but I'll add this answer anyhow since I think it clarifies closure under inverses, hence closure under division (the only tricky part). – 2017-01-28