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I will ask you some things about ideals.


Determine weither the ideal $(X^2+3) \subset \mathbb{F}_5[X]$ is maximal or prime.


Intuitively I'd say that the ideal is prime but not maximal. To prove that it's prime, let $fg \in (X^2+3)$ arbitrary. We know that $2 = \deg(fg)= deg(f)+deg(g)$. This implies, WLOG, that $\deg(f)=2$; $\deg(g)=0$ or $\deg(f)=\deg(g)=1$.

Let's consider the first option. In that case we need to solve the equation: $$ \bar{c}\cdot(\bar{a_0}+\bar{a_1}X+\bar{a_2}X^2) = X^2+\bar{3}$$ For this $ \bar{a_2}=0$. Moreover, we should choose $c$ such that $c\cdot a_0 \in 1 +5\mathbb{Z}$, and $c \cdot a_3 \in 3+5\mathbb{Z}$. Now I've actually found this, I don't precisely know what to do.

The other option: If $deg(f)=deg(g)=1$; we have to solve: $$(a_1X+a_0)(b_1X+b_0)=a_1b_1X^2+(a_0b_1+a_1b_0)X+a_0b_0= X^2+3$$ In this case we find that for some integers $k,m$ holds: $$a_1 = \frac{5m+1}{b_1} \quad \text{and} \quad a_0 = \frac{5k+3}{b_0} $$ This would satisfy the first and the third equation. to satisfy the second equation ass well, we we should choose our parameters such that $$ \frac{5k+3}{b_0}\cdot b_1+\frac{5m+1}{b_1}\cdot b_0=0$$

So how should I go on here? Is this a good approach? Thank you for your time.

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    $\begingroup$ Is this supposed to be $\mathbb{F}_5[x]$? If so, maximal and prime are the same thing... $\endgroup$ – Alex Youcis Aug 10 '13 at 21:58
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As alluded to in the comments, $k[x]$ is a PID if $k$ is a field, and in PID's, nonzero prime ideals are maximal, and are generated by irreducible elements. Hence to see if $(x^2+3)$ is a maximal ideal, and hence also prime, we need to check the irreducibility of $(x^2+3)$. This is what you are doing in the case of $\deg f=\deg g=1$.

Now, something nice about working with small degrees ($\le 3$) is that checking for irreducibility is the same as checking for roots. This follows because if polynomials of degree less than $4$ factor, they must have a linear term, which corresponds to a root. This fails for degree four polynomials, because they may factor into two quadratic terms.

Continuing with our problem, we've learned that $x^2+3$ is irreducible if and only if it has no roots in $\mathbb{F}_5$. Plugging in $x=0,1,2,3,4$ we see there are no roots, so that $x^2+3$ is indeed irreducible, and thus generates a maximal ideal.

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