- #26

- 13

- 3

I'll take a go at 8b.

Let ##p(t) = T^3-2##. Notice that ##p## has not roots over ##\mathbb{F}_7## which can be seen by simply evaluating ##p## at each element of ##\mathbb{F}_7##.

We claim then ##p## is irreducible. If not then ##p## can be factored as ##p = f g## where ##f##, ##g## are irreducible and ##deg(f) + deg(g) = 3##. Since ##f## and ##g## are irreducible, ##deg(f), deg(g) > 0## and so either ##deg(f) = 1## or ##deg(g) = 1##. Suppose without loss of generality that ##deg(f) = 1##, then ##f## is linear and hence has a root of ##\mathbb{F}_7## which means that ##p## has a root over ##\mathbb{F}_7## which is a contradiction.

Since ##p## is irreducible it then ##(p)## must be maximal. Otherwise suppose there exists some ##q \in \mathbb{F}_7[T]## such that ##(p) \subset (q)##. Hence there exists an ##f \in \mathbb{F}_7[T]## such that ##p = qf##. Since ##p## is irreducible either ##q## is a unit (in which case ##(q) = \mathbb{F}_7[T]##) or ##f## is a unit (in which case ##(f) = (q)##). So indeed ##(f)## is maximal.

We now use the fact that given any ring ##R## with a maximal ideal ##M##, ##R/M## is a field. This follows from one of the standard isomorphism theorems which states that there is a one-to-one correspondence between the ideals of ##R/M## and the ideals in ##R## containing ##M##. So if ##M## is maximal, ##R/M## can not contain any proper ideals, and so must be a field.

To compute the number of elements in ##\mathbb{F}_7[T]/(T^3-2)## we note that since ##3## is prime each cosets can be represented by elements of the form ## c_1 + c_2 T + c_3 T^2## where ##c_i \in \mathbb{F}_7[T]##, and there are ##7^3 = 343## such elements.

To compute ##(T^2+2T+4)(2T^2+5)## in ##\mathbb{F}_7[T]/(T^3-2)## we first note that ##(T^2+2T+4)(2T^2+5) = 2T^4 + 2T^3 + 6T^2 + 3T + 6## in ##\mathbb{F}_7## then reducing ##mod (T^3-2)## gives ## 2(2)T + 2(2) + 6T^2 + 3T + 6## and so $$(T^2+2T+4)(2T^2+5) = 6T^2 + 3$$

To compute ##\frac{1}{T+1}## you can do a sort of "brute force" method. Let ##\frac{1}{T+1} = P(T) = a_0 + a_1T + a_2T^2## then ##(T+1)P(T) = (a_1+a_2)T^2 + (a_1 + a_0)T + (a_0 + 2a_2) = 1## which gives a system $$a_0 + a_2 = 1, a_1 + a_0 = 0, a_1 + a_2 = 0$$

which has a solution ##a_0 = 5, a_1 = 2, a+2 = 5##. So $$\frac{1}{T+1} =5 + 2T + 5T^2$$

Let ##p(t) = T^3-2##. Notice that ##p## has not roots over ##\mathbb{F}_7## which can be seen by simply evaluating ##p## at each element of ##\mathbb{F}_7##.

We claim then ##p## is irreducible. If not then ##p## can be factored as ##p = f g## where ##f##, ##g## are irreducible and ##deg(f) + deg(g) = 3##. Since ##f## and ##g## are irreducible, ##deg(f), deg(g) > 0## and so either ##deg(f) = 1## or ##deg(g) = 1##. Suppose without loss of generality that ##deg(f) = 1##, then ##f## is linear and hence has a root of ##\mathbb{F}_7## which means that ##p## has a root over ##\mathbb{F}_7## which is a contradiction.

Since ##p## is irreducible it then ##(p)## must be maximal. Otherwise suppose there exists some ##q \in \mathbb{F}_7[T]## such that ##(p) \subset (q)##. Hence there exists an ##f \in \mathbb{F}_7[T]## such that ##p = qf##. Since ##p## is irreducible either ##q## is a unit (in which case ##(q) = \mathbb{F}_7[T]##) or ##f## is a unit (in which case ##(f) = (q)##). So indeed ##(f)## is maximal.

We now use the fact that given any ring ##R## with a maximal ideal ##M##, ##R/M## is a field. This follows from one of the standard isomorphism theorems which states that there is a one-to-one correspondence between the ideals of ##R/M## and the ideals in ##R## containing ##M##. So if ##M## is maximal, ##R/M## can not contain any proper ideals, and so must be a field.

To compute the number of elements in ##\mathbb{F}_7[T]/(T^3-2)## we note that since ##3## is prime each cosets can be represented by elements of the form ## c_1 + c_2 T + c_3 T^2## where ##c_i \in \mathbb{F}_7[T]##, and there are ##7^3 = 343## such elements.

To compute ##(T^2+2T+4)(2T^2+5)## in ##\mathbb{F}_7[T]/(T^3-2)## we first note that ##(T^2+2T+4)(2T^2+5) = 2T^4 + 2T^3 + 6T^2 + 3T + 6## in ##\mathbb{F}_7## then reducing ##mod (T^3-2)## gives ## 2(2)T + 2(2) + 6T^2 + 3T + 6## and so $$(T^2+2T+4)(2T^2+5) = 6T^2 + 3$$

To compute ##\frac{1}{T+1}## you can do a sort of "brute force" method. Let ##\frac{1}{T+1} = P(T) = a_0 + a_1T + a_2T^2## then ##(T+1)P(T) = (a_1+a_2)T^2 + (a_1 + a_0)T + (a_0 + 2a_2) = 1## which gives a system $$a_0 + a_2 = 1, a_1 + a_0 = 0, a_1 + a_2 = 0$$

which has a solution ##a_0 = 5, a_1 = 2, a+2 = 5##. So $$\frac{1}{T+1} =5 + 2T + 5T^2$$

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