How does my proof look?

We show that \(\displaystyle deg(f(x)+g(x))\leq max(deg (f(x)),deg (g(x))\). Let \(\displaystyle \mathbb{Q}\) be a polynomial ring.

Not a good idea to use the same symbol for two different things within the same 4-5 lines. Why not \(\displaystyle \mathbb{Q}[x]\)?

It will be the set of all polynomial functions \(\displaystyle f(x)=a_0+a_1x+a_2x^2+\cdots +a_nx^n\), \(\displaystyle a_i\in \mathbb{Q}\). If \(\displaystyle a_n\neq 0\) then we say \(\displaystyle n=deg(f(x))\). When adding polynomials we see that \(\displaystyle

\sum_{i=0}^h a_ix^i + \sum_{i=0}^k a_ix^i\leq \sum_{i=0}^{max(h,k)} (a_i+b_i)x^i\).

What does it mean that a polynomial is less or equal than other polynomial? This is meaningless unless properly defined and checked it works as a linear order...and again you use the same symbol (in this case, \(\displaystyle a_i\) ) to denote two different things: the coefficients of polynomial \(\displaystyle f(x)\) and those of pol. \(\displaystyle g(x)\)...!

Thus, \(\displaystyle deg(f(x)+g(x))\leq max(deg (f(x)),deg (g(x))\) as desired.

You haven't yet proved anything: a good idea would be to show that in \(\displaystyle f(x)+g(x)\) we have the monomials \(\displaystyle a_hx^h\,,\,b_kx^k\) of \(\displaystyle f(x)\,,\,g(x)\) resp., and thus...