Prove that for each $f \in \mathbb{Q}[x]$ there exists $a+bx \in \mathbb{Q}[x]$ such that $f + \langle x^2 \rangle = (a+bx) + \langle x^2 \rangle$

Question

Usually when I begin with proofs, I like to define all terms to ensure I have a clear understanding of what is being asked of me.

I know that $\langle x^2 \rangle \subset \mathbb{Q}[x]$ is the principal ideal generated by $x^2$, which is:

{$ax^2 : a \in \mathbb{Q}[x]$}.

Next, I know that given $\langle x^2 \rangle \subset \mathbb{Q}[x]$ and $f \in \mathbb{Q}[x]$, the coset of $f$ with respect to $\langle x^2\rangle$ is:

$f + \langle x^2 \rangle = [a]_{\langle x^2 \rangle} =$ {$f + a : a \in \langle x^2 \rangle$}.

Thus, for this proof, I understand that I must show that given a polynomial $f = a_0 +a_1x + ... + a_nx^n \in \mathbb{Q}[x]$, we want to find a polynomial $a +bx \in \mathbb{Q}[x]$ such that $f$ and $a+bx$ are equivalent in $\mathbb{Q}[x]/x^2$.

But with this, I am still not sure how to proceed. Any guidance would be much appreciated.

Answer 1

With

$f(x) = \displaystyle \sum_0^n a_i x^i, \tag 1$

we may write

$f(x) = a_0 + a_1x + \displaystyle \sum_2^n a_i x^i = a_0 + a_1 x + x^2 \sum_2^n a_i x^{i - 2}; \tag 2$

then we have the coset of the principal ideal $\langle x^2 \rangle$:

$f(x) + \langle x^2 \rangle = (a_0 + a_1x) + x^2 \displaystyle \sum_2^n a_ix^{i - 2} + \langle x^2 \rangle$ $= ((a_0 + a_1x) + \langle x^2 \rangle) + ((x^2 \displaystyle \sum_2^n a_i x^{i - 2} + \langle x^2 \rangle); \tag 3$

but

$x^2 \displaystyle \sum_2^n a_i x^{i - 2} \in \langle x^2 \rangle \Longrightarrow x^2 \sum_2^n a_i x^{i - 2} + \langle x^2 \rangle = \langle x^2 \rangle; \tag 4$

thus (3) becomes

$f(x) + \langle x^2 \rangle = (a_0 + a_1x) + x^2 \displaystyle \sum_2^n a_ix^{i - 2} + \langle x^2 \rangle = ((a_0 + a_1x) + \langle x^2 \rangle) + \langle x^2 \rangle$ $= a_0 + a_1x + \langle x^2 \rangle, \tag 5$

$OE\Delta$.

Answer 2

Hint:

Can you show that $f= a_0 +a_1x + ... + a_nx^n$ and $a_0+a_1x$ are equivalent in $\mathbb Q[x]/\langle x^2\rangle$?