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A function $f:\mathbb{R}\rightarrow \mathbb{R}$ is said to be a $polynomial function$ if there is a natural number $n$ and real numbers $f_0,f_1,\dots,f_n$ such that $$f(x)=f_nx^n+\dots f_1x+f_0$$ for x in $\mathbb{R}$. Show that the set of all polynomial functions forms a monoid of functions on $\mathbb{R}$.
My attempt: So im new to abstract algebra and these type of proofs, so I apologize if this is a very easy question. I know that we need to show that it is closed under composition and there exist's an identity element. I'm unsure how to go about that, I'm only necessarily looking for a hint to get me started, however full awnsers will be accepted too!

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    what is the operation? – 2017-02-01
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    What do you mean what is the operation? – 2017-02-01
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    yes, the operation for the monoid. – 2017-02-01
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    Is that something that should be given in the problem? I have heard of monoids having an operation on them, however so far we haven't learned anything about them having an operation. We just prove that it is closed under composition and that there exists an identity element. – 2017-02-01
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    ok, so then the operation is most likely composition – 2017-02-01
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    Okay so in general it's not alway's going to be composition, the way our teacher made it seem is that that was the definition of a monoid being closed under composition, and having an identity element. – 2017-02-01
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    no, the definition allows for any operation. However if you are inside a category then you can define a monoid as a subset of the endomorfisms of a fixed object that contains the identity of the object. In this context the operation is always morfism composition. But the classic definition of monoid is usually taught first, and in the classic definition the operation can vary. – 2017-02-01

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The polynomial functions under composition form a mnoid.

Notice that the identity function is a polynomial, it is $f(x)=x$.

Now we must only prove that if $f$ and $g$ are polynomials then $f\circ g$ is a polynomial.

Let $f(x)=a_nx^n+\dots+a_0$ and $g(x)=b_mx^m+\dots+b_0$.

Notice that $f\circ g(c)=f(b_mx^m+\dots+b_0)=a_n(b_mx^m+\dots+b_0)^n+a_{n-1}(b_mx^m+\dots+b_0)^{n-1}+\dots+ a_0$.

You must prove that each summand is a polynomial then you would be done ( prove that the sum of polynomials is a polynomial).

The reason why each summand is a polynomial is that the product of polynomials is also a polynomial (you should also prove this).

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    Thank-you for this response, it clarified a lot for me! – 2017-02-01
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    sure no probz, glad to help. – 2017-02-01